Toniolo and Linder, Equation (3b), real

Percentage Accurate: 93.8% → 99.7%

Time: 21.1s

Alternatives: 19

Speedup: 1.4×

Specification

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = (sin(ky) / sqrt(((sin(kx) ** 2.0d0) + (sin(ky) ** 2.0d0)))) * sin(th)
end function

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Initial Program: 93.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = (sin(ky) / sqrt(((sin(kx) ** 2.0d0) + (sin(ky) ** 2.0d0)))) * sin(th)
end function

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.7% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (hypot (sin ky) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / hypot(sin(ky), sin(kx))) * sin(th);
}

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 94.0%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Step-by-step derivation

   1. +-commutative 94.0%

      \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

   2. unpow2 94.0%

      \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

   3. unpow2 94.0%

      \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

   4. hypot-def 99.7%

      \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

3. Simplified 99.7%

   \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

4. Final simplification 99.7%

   \[\leadsto \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th \]

Alternative 2: 76.9% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -5e-5)
   (* (sin ky) (/ th (hypot (sin ky) (sin kx))))
   (if (<= (sin ky) 5e-8)
     (* ky (/ (sin th) (hypot (sin kx) (sin ky))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -5e-5) {
		tmp = sin(ky) * (th / hypot(sin(ky), sin(kx)));
	} else if (sin(ky) <= 5e-8) {
		tmp = ky * (sin(th) / hypot(sin(kx), sin(ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -5e-5) {
		tmp = Math.sin(ky) * (th / Math.hypot(Math.sin(ky), Math.sin(kx)));
	} else if (Math.sin(ky) <= 5e-8) {
		tmp = ky * (Math.sin(th) / Math.hypot(Math.sin(kx), Math.sin(ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -5e-5:
		tmp = math.sin(ky) * (th / math.hypot(math.sin(ky), math.sin(kx)))
	elif math.sin(ky) <= 5e-8:
		tmp = ky * (math.sin(th) / math.hypot(math.sin(kx), math.sin(ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -5e-5)
		tmp = Float64(sin(ky) * Float64(th / hypot(sin(ky), sin(kx))));
	elseif (sin(ky) <= 5e-8)
		tmp = Float64(ky * Float64(sin(th) / hypot(sin(kx), sin(ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -5e-5)
		tmp = sin(ky) * (th / hypot(sin(ky), sin(kx)));
	elseif (sin(ky) <= 5e-8)
		tmp = ky * (sin(th) / hypot(sin(kx), sin(ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -5e-5], N[(N[Sin[ky], $MachinePrecision] * N[(th / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 5e-8], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -5.00000000000000024e-5

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in th around 0 56.2%

      \[\leadsto \frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. *-commutative 56.2%

         \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

   6. Simplified 56.2%

      \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   7. Step-by-step derivation

      1. expm1-log1p-u 55.6%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right)} \]

      2. expm1-udef 4.6%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)} - 1} \]

      3. *-commutative 4.6%

         \[\leadsto e^{\mathsf{log1p}\left(\frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)} - 1 \]

      4. associate-/l* 4.6%

         \[\leadsto e^{\mathsf{log1p}\left(\color{blue}{\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}}\right)} - 1 \]

   8. Applied egg-rr 4.6%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}\right)} - 1} \]
   9. Step-by-step derivation

      1. expm1-def 55.7%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}\right)\right)} \]

      2. expm1-log1p 56.3%

         \[\leadsto \color{blue}{\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}} \]

      3. associate-/r/ 56.2%

         \[\leadsto \color{blue}{\frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin ky} \]

      4. *-commutative 56.2%

         \[\leadsto \color{blue}{\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   10. Simplified 56.2%

       \[\leadsto \color{blue}{\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   if -5.00000000000000024e-5 < (sin.f64 ky) < 4.9999999999999998e-8

   1. Initial program 88.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 83.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 90.9%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 90.9%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in ky around 0 90.7%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. associate-/l* 99.4%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      2. div-inv 99.3%

         \[\leadsto \color{blue}{ky \cdot \frac{1}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      3. clear-num 99.5%

         \[\leadsto ky \cdot \color{blue}{\frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

      4. hypot-udef 87.7%

         \[\leadsto ky \cdot \frac{\sin th}{\color{blue}{\sqrt{\sin ky \cdot \sin ky + \sin kx \cdot \sin kx}}} \]

      5. +-commutative 87.7%

         \[\leadsto ky \cdot \frac{\sin th}{\sqrt{\color{blue}{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

      6. hypot-udef 99.5%

         \[\leadsto ky \cdot \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   if 4.9999999999999998e-8 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 62.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 78.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 3: 77.0% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -5e-5)
   (* (sin ky) (/ th (hypot (sin ky) (sin kx))))
   (if (<= (sin ky) 5e-8)
     (* (sin th) (/ ky (hypot (sin kx) (sin ky))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -5e-5) {
		tmp = sin(ky) * (th / hypot(sin(ky), sin(kx)));
	} else if (sin(ky) <= 5e-8) {
		tmp = sin(th) * (ky / hypot(sin(kx), sin(ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -5e-5) {
		tmp = Math.sin(ky) * (th / Math.hypot(Math.sin(ky), Math.sin(kx)));
	} else if (Math.sin(ky) <= 5e-8) {
		tmp = Math.sin(th) * (ky / Math.hypot(Math.sin(kx), Math.sin(ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -5e-5:
		tmp = math.sin(ky) * (th / math.hypot(math.sin(ky), math.sin(kx)))
	elif math.sin(ky) <= 5e-8:
		tmp = math.sin(th) * (ky / math.hypot(math.sin(kx), math.sin(ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -5e-5)
		tmp = Float64(sin(ky) * Float64(th / hypot(sin(ky), sin(kx))));
	elseif (sin(ky) <= 5e-8)
		tmp = Float64(sin(th) * Float64(ky / hypot(sin(kx), sin(ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -5e-5)
		tmp = sin(ky) * (th / hypot(sin(ky), sin(kx)));
	elseif (sin(ky) <= 5e-8)
		tmp = sin(th) * (ky / hypot(sin(kx), sin(ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -5e-5], N[(N[Sin[ky], $MachinePrecision] * N[(th / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 5e-8], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -5.00000000000000024e-5

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in th around 0 56.2%

      \[\leadsto \frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. *-commutative 56.2%

         \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

   6. Simplified 56.2%

      \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   7. Step-by-step derivation

      1. expm1-log1p-u 55.6%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right)} \]

      2. expm1-udef 4.6%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)} - 1} \]

      3. *-commutative 4.6%

         \[\leadsto e^{\mathsf{log1p}\left(\frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)} - 1 \]

      4. associate-/l* 4.6%

         \[\leadsto e^{\mathsf{log1p}\left(\color{blue}{\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}}\right)} - 1 \]

   8. Applied egg-rr 4.6%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}\right)} - 1} \]
   9. Step-by-step derivation

      1. expm1-def 55.7%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}\right)\right)} \]

      2. expm1-log1p 56.3%

         \[\leadsto \color{blue}{\frac{th}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin ky}}} \]

      3. associate-/r/ 56.2%

         \[\leadsto \color{blue}{\frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin ky} \]

      4. *-commutative 56.2%

         \[\leadsto \color{blue}{\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   10. Simplified 56.2%

       \[\leadsto \color{blue}{\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   if -5.00000000000000024e-5 < (sin.f64 ky) < 4.9999999999999998e-8

   1. Initial program 88.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 83.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 90.9%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 90.9%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in ky around 0 90.7%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. associate-/l* 99.4%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      2. associate-/r/ 99.5%

         \[\leadsto \color{blue}{\frac{ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

      3. hypot-udef 87.8%

         \[\leadsto \frac{ky}{\color{blue}{\sqrt{\sin ky \cdot \sin ky + \sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. +-commutative 87.8%

         \[\leadsto \frac{ky}{\sqrt{\color{blue}{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \cdot \sin th \]

      5. hypot-udef 99.5%

         \[\leadsto \frac{ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \cdot \sin th \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

   if 4.9999999999999998e-8 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 62.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 78.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\sin ky \cdot \frac{th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 4: 77.0% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -5e-5)
   (* (/ (sin ky) (hypot (sin ky) (sin kx))) th)
   (if (<= (sin ky) 5e-8)
     (* (sin th) (/ ky (hypot (sin kx) (sin ky))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -5e-5) {
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	} else if (sin(ky) <= 5e-8) {
		tmp = sin(th) * (ky / hypot(sin(kx), sin(ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -5e-5) {
		tmp = (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * th;
	} else if (Math.sin(ky) <= 5e-8) {
		tmp = Math.sin(th) * (ky / Math.hypot(Math.sin(kx), Math.sin(ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -5e-5:
		tmp = (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * th
	elif math.sin(ky) <= 5e-8:
		tmp = math.sin(th) * (ky / math.hypot(math.sin(kx), math.sin(ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -5e-5)
		tmp = Float64(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * th);
	elseif (sin(ky) <= 5e-8)
		tmp = Float64(sin(th) * Float64(ky / hypot(sin(kx), sin(ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -5e-5)
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	elseif (sin(ky) <= 5e-8)
		tmp = sin(th) * (ky / hypot(sin(kx), sin(ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -5e-5], N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * th), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 5e-8], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -5.00000000000000024e-5

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in th around 0 56.2%

      \[\leadsto \frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. *-commutative 56.2%

         \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

   6. Simplified 56.2%

      \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   7. Step-by-step derivation

      1. associate-/l* 56.1%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}} \]

      2. associate-/r/ 56.2%

         \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th} \]

   8. Applied egg-rr 56.2%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th} \]

   if -5.00000000000000024e-5 < (sin.f64 ky) < 4.9999999999999998e-8

   1. Initial program 88.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 83.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 90.9%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 90.9%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in ky around 0 90.7%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. associate-/l* 99.4%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      2. associate-/r/ 99.5%

         \[\leadsto \color{blue}{\frac{ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

      3. hypot-udef 87.8%

         \[\leadsto \frac{ky}{\color{blue}{\sqrt{\sin ky \cdot \sin ky + \sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. +-commutative 87.8%

         \[\leadsto \frac{ky}{\sqrt{\color{blue}{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \cdot \sin th \]

      5. hypot-udef 99.5%

         \[\leadsto \frac{ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \cdot \sin th \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

   if 4.9999999999999998e-8 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 62.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 78.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 5: 77.1% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -5e-5)
   (/
    (sin ky)
    (* (hypot (sin ky) (sin kx)) (+ (/ 1.0 th) (* th 0.16666666666666666))))
   (if (<= (sin ky) 5e-8)
     (* (sin th) (/ ky (hypot (sin kx) (sin ky))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -5e-5) {
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (sin(ky) <= 5e-8) {
		tmp = sin(th) * (ky / hypot(sin(kx), sin(ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -5e-5) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(ky), Math.sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (Math.sin(ky) <= 5e-8) {
		tmp = Math.sin(th) * (ky / Math.hypot(Math.sin(kx), Math.sin(ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -5e-5:
		tmp = math.sin(ky) / (math.hypot(math.sin(ky), math.sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)))
	elif math.sin(ky) <= 5e-8:
		tmp = math.sin(th) * (ky / math.hypot(math.sin(kx), math.sin(ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -5e-5)
		tmp = Float64(sin(ky) / Float64(hypot(sin(ky), sin(kx)) * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	elseif (sin(ky) <= 5e-8)
		tmp = Float64(sin(th) * Float64(ky / hypot(sin(kx), sin(ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -5e-5)
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)));
	elseif (sin(ky) <= 5e-8)
		tmp = sin(th) * (ky / hypot(sin(kx), sin(ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -5e-5], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] * N[(N[(1.0 / th), $MachinePrecision] + N[(th * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 5e-8], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -5.00000000000000024e-5

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-/r/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      2. +-commutative 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}}{\sin th}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\sin kx \cdot \sin kx}}}{\sin th}} \]

      4. sqr-neg 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\left(-\sin kx\right) \cdot \left(-\sin kx\right)}}}{\sin th}} \]

      5. sin-neg 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\sin \left(-kx\right)} \cdot \left(-\sin kx\right)}}{\sin th}} \]

      6. sin-neg 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \sin \left(-kx\right) \cdot \color{blue}{\sin \left(-kx\right)}}}{\sin th}} \]

      7. unpow2 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{{\sin \left(-kx\right)}^{2}}}}{\sin th}} \]

      8. +-commutative 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin \left(-kx\right)}^{2} + {\sin ky}^{2}}}}{\sin th}} \]

   3. Simplified 99.5%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

   4. Taylor expanded in th around 0 56.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right) + \frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   5. Step-by-step derivation

      1. +-commutative 56.4%

         \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)}} \]

      2. unpow2 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      3. unpow2 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      4. hypot-def 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      5. associate-*r* 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \color{blue}{\left(0.16666666666666666 \cdot th\right) \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      6. unpow2 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}} \]

      7. unpow2 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}} \]

      8. hypot-def 56.4%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      9. distribute-rgt-out 56.4%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   6. Simplified 56.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   if -5.00000000000000024e-5 < (sin.f64 ky) < 4.9999999999999998e-8

   1. Initial program 88.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 83.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 83.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 90.9%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 90.9%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in ky around 0 90.7%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. associate-/l* 99.4%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      2. associate-/r/ 99.5%

         \[\leadsto \color{blue}{\frac{ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

      3. hypot-udef 87.8%

         \[\leadsto \frac{ky}{\color{blue}{\sqrt{\sin ky \cdot \sin ky + \sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. +-commutative 87.8%

         \[\leadsto \frac{ky}{\sqrt{\color{blue}{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \cdot \sin th \]

      5. hypot-udef 99.5%

         \[\leadsto \frac{ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \cdot \sin th \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

   if 4.9999999999999998e-8 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 62.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 78.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 6: 46.2% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-308}:\\ \;\;\;\;\frac{ky}{\frac{\sin kx}{\sin th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -2e-308)
   (/ ky (/ (sin kx) (sin th)))
   (if (<= (sin ky) 1e-69) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -2e-308) {
		tmp = ky / (sin(kx) / sin(th));
	} else if (sin(ky) <= 1e-69) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= (-2d-308)) then
        tmp = ky / (sin(kx) / sin(th))
    else if (sin(ky) <= 1d-69) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -2e-308) {
		tmp = ky / (Math.sin(kx) / Math.sin(th));
	} else if (Math.sin(ky) <= 1e-69) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -2e-308:
		tmp = ky / (math.sin(kx) / math.sin(th))
	elif math.sin(ky) <= 1e-69:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -2e-308)
		tmp = Float64(ky / Float64(sin(kx) / sin(th)));
	elseif (sin(ky) <= 1e-69)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -2e-308)
		tmp = ky / (sin(kx) / sin(th));
	elseif (sin(ky) <= 1e-69)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -2e-308], N[(ky / N[(N[Sin[kx], $MachinePrecision] / N[Sin[th], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-69], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -1.9999999999999998e-308

   1. Initial program 92.1%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 21.5%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 22.8%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

   4. Simplified 22.8%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

   if -1.9999999999999998e-308 < (sin.f64 ky) < 9.9999999999999996e-70

   1. Initial program 90.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 61.9%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 48.0%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{ky}{\sin kx}} \cdot \sqrt{\frac{ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 68.9%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 68.9%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   6. Applied egg-rr 68.9%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. unpow2 68.9%

         \[\leadsto \sqrt{\color{blue}{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 94.7%

         \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   8. Simplified 94.7%

      \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   if 9.9999999999999996e-70 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 61.6%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 45.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-308}:\\ \;\;\;\;\frac{ky}{\frac{\sin kx}{\sin th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 7: 47.4% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-308}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -2e-308)
   (* (sin th) (/ (sin ky) (sin kx)))
   (if (<= (sin ky) 1e-69) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -2e-308) {
		tmp = sin(th) * (sin(ky) / sin(kx));
	} else if (sin(ky) <= 1e-69) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= (-2d-308)) then
        tmp = sin(th) * (sin(ky) / sin(kx))
    else if (sin(ky) <= 1d-69) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -2e-308) {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.sin(kx));
	} else if (Math.sin(ky) <= 1e-69) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -2e-308:
		tmp = math.sin(th) * (math.sin(ky) / math.sin(kx))
	elif math.sin(ky) <= 1e-69:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -2e-308)
		tmp = Float64(sin(th) * Float64(sin(ky) / sin(kx)));
	elseif (sin(ky) <= 1e-69)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -2e-308)
		tmp = sin(th) * (sin(ky) / sin(kx));
	elseif (sin(ky) <= 1e-69)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -2e-308], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-69], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -1.9999999999999998e-308

   1. Initial program 92.1%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 24.5%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]

   if -1.9999999999999998e-308 < (sin.f64 ky) < 9.9999999999999996e-70

   1. Initial program 90.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 61.9%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 48.0%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{ky}{\sin kx}} \cdot \sqrt{\frac{ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 68.9%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 68.9%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   6. Applied egg-rr 68.9%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. unpow2 68.9%

         \[\leadsto \sqrt{\color{blue}{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 94.7%

         \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   8. Simplified 94.7%

      \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   if 9.9999999999999996e-70 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 61.6%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 46.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-308}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 8: 46.4% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-270}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 2e-270)
   (* (sin ky) (/ (sin th) (sin kx)))
   (if (<= (sin ky) 1e-69) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 2e-270) {
		tmp = sin(ky) * (sin(th) / sin(kx));
	} else if (sin(ky) <= 1e-69) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 2d-270) then
        tmp = sin(ky) * (sin(th) / sin(kx))
    else if (sin(ky) <= 1d-69) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 2e-270) {
		tmp = Math.sin(ky) * (Math.sin(th) / Math.sin(kx));
	} else if (Math.sin(ky) <= 1e-69) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 2e-270:
		tmp = math.sin(ky) * (math.sin(th) / math.sin(kx))
	elif math.sin(ky) <= 1e-69:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 2e-270)
		tmp = Float64(sin(ky) * Float64(sin(th) / sin(kx)));
	elseif (sin(ky) <= 1e-69)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 2e-270)
		tmp = sin(ky) * (sin(th) / sin(kx));
	elseif (sin(ky) <= 1e-69)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 2e-270], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-69], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < 2.0000000000000001e-270

   1. Initial program 91.9%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-/r/ 91.7%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      2. +-commutative 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}}{\sin th}} \]

      3. unpow2 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\sin kx \cdot \sin kx}}}{\sin th}} \]

      4. sqr-neg 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\left(-\sin kx\right) \cdot \left(-\sin kx\right)}}}{\sin th}} \]

      5. sin-neg 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\sin \left(-kx\right)} \cdot \left(-\sin kx\right)}}{\sin th}} \]

      6. sin-neg 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \sin \left(-kx\right) \cdot \color{blue}{\sin \left(-kx\right)}}}{\sin th}} \]

      7. unpow2 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{{\sin \left(-kx\right)}^{2}}}}{\sin th}} \]

      8. +-commutative 91.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin \left(-kx\right)}^{2} + {\sin ky}^{2}}}}{\sin th}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]
   4. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto \color{blue}{\frac{1}{\frac{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}{\sin ky}}} \]

      2. associate-/r/ 99.5%

         \[\leadsto \color{blue}{\frac{1}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}} \cdot \sin ky} \]

      3. clear-num 99.7%

         \[\leadsto \color{blue}{\frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin ky \]

      4. hypot-udef 91.8%

         \[\leadsto \frac{\sin th}{\color{blue}{\sqrt{\sin ky \cdot \sin ky + \sin kx \cdot \sin kx}}} \cdot \sin ky \]

      5. unpow2 91.8%

         \[\leadsto \frac{\sin th}{\sqrt{\color{blue}{{\sin ky}^{2}} + \sin kx \cdot \sin kx}} \cdot \sin ky \]

      6. unpow2 91.8%

         \[\leadsto \frac{\sin th}{\sqrt{{\sin ky}^{2} + \color{blue}{{\sin kx}^{2}}}} \cdot \sin ky \]

      7. +-commutative 91.8%

         \[\leadsto \frac{\sin th}{\sqrt{\color{blue}{{\sin kx}^{2} + {\sin ky}^{2}}}} \cdot \sin ky \]

      8. unpow2 91.8%

         \[\leadsto \frac{\sin th}{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}} \cdot \sin ky \]

      9. unpow2 91.8%

         \[\leadsto \frac{\sin th}{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}} \cdot \sin ky \]

      10. hypot-def 99.7%

          \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \cdot \sin ky \]

   5. Applied egg-rr 99.7%

      \[\leadsto \color{blue}{\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin ky} \]

   6. Taylor expanded in ky around 0 26.4%

      \[\leadsto \color{blue}{\frac{\sin th}{\sin kx}} \cdot \sin ky \]

   if 2.0000000000000001e-270 < (sin.f64 ky) < 9.9999999999999996e-70

   1. Initial program 90.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 90.6%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 90.6%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 90.6%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.8%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.8%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 63.2%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 49.1%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{ky}{\sin kx}} \cdot \sqrt{\frac{ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 72.8%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 72.8%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   6. Applied egg-rr 72.8%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. unpow2 72.8%

         \[\leadsto \sqrt{\color{blue}{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 93.3%

         \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   8. Simplified 93.3%

      \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   if 9.9999999999999996e-70 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 61.6%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 45.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-270}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 9: 47.4% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-308}:\\ \;\;\;\;\frac{\sin ky}{\frac{\sin kx}{\sin th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -2e-308)
   (/ (sin ky) (/ (sin kx) (sin th)))
   (if (<= (sin ky) 1e-69) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -2e-308) {
		tmp = sin(ky) / (sin(kx) / sin(th));
	} else if (sin(ky) <= 1e-69) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= (-2d-308)) then
        tmp = sin(ky) / (sin(kx) / sin(th))
    else if (sin(ky) <= 1d-69) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -2e-308) {
		tmp = Math.sin(ky) / (Math.sin(kx) / Math.sin(th));
	} else if (Math.sin(ky) <= 1e-69) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -2e-308:
		tmp = math.sin(ky) / (math.sin(kx) / math.sin(th))
	elif math.sin(ky) <= 1e-69:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -2e-308)
		tmp = Float64(sin(ky) / Float64(sin(kx) / sin(th)));
	elseif (sin(ky) <= 1e-69)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -2e-308)
		tmp = sin(ky) / (sin(kx) / sin(th));
	elseif (sin(ky) <= 1e-69)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -2e-308], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sin[kx], $MachinePrecision] / N[Sin[th], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-69], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -1.9999999999999998e-308

   1. Initial program 92.1%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-/r/ 91.9%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      2. +-commutative 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}}{\sin th}} \]

      3. unpow2 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\sin kx \cdot \sin kx}}}{\sin th}} \]

      4. sqr-neg 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\left(-\sin kx\right) \cdot \left(-\sin kx\right)}}}{\sin th}} \]

      5. sin-neg 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{\sin \left(-kx\right)} \cdot \left(-\sin kx\right)}}{\sin th}} \]

      6. sin-neg 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \sin \left(-kx\right) \cdot \color{blue}{\sin \left(-kx\right)}}}{\sin th}} \]

      7. unpow2 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin ky}^{2} + \color{blue}{{\sin \left(-kx\right)}^{2}}}}{\sin th}} \]

      8. +-commutative 91.9%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin \left(-kx\right)}^{2} + {\sin ky}^{2}}}}{\sin th}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

   4. Taylor expanded in ky around 0 24.5%

      \[\leadsto \frac{\sin ky}{\color{blue}{\frac{\sin kx}{\sin th}}} \]

   if -1.9999999999999998e-308 < (sin.f64 ky) < 9.9999999999999996e-70

   1. Initial program 90.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 90.2%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 61.9%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 48.0%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{ky}{\sin kx}} \cdot \sqrt{\frac{ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 68.9%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 68.9%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   6. Applied egg-rr 68.9%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. unpow2 68.9%

         \[\leadsto \sqrt{\color{blue}{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 94.7%

         \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   8. Simplified 94.7%

      \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   if 9.9999999999999996e-70 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 61.6%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 46.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-308}:\\ \;\;\;\;\frac{\sin ky}{\frac{\sin kx}{\sin th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 10: 46.4% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-270}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 2e-270)
   (/ (sin th) (/ (sin kx) (sin ky)))
   (if (<= (sin ky) 1e-69) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 2e-270) {
		tmp = sin(th) / (sin(kx) / sin(ky));
	} else if (sin(ky) <= 1e-69) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 2d-270) then
        tmp = sin(th) / (sin(kx) / sin(ky))
    else if (sin(ky) <= 1d-69) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 2e-270) {
		tmp = Math.sin(th) / (Math.sin(kx) / Math.sin(ky));
	} else if (Math.sin(ky) <= 1e-69) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 2e-270:
		tmp = math.sin(th) / (math.sin(kx) / math.sin(ky))
	elif math.sin(ky) <= 1e-69:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 2e-270)
		tmp = Float64(sin(th) / Float64(sin(kx) / sin(ky)));
	elseif (sin(ky) <= 1e-69)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 2e-270)
		tmp = sin(th) / (sin(kx) / sin(ky));
	elseif (sin(ky) <= 1e-69)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 2e-270], N[(N[Sin[th], $MachinePrecision] / N[(N[Sin[kx], $MachinePrecision] / N[Sin[ky], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-69], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < 2.0000000000000001e-270

   1. Initial program 91.9%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 26.5%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]

   3. Taylor expanded in ky around inf 24.7%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sin kx}} \]
   4. Step-by-step derivation

      1. *-commutative 24.7%

         \[\leadsto \frac{\color{blue}{\sin th \cdot \sin ky}}{\sin kx} \]

      2. associate-/l* 26.5%

         \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sin kx}{\sin ky}}} \]

   5. Simplified 26.5%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sin kx}{\sin ky}}} \]

   if 2.0000000000000001e-270 < (sin.f64 ky) < 9.9999999999999996e-70

   1. Initial program 90.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 90.6%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 90.6%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 90.6%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.8%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.8%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 63.2%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 49.1%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{ky}{\sin kx}} \cdot \sqrt{\frac{ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 72.8%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 72.8%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   6. Applied egg-rr 72.8%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. unpow2 72.8%

         \[\leadsto \sqrt{\color{blue}{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 93.3%

         \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   8. Simplified 93.3%

      \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   if 9.9999999999999996e-70 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 61.6%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 45.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-270}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq 10^{-69}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 11: 65.8% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 5e-8)
   (* ky (/ (sin th) (hypot (sin kx) (sin ky))))
   (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 5e-8) {
		tmp = ky * (sin(th) / hypot(sin(kx), sin(ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 5e-8) {
		tmp = ky * (Math.sin(th) / Math.hypot(Math.sin(kx), Math.sin(ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 5e-8:
		tmp = ky * (math.sin(th) / math.hypot(math.sin(kx), math.sin(ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 5e-8)
		tmp = Float64(ky * Float64(sin(th) / hypot(sin(kx), sin(ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 5e-8)
		tmp = ky * (sin(th) / hypot(sin(kx), sin(ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 5e-8], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 4.9999999999999998e-8

   1. Initial program 92.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 89.4%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 89.4%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 89.4%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 89.4%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 94.1%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 94.1%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in ky around 0 59.4%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. associate-/l* 64.9%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      2. div-inv 64.8%

         \[\leadsto \color{blue}{ky \cdot \frac{1}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{\sin th}}} \]

      3. clear-num 64.9%

         \[\leadsto ky \cdot \color{blue}{\frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

      4. hypot-udef 57.4%

         \[\leadsto ky \cdot \frac{\sin th}{\color{blue}{\sqrt{\sin ky \cdot \sin ky + \sin kx \cdot \sin kx}}} \]

      5. +-commutative 57.4%

         \[\leadsto ky \cdot \frac{\sin th}{\sqrt{\color{blue}{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

      6. hypot-udef 64.9%

         \[\leadsto ky \cdot \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   6. Applied egg-rr 64.9%

      \[\leadsto \color{blue}{ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   if 4.9999999999999998e-8 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 62.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 64.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 5 \cdot 10^{-8}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 12: 33.9% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-73) (* (sin ky) (/ (sin th) kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-73) {
		tmp = sin(ky) * (sin(th) / kx);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 1d-73) then
        tmp = sin(ky) * (sin(th) / kx)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 1e-73) {
		tmp = Math.sin(ky) * (Math.sin(th) / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-73:
		tmp = math.sin(ky) * (math.sin(th) / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-73)
		tmp = Float64(sin(ky) * Float64(sin(th) / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 1e-73)
		tmp = sin(ky) * (sin(th) / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-73], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 9.99999999999999997e-74

   1. Initial program 91.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 32.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]

   3. Taylor expanded in kx around 0 21.3%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{kx}} \]
   4. Step-by-step derivation

      1. associate-/l* 23.1%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{kx}{\sin th}}} \]

      2. div-inv 23.1%

         \[\leadsto \color{blue}{\sin ky \cdot \frac{1}{\frac{kx}{\sin th}}} \]

   5. Applied egg-rr 23.1%

      \[\leadsto \color{blue}{\sin ky \cdot \frac{1}{\frac{kx}{\sin th}}} \]

   6. Taylor expanded in kx around 0 23.1%

      \[\leadsto \sin ky \cdot \color{blue}{\frac{\sin th}{kx}} \]

   if 9.99999999999999997e-74 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 60.7%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 34.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 13: 40.9% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-73) (* (sin th) (/ ky (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-73) {
		tmp = sin(th) * (ky / sin(kx));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 1d-73) then
        tmp = sin(th) * (ky / sin(kx))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 1e-73) {
		tmp = Math.sin(th) * (ky / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-73:
		tmp = math.sin(th) * (ky / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-73)
		tmp = Float64(sin(th) * Float64(ky / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 1e-73)
		tmp = sin(th) * (ky / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-73], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 9.99999999999999997e-74

   1. Initial program 91.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 31.1%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]

   if 9.99999999999999997e-74 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 60.7%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 40.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 14: 40.9% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-73) (* ky (/ (sin th) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-73) {
		tmp = ky * (sin(th) / sin(kx));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 1d-73) then
        tmp = ky * (sin(th) / sin(kx))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 1e-73) {
		tmp = ky * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-73:
		tmp = ky * (math.sin(th) / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-73)
		tmp = Float64(ky * Float64(sin(th) / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 1e-73)
		tmp = ky * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-73], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 9.99999999999999997e-74

   1. Initial program 91.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 31.1%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]
   5. Step-by-step derivation

      1. *-commutative 31.1%

         \[\leadsto \color{blue}{\sin th \cdot \frac{ky}{\sin kx}} \]

      2. clear-num 31.1%

         \[\leadsto \sin th \cdot \color{blue}{\frac{1}{\frac{\sin kx}{ky}}} \]

      3. un-div-inv 31.1%

         \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sin kx}{ky}}} \]

   6. Applied egg-rr 31.1%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sin kx}{ky}}} \]
   7. Step-by-step derivation

      1. associate-/r/ 31.1%

         \[\leadsto \color{blue}{\frac{\sin th}{\sin kx} \cdot ky} \]

   8. Simplified 31.1%

      \[\leadsto \color{blue}{\frac{\sin th}{\sin kx} \cdot ky} \]

   if 9.99999999999999997e-74 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 60.7%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 40.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 15: 30.8% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin ky \cdot \frac{th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-73) (* (sin ky) (/ th kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-73) {
		tmp = sin(ky) * (th / kx);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 1d-73) then
        tmp = sin(ky) * (th / kx)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 1e-73) {
		tmp = Math.sin(ky) * (th / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-73:
		tmp = math.sin(ky) * (th / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-73)
		tmp = Float64(sin(ky) * Float64(th / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 1e-73)
		tmp = sin(ky) * (th / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-73], N[(N[Sin[ky], $MachinePrecision] * N[(th / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 9.99999999999999997e-74

   1. Initial program 91.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 32.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]

   3. Taylor expanded in kx around 0 21.3%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{kx}} \]
   4. Step-by-step derivation

      1. associate-/l* 23.1%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{kx}{\sin th}}} \]

      2. div-inv 23.1%

         \[\leadsto \color{blue}{\sin ky \cdot \frac{1}{\frac{kx}{\sin th}}} \]

   5. Applied egg-rr 23.1%

      \[\leadsto \color{blue}{\sin ky \cdot \frac{1}{\frac{kx}{\sin th}}} \]

   6. Taylor expanded in th around 0 20.1%

      \[\leadsto \sin ky \cdot \color{blue}{\frac{th}{kx}} \]

   if 9.99999999999999997e-74 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 60.7%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 32.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin ky \cdot \frac{th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 16: 33.7% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-73) (* (sin th) (/ ky kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-73) {
		tmp = sin(th) * (ky / kx);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= 1d-73) then
        tmp = sin(th) * (ky / kx)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 1e-73) {
		tmp = Math.sin(th) * (ky / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-73:
		tmp = math.sin(th) * (ky / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-73)
		tmp = Float64(sin(th) * Float64(ky / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 1e-73)
		tmp = sin(th) * (ky / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-73], N[(N[Sin[th], $MachinePrecision] * N[(ky / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 9.99999999999999997e-74

   1. Initial program 91.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 91.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 31.1%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]

   5. Taylor expanded in kx around 0 23.0%

      \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]

   if 9.99999999999999997e-74 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 60.7%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 34.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-73}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 17: 30.4% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq -1000:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq 9 \cdot 10^{-74}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky -1000.0) (sin th) (if (<= ky 9e-74) (/ ky (/ kx th)) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= -1000.0) {
		tmp = sin(th);
	} else if (ky <= 9e-74) {
		tmp = ky / (kx / th);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= (-1000.0d0)) then
        tmp = sin(th)
    else if (ky <= 9d-74) then
        tmp = ky / (kx / th)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= -1000.0) {
		tmp = Math.sin(th);
	} else if (ky <= 9e-74) {
		tmp = ky / (kx / th);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= -1000.0:
		tmp = math.sin(th)
	elif ky <= 9e-74:
		tmp = ky / (kx / th)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= -1000.0)
		tmp = sin(th);
	elseif (ky <= 9e-74)
		tmp = Float64(ky / Float64(kx / th));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= -1000.0)
		tmp = sin(th);
	elseif (ky <= 9e-74)
		tmp = ky / (kx / th);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, -1000.0], N[Sin[th], $MachinePrecision], If[LessEqual[ky, 9e-74], N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if ky < -1e3 or 8.9999999999999998e-74 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 33.5%

      \[\leadsto \color{blue}{\sin th} \]

   if -1e3 < ky < 8.9999999999999998e-74

   1. Initial program 86.1%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 86.1%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 86.1%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 86.1%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 49.4%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]

   5. Taylor expanded in kx around 0 35.9%

      \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]

   6. Taylor expanded in th around 0 27.8%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{kx}} \]
   7. Step-by-step derivation

      1. associate-/l* 30.8%

         \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

   8. Simplified 30.8%

      \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 32.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq -1000:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq 9 \cdot 10^{-74}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 18: 21.6% accurate, 77.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq -6 \cdot 10^{+34}:\\ \;\;\;\;th\\ \mathbf{elif}\;ky \leq 1.85 \cdot 10^{-73}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky -6e+34) th (if (<= ky 1.85e-73) (/ ky (/ kx th)) th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= -6e+34) {
		tmp = th;
	} else if (ky <= 1.85e-73) {
		tmp = ky / (kx / th);
	} else {
		tmp = th;
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= (-6d+34)) then
        tmp = th
    else if (ky <= 1.85d-73) then
        tmp = ky / (kx / th)
    else
        tmp = th
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= -6e+34) {
		tmp = th;
	} else if (ky <= 1.85e-73) {
		tmp = ky / (kx / th);
	} else {
		tmp = th;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= -6e+34:
		tmp = th
	elif ky <= 1.85e-73:
		tmp = ky / (kx / th)
	else:
		tmp = th
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= -6e+34)
		tmp = th;
	elseif (ky <= 1.85e-73)
		tmp = Float64(ky / Float64(kx / th));
	else
		tmp = th;
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= -6e+34)
		tmp = th;
	elseif (ky <= 1.85e-73)
		tmp = ky / (kx / th);
	else
		tmp = th;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, -6e+34], th, If[LessEqual[ky, 1.85e-73], N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision], th]]

Derivation

1. Split input into 2 regimes

2. if ky < -6.00000000000000037e34 or 1.85e-73 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. +-commutative 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      4. unpow2 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   4. Taylor expanded in th around 0 53.9%

      \[\leadsto \frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
   5. Step-by-step derivation

      1. *-commutative 53.9%

         \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

   6. Simplified 53.9%

      \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

   7. Taylor expanded in kx around 0 18.8%

      \[\leadsto \color{blue}{th} \]

   if -6.00000000000000037e34 < ky < 1.85e-73

   1. Initial program 87.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. +-commutative 87.0%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

      2. unpow2 87.0%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

      3. unpow2 87.0%

         \[\leadsto \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

      4. hypot-def 99.7%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

   3. Simplified 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

   4. Taylor expanded in ky around 0 46.4%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx}} \cdot \sin th \]

   5. Taylor expanded in kx around 0 33.6%

      \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]

   6. Taylor expanded in th around 0 26.0%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{kx}} \]
   7. Step-by-step derivation

      1. associate-/l* 28.8%

         \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

   8. Simplified 28.8%

      \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 23.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq -6 \cdot 10^{+34}:\\ \;\;\;\;th\\ \mathbf{elif}\;ky \leq 1.85 \cdot 10^{-73}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \]

Alternative 19: 13.8% accurate, 709.0× speedup

Math

\[\begin{array}{l} \\ th \end{array} \]

FPCore

(FPCore (kx ky th) :precision binary64 th)

C

double code(double kx, double ky, double th) {
	return th;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = th
end function

Java

public static double code(double kx, double ky, double th) {
	return th;
}

Python

def code(kx, ky, th):
	return th

Julia

function code(kx, ky, th)
	return th
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = th;
end

Wolfram

code[kx_, ky_, th_] := th

Derivation

1. Initial program 94.0%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Step-by-step derivation

   1. associate-*l/ 91.8%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   2. +-commutative 91.8%

      \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

   3. unpow2 91.8%

      \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

   4. unpow2 91.8%

      \[\leadsto \frac{\sin ky \cdot \sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

   5. hypot-def 95.4%

      \[\leadsto \frac{\sin ky \cdot \sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

3. Simplified 95.4%

   \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

4. Taylor expanded in th around 0 49.1%

   \[\leadsto \frac{\color{blue}{th \cdot \sin ky}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
5. Step-by-step derivation

   1. *-commutative 49.1%

      \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

6. Simplified 49.1%

   \[\leadsto \frac{\color{blue}{\sin ky \cdot th}}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

7. Taylor expanded in kx around 0 12.1%

   \[\leadsto \color{blue}{th} \]

8. Final simplification 12.1%

   \[\leadsto th \]

Reproduce

herbie shell --seed 2023291 
(FPCore (kx ky th)
  :name "Toniolo and Linder, Equation (3b), real"
  :precision binary64
  (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))