ab-angle->ABCF C

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Percentage Accurate: 79.7% → 79.6%
Time: 20.3s
Precision: binary64
Cost: 26240

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\[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
↓
\[{a}^{2} + {\left(b \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} \]
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (cos (* PI (/ angle 180.0)))) 2.0)
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))
↓
(FPCore (a b angle)
 :precision binary64
 (+ (pow a 2.0) (pow (* b (sin (/ PI (/ 180.0 angle)))) 2.0)))
double code(double a, double b, double angle) {
	return pow((a * cos((((double) M_PI) * (angle / 180.0)))), 2.0) + pow((b * sin((((double) M_PI) * (angle / 180.0)))), 2.0);
}
↓
double code(double a, double b, double angle) {
	return pow(a, 2.0) + pow((b * sin((((double) M_PI) / (180.0 / angle)))), 2.0);
}

public static double code(double a, double b, double angle) {
	return Math.pow((a * Math.cos((Math.PI * (angle / 180.0)))), 2.0) + Math.pow((b * Math.sin((Math.PI * (angle / 180.0)))), 2.0);
}
↓
public static double code(double a, double b, double angle) {
	return Math.pow(a, 2.0) + Math.pow((b * Math.sin((Math.PI / (180.0 / angle)))), 2.0);
}

def code(a, b, angle):
	return math.pow((a * math.cos((math.pi * (angle / 180.0)))), 2.0) + math.pow((b * math.sin((math.pi * (angle / 180.0)))), 2.0)
↓
def code(a, b, angle):
	return math.pow(a, 2.0) + math.pow((b * math.sin((math.pi / (180.0 / angle)))), 2.0)

function code(a, b, angle)
	return Float64((Float64(a * cos(Float64(pi * Float64(angle / 180.0)))) ^ 2.0) + (Float64(b * sin(Float64(pi * Float64(angle / 180.0)))) ^ 2.0))
end
↓
function code(a, b, angle)
	return Float64((a ^ 2.0) + (Float64(b * sin(Float64(pi / Float64(180.0 / angle)))) ^ 2.0))
end

function tmp = code(a, b, angle)
	tmp = ((a * cos((pi * (angle / 180.0)))) ^ 2.0) + ((b * sin((pi * (angle / 180.0)))) ^ 2.0);
end
↓
function tmp = code(a, b, angle)
	tmp = (a ^ 2.0) + ((b * sin((pi / (180.0 / angle)))) ^ 2.0);
end

code[a_, b_, angle_] := N[(N[Power[N[(a * N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
↓
code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2}
↓
{a}^{2} + {\left(b \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2}

Local Percentage Accuracy?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 7 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each of Herbie's proposed alternatives. Up and to the right is better. Each dot represents an alternative program; the red square represents the initial program.

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 81.3%

    \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

  2. Taylor expanded in angle around 0 82.0%

    \[\leadsto {\left(a \cdot \color{blue}{1}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

  3. Applied egg-rr82.1%

    \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} \]
    Step-by-step derivation

    [Start]82.0

    \[ {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

    clear-num [=>]82.0

    \[ {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \color{blue}{\frac{1}{\frac{180}{angle}}}\right)\right)}^{2} \]

    un-div-inv [=>]82.1

    \[ {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} \]

  4. Final simplification82.1%

    \[\leadsto {a}^{2} + {\left(b \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} \]

Alternatives

Alternative 1
Accuracy79.6%
Cost26240
\[{a}^{2} + {\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2} \]
Alternative 2
Accuracy76.7%
Cost20360
\[\begin{array}{l} \mathbf{if}\;b \leq -4 \cdot 10^{-107}:\\ \;\;\;\;{a}^{2} + {\left(0.005555555555555556 \cdot \left(\pi \cdot \left(b \cdot angle\right)\right)\right)}^{2}\\ \mathbf{elif}\;b \leq 3.8 \cdot 10^{-174}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(angle \cdot \left(angle \cdot \left(b \cdot \left(b \cdot {\pi}^{2}\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \end{array} \]
Alternative 3
Accuracy76.6%
Cost20360
\[\begin{array}{l} \mathbf{if}\;b \leq -2 \cdot 10^{+48}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(b \cdot \pi\right)\right)}^{2}\\ \mathbf{elif}\;b \leq 10^{-174}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(angle \cdot \left(angle \cdot \left({\pi}^{2} \cdot \left(b \cdot b\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \end{array} \]
Alternative 4
Accuracy75.8%
Cost20360
\[\begin{array}{l} \mathbf{if}\;b \leq -4.6 \cdot 10^{+178}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left({\pi}^{2} \cdot \left(\left(b \cdot angle\right) \cdot \left(b \cdot angle\right)\right)\right)\\ \mathbf{elif}\;b \leq 3.8 \cdot 10^{-174}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(angle \cdot \left(angle \cdot \left({\pi}^{2} \cdot \left(b \cdot b\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \end{array} \]
Alternative 5
Accuracy74.3%
Cost19840
\[{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(b \cdot \pi\right)\right)}^{2} \]
Alternative 6
Accuracy74.4%
Cost19840
\[{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2} \]

Reproduce?

herbie shell --seed 2023160 
(FPCore (a b angle)
  :name "ab-angle->ABCF C"
  :precision binary64
  (+ (pow (* a (cos (* PI (/ angle 180.0)))) 2.0) (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))