从正弦计算余弦的性能和准确性，反之亦然

Question

Consider a right triangle of which you know the hypotenuse and one of the catheti - let's say the one opposite to the angle alpha in which you are interested. The value sin(alpha) can be easily calculated as

sin(alpha) = a / c,

with a being the opposite cathetus and c the hypotenuse. I do not know the length of the adjacent cathetus b. What would be the faster and/or more accurate way to calculate cos(alpha) ?

One can either use

   sin²(alpha) + cos²(alpha) = 1
=> cos(alpha) = Sqrt(1 - sin²(alpha)),

where you have one multiplication, one subtraction and one square root operation, or

alpha = asin(a / c)
=> cos(alpha) = cos(asin(a / c)),

where you have one inverse sine operation and the cosine operation (the quotient a / c has already been calculated).

I'm interested in the performance and the accuracy of both methods, and if there might be better methods.

Answer 1

I tested the performance with the following code in C#, on a Core i7-6700 @ 3.40GHz, 8 GB RAM, running Windows 10 and Visual Studio 2013:

  var stopwatch = new Stopwatch();
  var random = new Random();

  var numberOfValues = 1000000;
  var repetitions = 100;
  var quotients = new double[numberOfValues];
  var sineValues = new double[numberOfValues];
  var results = new double[numberOfValues];


  // Preparing values for the measurement.
  for (var i = 0; i < numberOfValues; i++)
  {
    quotients[i] = random.NextDouble() / random.NextDouble();
    sineValues[i] = Math.Sin(quotients[i]);
  }


  // First method: Squaring and taking square root.
  stopwatch.Start();
  for (var j = 0; j < repetitions; j++)
  {
    for (var i = 0; i < numberOfValues; i++)
    {
      results[i] = Math.Sqrt(1 - Math.Pow(sineValues[i], 2));
    }
  }
  stopwatch.Stop();
  Console.WriteLine(stopwatch.Elapsed);


  stopwatch.Reset();


  // Second method: Arcsine and cosine.
  stopwatch.Start();
  for (var j = 0; j < repetitions; j++)
  {
    for (var i = 0; i < numberOfValues; i++)
    {
      results[i] = Math.Cos(Math.Asin(quotients[i]));
    }
  }
  stopwatch.Stop();
  Console.WriteLine(stopwatch.Elapsed);

Results:

04.7030170 sec
10.4038198 sec,

which is only a factor of two difference. However, if Math.Pow is replaced with direct multiplication, the values change to:

00.4991018 sec
10.3393635 sec,

which yields a factor of about 20!