how to handle result of cos(M_PI/2)

Question

M_PI is a macro which is defined as 3.14159265358979323846, which seems more precise than a double. I'm trying to get the answer in a float and no amount of casting will help me, the result is always 6.12323426e-017 or -4.37113883e-008 if i try and cast M_PI to a float.

The answer should be 0 and I would like to store that in a float.

Answer 1

That's impossible. There's no way to represent the exact value of pi in a finite floating-point data type.

In general, you can't expect any operation on floating-point values to be exactly equal to the corresponding mathematical operation on real values. The usual approach is to compare with a tolerance, in this case perhaps something like

if (std::abs(result) < tolerance) {
    // treat it as zero
}

Choosing a suitable value of tolerance to get the accuracy you need for your particular problem is the hard part.

Answer 2

Since the floating point representation of pi will never be any closer to pi than machine epsilon, the cosine of that value over two will be similarly different from the expected result of 0. Mathematically, the floating point value of pi can be represented as pi + eps, and thus the value of the function call will be

  cos(pi/2 + eps/2) = -sin(eps/2) ~= -eps/2.

Answer 3

The fact that the M_PI macro uses more precision than a double has, does not change the fact that the argument of the cos function, M_PI/2 is calculated as a double.

So, instead of calculating the cosine of the exact number 'pi/2', which is 0, you're calculating the cosine of the floating-point number closest to 'pi/2'. The result will therefore be not exactly 0.

No amount of casting will change that.