C++ (and maths) : fast approximation of a trigonometric function

Question

I know this is a recurring question, but I haven\'t really found a useful answer yet. I\'m basically looking for a fast approximation of the function acos in C+

Answer 1

sin( acos(t1) + acos(t2) + ... + acos(tn) )

boils down to the calculation of

sin( acos(x) ) and cos(acos(x))=x

because

sin(a+b) = cos(a)sin(b)+sin(a)cos(b).

The first thing is

sin( acos(x) )  = sqrt(1-x*x)

Taylor series expansion for the sqrt reduces the problem to polynomial calculations.

To clarify, here's the expansion to n=2, n=3:

sin( acos(t1) + acos(t2) ) = sin(acos(t1))cos(acos(t2)) + sin(acos(t2))cos(acos(t1) = sqrt(1-t1*t1) * t2 + sqrt(1-t2*t2) * t1

cos( acos(t2) + acos(t3) ) = cos(acos(t2)) cos(acos(t3)) - sin(acos(t2))sin(acos(t3)) = t2*t3 - sqrt(1-t2*t2)*sqrt(1-t3*t3)

sin( acos(t1) + acos(t2) + acos(t3)) = 
sin(acos(t1))cos(acos(t2) + acos(t3)) + sin(acos(t2)+acos(t3) )cos(acos(t1)=
sqrt(1-t1*t1) * (t2*t3 - sqrt(1-t2*t2)*sqrt(1-t3*t3)) + (sqrt(1-t2*t2) * t3 + sqrt(1-t3*t3) * t2 ) * t1

and so on.

The sqrt() for x in (-1,1) can be computed using

x_0 is some approximation, say, zero

x_(n+1) = 0.5 * (x_n + S/x_n)  where S is the argument.

EDIT: I mean the "Babylonian method", see Wikipedia's article for details. You will need not more than 5-6 iterations to achieve 1e-10 with x in (0,1).