Please help! When evaluating sin(2Cos^-1(4/5)) what is an identity used in the solution?
a. cos2x = 1 - 2cos^2x
b. cos2x = sin^2x - cos^2x
c. sin2x = 1 - 2sin^2x
d. sin2x = 2sin x cos x

Question

Please help! When evaluating sin(2Cos^-1(4/5)) what is an identity used in the solution?
a. cos2x = 1 - 2cos^2x
b. cos2x = sin^2x - cos^2x
c. sin2x = 1 - 2sin^2x
d. sin2x = 2sin x cos x

anonymous · Answer

You probably want to double check this, but I'm thinking that sin(2theta) where theta = arccos(4/5).  The we could derive or just look at a trig book and see sin2x = 2sinxcosx. I'd go with answer d.

It doesn't ask for it, but you could also just plug the question into a scientific calculator exactly as it is to get 24/25 (if you had to evaluate it).  By hand isn't much harder (and probably better for you).