Trignometric Identities

Question

1-2sin^2x x x

------------ + 2sin ____ cos ___ = cosx

cosx + sinx 2 2

Answer 1

this might be "faster"

2sinAcosA=sin2A

have 1-2sin²x=(cosx+sinx)(cosx-sinx)=cos²x-sin²x=1-sin²x-sin²=1-2sin²x

Answer 2

I'm going to assume that the problem is to prove the identity.

Start with the more complicated side (left side) and try to morph it into the other side using the basic identities you have learned.

I'll first use:

1. 1 - 2sin²x = cos2x and cos2x = cos²x - sin²x therefore 1 - 2sin²x = cos²x - sin²x
2. sin(x/2) = √((1-cosx)/2)
3. cos(x/2) = √((1+cosx)/2)

Substitute this and simplify, get a common denominator, cancel and you've got it:

I didn't perform the last few steps but you should be able to do it from here. Combine the numerators, simplify, then factor out cos(x) and cancel