Verify the identity

Question

How would I verify this identify. Thank you.

2 sin (𝜋/4 + 𝑥−𝑦/2) cos (𝜋/4 + 𝑥+𝑦/2) = cos𝑥−sin𝑦

Answer 1

2 sin A cos B = sin(A + B) + sin(A - B) (it is easier to prove this going right to left, as you use the sin addition formula, or look it up.

I am assuming that x - y and x + y are in numerators.

Thus, 2 sin (𝜋/4 + (𝑥−𝑦)/2) cos (𝜋/4 + (𝑥+𝑦)/2) = (substituting A with the first term and B with the second)

sin(𝜋/4 + (𝑥−𝑦)/2 + 𝜋/4 + (𝑥+𝑦)/2) + sin(𝜋/4 + (𝑥−𝑦)/2 - (𝜋/4 + (𝑥+𝑦)/2)) =

sin(𝜋/2 + x) + sin(-y) =

sin 𝜋/2 cos x + cos 𝜋/2 sin x - sin y =

1 cos x + 0 sin x - sin y =

cos x - sin y

Answer 2

Just a guess, but maybe there is a typo or two in the problem, such as a sign and missing exponents?

2sin(pi/4 + x + y/2)cos(pi/4 + x + y/2) by the double angle formula

= sin[2(pi/4 + x + y/2)] = sin(pi/2 + 2x + y)

use addition formulas

= sin(pi/2)cos(2x+y) + cos(pi/2)sin(2x+y) = (1)(cos(2x+y) + (0)sin(2x+y)

= cos(2x+y)

use another addition formula

= cos^2(2x) - sin^2(y) which looks like the right side of the problem you have, except for 2 exponents