Prove the identities:

Question

(sin^2x-1)^2 = cos(2x)+sin^4x

Answer 1

(sin^2 (x) - 1)^2 = cos(2x) + sin^4 (x)
We'll only modify the left side. Expand the left side:
(sin^2 (x) - 1)^2
sin^4 (x) - 2 sin^2 (x) + 1

Recall that 1 = cos^2 (x) + sin^2 (x) : Pythagorean identity.
sin^4 (x) - 2 sin^2 (x) + cos^2 (x) + sin^2 (x)
sin^4 (x) + cos^2 (x) - sin^2 (x)

Recall the sum of cosines: cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B).
Thus, cos^2 (x) - sin^2 (x) = cos(x + x):
sin^4 (x) + cos(x + x)
sin^4 (x) + cos(2x)

Hope this helps~ `u`