(cosx + sin2x)^2 + (sinx - cos2x)^2 [is identically equal to] 2(1 + sinx)?

Question

the first half is identically equal to the second half and NOT equal to the second half

Answer 1

Prove the identity.

(cosx + sin 2x)² + (sinx - cos 2x)² = 2(1 + sinx)
_____________

Left Hand Side = (cosx + sin 2x)² + (sinx - cos 2x)²

= cos²x + 2(cosx)(sin 2x) + sin²2x
+ sin²x - 2(sinx)(cos 2x) + cos²2x

= (cos²x + sin²x) + (cos²2x + sin²2x)
+ 2(cosx)(sin 2x) - 2(sinx)(cos 2x)

= 1 + 1 + 2(cosx)[2(sinx)(cosx)] - 2(sinx)(2cos²x - 1)

= 2 + 4(cos²x)(sinx) - 4(cos²x)(sinx) + 2sinx

= 2 + 2sinx = 2(1 + sinx) = Right Hand Side