Question

(cosx+sinx)^2-(cosx-sinx)^2=2sin(2x) proof

Answer 1

Oops....a long one. Let's simplify first by using the convention sinx = S and cosx = C.

This reduces the problem to:
(S + C)^2 - (C - S)^2) =
S^2 + 2*S*C + C^2 -[C^2 -2*S*C + S^2] =
S^2 + 2SC + C^2 -C^2 +2SC-S^2 =
4SC = 4* sinx * cosx

We also know : sin(2x) = 2*sinx*cosx, so 4*sinx*cosx = 2*sin(2x),

Hence manipulating the left hand side results in 2*sin(2x). Comparing this with the righthand side in the problem description shows that both are identical. Not really proof, but understandable.

Answer 2

I like the way NLCircle did it, but this might be easier....

First expand on the left hand side\[(cosx+sinx)^2-(cosx-sinx)^2=2\sin(2x) \]\[4cosxsinx=\sin(2x)\]Cancel your common terms.\[2cosxsinx = 2cosxsinx\]PROVED!