Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$ 
My question is, is this a double angle formula? And if so, how would I go about to solve it? 
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$ 
$$=2\sin{\pi \over 4}+\left(1-2\sin{\pi \over 4}\right)$$ 
Have I done this right so far? I feel I have not.

It is related to a double-angle identity.  The relevant identities you need are:
$$\sin^2\theta + \cos^2\theta = 1$$
and
$$\sin 2\theta = 2\sin\theta\cos\theta$$
You will also need to expand your binomial, using the basic algebraic identity
$$(A+B)^2 = A^2 + 2AB + B^2$$.
So, start by expanding the binomial; then use the two trig identities to simplify.
