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.

Yes, you can use double angle identity to simplify as follows $$\left(\sin\frac{\pi}{8}+\cos\frac{\pi}{8}\right)^2=\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
$$=\left(\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}\right)+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
$$=1+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
using double angle identity, $2\sin A\cos A=\sin 2A$
$$=1+\sin2\left(\frac{\pi}{8}\right)$$
$$=1+\sin\frac{\pi}{4}$$
$$=1+\frac{1}{\sqrt 2}=\color{red}{\frac{2+\sqrt 2}{2}}$$
