<s>[INST] 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.
 [/INST] Notice, $$\left(\sin\frac{\pi}{8}+\cos\frac{\pi}{8}\right)^2$$
$$=2\left(\frac{1}{\sqrt2}\sin\frac{\pi}{8}+\frac{1}{\sqrt2}\cos\frac{\pi}{8}\right)^2$$
$$=2\left(\sin\frac{\pi}{8}\cos\frac{\pi}{4}+\cos\frac{\pi}{8}\sin\frac{\pi}{4}\right)^2$$ 
Using trig identity $\sin A\cos B+\cos A\sin B=\sin(A+B)$ $$=2\sin^2\left(\frac{\pi}{8}+\frac{\pi}{4}\right)$$
$$=2\sin^2\left(\frac{3\pi}{8}\right)$$
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