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.

In general, you can write expressions of the form $a\sin \theta + b\cos\theta$ in the form $c\sin(\theta + \psi)$ for suitable choices of $c$ and $\psi$. To see this, look at the double-angle formula for sine; you obtain
$$a\sin\theta + b\cos\theta = (c\cos \psi)\sin\theta + (c\sin\psi)\cos\theta$$
Comparing coefficients gives $a=c\cos\psi$ and $b=c\sin\psi$.
Dividing the second equation by the first gives $\tan\psi = \frac{b}{a}$; and squaring both equations and adding them together gives
$$a^2+b^2 = c^2(\cos^2\psi+\sin^2\psi) = c^2$$
So you can work out the values of $c$ and $\psi$ from here.
Here, $a=b=1$, so we get
$$\tan\psi = 1 \quad \text{and} \quad c^2=1^2+1^2=2$$
so $\psi = \frac{\pi}{4}$ and $c = \sqrt{2}$ will work. Hence
$$\boxed{\sin \frac{\pi}{8} + \cos \frac{\pi}{8} = \sqrt{2}\sin\left(\frac{\pi}{8} + \frac{\pi}{4}\right)}$$
You can probably take it from here.
