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.
 A: 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.
A: HINT:
Use the identity $\sin x+\cos x=\sqrt{2}\sin(x+\pi/4)$ with $x=\pi/8$
A: 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, $2\sin A\cos A=\sin 2A$
$$=1+\sin2\left(\frac{\pi}{8}\right)$$
$$=1+\sin\frac{\pi}{4}$$
$$=1+\frac{1}{\sqrt 2}=\color{red}{\frac{2+\sqrt 2}{2}}$$
A: We have identity here to directly solve. Its $\left(\sin(x)+\cos(x)\right)^2=1+\sin(2x)$ thus here $\theta =\frac{π}{8}$ so $2\theta=\frac{π}{4}$ hence answer is $$1+\sin\left(\frac{π}{4}\right)=1+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}}$$ hope it's clear.
A: 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 $\sin A\cos B+\cos A\sin B=\sin(A+B)$ $$=2\sin^2\left(\frac{\pi}{8}+\frac{\pi}{4}\right)$$
$$=2\sin^2\left(\frac{3\pi}{8}\right)$$
A: 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.
A: We can solve an equation, but not a number, so I interpret "solve" as "compute". Then: $$(sin{\pi \over 8}+cos{\pi \over 8})^2=\sin^2(\frac{\pi}{8})+\cos^2(\frac{\pi}{8})+2\sin(\frac{\pi}{8})\cos(\frac{\pi}{8})=1+\sin(\frac{\pi}{4})=1+\frac{\sqrt{2}}{2}$$
using the Double Angle indentity $\sin(2\theta)=2\sin(\theta)\cos(\theta)$.
If the intended question was to compute $\sin(\frac{\pi}{8})+\cos(\frac{\pi}{8})$, then $\sin(\frac{\pi}{8})+\cos(\frac{\pi}{8})=\sqrt{1+\frac{\sqrt{2}}{2}}$, since $\frac{\pi}{8}$ is in the first quadrant, which means that $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$ are both positive.
