Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$ 
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$

The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calculator, 
I was thinking more along the lines of using the Taylor polynomial series but that really didn't work out for me though.
 A: An easy way to find the value is to use the identity
$$\sin A + \sin B = 2 \sin\frac{(A+B)}{2} \cos\frac{(A-B)}{2}$$
which is equivalent to product the formula.
Plugging in the values for $A$ and $B$, we can easily see it to be
$$ 2\sin\frac{\pi}{4} \cos\frac{\pi}{6} = \sqrt\frac{3}{2}$$
A: Use the product formula:
$$\sin(a)\cos(b) = \frac{1}{2}(\sin(a+b) + \sin(a-b))$$
let $a = \frac{3\pi}{12}$ and $b = \frac{2\pi}{12}$
Then 
$$\sin(\frac{5\pi}{12}) + \sin(\frac{\pi}{12})=
\sin(\frac{3\pi}{12}+\frac{2\pi}{12}) + sin(\frac{3\pi}{12}-\frac{2\pi}{12}) $$
$$=2\sin(\frac{3\pi}{12})\cos(\frac{2\pi}{12})=2\sin(\frac{\pi}{4})\cos(\frac{\pi}{6})$$
A: $$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}}+\sqrt{\frac{1-\cos\frac{\pi}{6}}{2}}=$$
$$=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}+\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}}{2\sqrt2}=$$
$$=\frac{\sqrt3+1+\sqrt3-1}{2\sqrt2}=\sqrt{\frac{3}{2}}.$$
A: One more method:
\begin{align*}
\left(\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}\right)^2&=\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)^2\\
&=\cos^2\frac{\pi}{12}+\sin^2\frac{\pi}{12}+2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\\
&=1+\sin\frac{\pi}{6}\\
&=\frac{3}{2}
\end{align*}
As $\displaystyle \sin\frac{5\pi}{12}>0$ and $\displaystyle \sin\frac{\pi}{12}>0$,
$$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\displaystyle \frac{3}{2}}=\frac{\sqrt{6}}{2}$$
