How to relate $2\sin(3\pi/8)-2\sin(7\pi/8)$ and $\csc(3\pi/8)$? Trying to simplify $2\sin(3\pi/8)-2\sin(7\pi/8)$ down to $\csc(3\pi/8)$. The two expressions have equal decimal approximations but I'm literally at my wit's end trying to relate them based on trigonometric identities.
 A: $$\sin(7 \pi/8) = \sin(\pi/2 + 3 \pi/8) = \cos(3 \pi/8)$$
\begin{align}
\dfrac{\sin(3 \pi/8) - \sin(7 \pi/8)}{\csc(3 \pi/8)} & = \sin(3 \pi/8) (\sin(3 \pi/8) - \sin(7 \pi/8))\\
& = \sin(3 \pi/8) (\sin(3 \pi/8) - \cos(3 \pi/8))\\
& = \sin^2(3 \pi/8) - \sin(3 \pi/8) \cos(3 \pi/8)\\
& = \underbrace{\dfrac{1 - \cos(3 \pi/4)}{2}}_{\sin^2(\theta) = \frac{1 - \cos(2 \theta)}2} - \underbrace{\dfrac{\sin(3 \pi/4)}2}_{\sin(\theta) \cos(\theta) = \frac{\sin(2 \theta)}2}\\
& = \dfrac{1 + \dfrac1{\sqrt{2}} - \dfrac1{\sqrt{2}}}2\\
& = \dfrac12
\end{align}
which is what we want.
A: \begin{align}
\sin\left(\pi - x\right) & = \sin x; \\[8pt]
\text{Hence } \sin(7\pi/8) & = \sin(\pi/8). \\[12pt]
\sin\left(\frac\pi2 - x\right) & = \cos x \\[8pt]
\text{Hence } \sin(3\pi/8) & = \cos(\pi/8)
\end{align}
$$
\sin(3\pi/8) - \sin(7\pi/8) = \cos(\pi/8) - \sin(\pi/8)
$$
$$
= \sqrt{2}\left( \frac{\sqrt{2}}{2}\cos(\pi/8) - \frac{\sqrt{2}}{2}\sin(\pi/8) \right)
$$
$$
= \sqrt{2}\left( \cos(\pi/4)\cos(\pi/8) -\sin(\pi/4)\sin(\pi/8)  \right)
$$
$$
=\sqrt{2}\cos(\pi/4 + \pi/8)
$$
