# 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.

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## 2 Answers

\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} $$\sin(3\pi/8) - \sin(7\pi/8) = \cos(\pi/8) - \sin(\pi/8)$$ $$= \sqrt{2}\left( \frac{\sqrt{2}}{2}\cos(\pi/8) - \frac{\sqrt{2}}{2}\sin(\pi/8) \right)$$ $$= \sqrt{2}\left( \cos(\pi/4)\cos(\pi/8) -\sin(\pi/4)\sin(\pi/8) \right)$$ $$=\sqrt{2}\cos(\pi/4 + \pi/8)$$

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Thanks for the answer! According the textbook solution, cosecant is correct, so that must be wrong. – user44184 Oct 27 '12 at 20:35
@user44184 See my answer above. Both are equal as you guessed. – user17762 Oct 27 '12 at 20:36
OK, exercise: Find the error, if any, above. Maybe I'll look at it later....... (BTW, textbooks are not infallible.) – Michael Hardy Oct 27 '12 at 20:37

$$\sin(7 \pi/8) = \sin(\pi/2 + 3 \pi/8) = \cos(3 \pi/8)$$ \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.

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