Chapter 7.5, Problem 19AYU

To determine

Find the exact value of expression.

Answer to Problem 19AYU

  $\boxed{\sin \frac{17\pi }{12}=-\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)}$

Explanation of Solution

Given information:

Find the exact value of expression.

  $\sin \frac{17\pi }{12}$

Calculation:

Consider the given repression,

  $\sin \frac{17\pi }{12}$

Rewrite $\sin \frac{17\pi }{12}$ as

  $\sin \frac{17\pi }{12}=\sin \frac{8\pi }{12}+\frac{9\pi }{12}or\sin \frac{2\pi }{3}+\frac{3\pi }{4}$

Apply identity

  $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$

  $\sin \frac{2\pi }{3}+\frac{3\pi }{4}=\sin \frac{2\pi }{3}\cos \frac{3\pi }{4}+\cos \frac{2\pi }{3}\sin \frac{3\pi }{4}$

The angles $\frac{2\pi }{3}\&\frac{3\pi }{4}$ are not in the form of $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ then find their reference angles.

  $\begin{array}{l}\sin \frac{2\pi }{3}\cos \frac{3\pi }{4}+\cos \frac{2\pi }{3}\sin \frac{3\pi }{4}=\sin \left(\frac{\pi }{2}-\frac{2\pi }{3}\right)\cos \left(\frac{\pi }{2}-\frac{3\pi }{4}\right)+\cos \left(\frac{\pi }{2}-\frac{2\pi }{3}\right)\sin \left(\frac{\pi }{2}-\frac{3\pi }{4}\right)\\ use,identity\cos \left(\frac{\pi }{2}-\theta \right)=\sin \theta and\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta \\ \cos \left(\frac{\pi }{6}\right)\sin \left(-\frac{\pi }{4}\right)+\sin \left(\frac{\pi }{6}\right)\cos \left(-\frac{\pi }{4}\right)\end{array}$

Use the odd ,even identities.

  $\begin{array}{l}\cos \left(\frac{\pi }{6}\right)\sin \left(-\frac{\pi }{4}\right)+\sin \left(\frac{\pi }{6}\right)\cos \left(-\frac{\pi }{4}\right)=-\cos \left(\frac{\pi }{6}\right)\sin \left(\frac{\pi }{4}\right)-\sin \left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{4}\right)\\ =-\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}+\frac{1}{2}\times \frac{\sqrt{2}}{2}\right)\end{array}$

  $\sin \frac{17\pi }{12}=-\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)$

Hence, $\boxed{\sin \frac{17\pi }{12}=-\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)}$ .