Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 221: 56

Answer

$\dfrac{\sqrt6-\sqrt2}{4}$

Work Step by Step

Note that $-\frac{13\pi}{12}=-\frac{\pi}{3}+\left(-\frac{3\pi}{4}\right)$. Hence, the given expression is equivalent to $\sin{\left[\left(-\frac{\pi}{3}\right)+\left(-\frac{3\pi}{4}\right)\right]}$. RECALL: $\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}$ Use the identity above with $A=-\frac{\pi}{3}$ and $B=-\frac{3\pi}{4}$ to obtain: \begin{align*} \sin{\left(-\frac{13\pi}{12}\right)}&=\sin{\left[\left(-\frac{\pi}{3}\right)+\left(-\frac{3\pi}{4}\right)\right]}\\\\ &=\sin{\left(-\frac{\pi}{3}\right)}\cos{\left(-\frac{3\pi}{4}\right)}+\cos{\left(-\frac{\pi}{3}\right)}\sin{\left(-\frac{3\pi}{4}\right)}\\\\ &=\left(-\frac{\sqrt3}{2}\right)\left(\frac{-\sqrt2}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{-\sqrt2}{2}\right)\\\\ &=\frac{\sqrt6}{4}+\left(-\frac{\sqrt2}{4}\right)\\\\ &=\frac{\sqrt6-\sqrt2}{4} \end{align*} Thus, $\sin{\left(-\frac{13\pi}{12}\right)}=\dfrac{\sqrt6-\sqrt2}{4}$
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