Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 576: 92


The exact value of the expression is $-\frac{3\sqrt{3}}{2}$.

Work Step by Step

Consider the following equation. $x=\sin \frac{3\pi }{2}\tan \left( -\frac{8\pi }{3} \right)+\cos \left( -\frac{5\pi }{6} \right)$ …… (1) In the given expression, if there are angles greater than $2\pi $ or less than $-2\pi $, do the following: first, find the positive co-terminal angle of that angle. Then, find the reference angle of the co-terminal angle. Finally, evaluate the trigonometric function at the reference angle with the correct sign. The measure of $\left( -\frac{8\pi }{3} \right)$ is less than $-2\pi $. Add $4\pi $ to $\left( -\frac{8\pi }{3} \right)$ , to get a positive co-terminal angle less than $2\pi $. Consider $\alpha $ to be the positive co-terminal angle. $\begin{align} & \alpha =\left( -\frac{8\pi }{3} \right)+4\pi \\ & =\frac{-8\pi +12\pi }{3} \\ & =\frac{4\pi }{3} \end{align}$ We know that $\frac{4\pi }{3}$ lies in quadrant III. Subtract $\pi $ form $\frac{4\pi }{3}$, to find the reference angle. Consider ${\alpha }'$ to be the reference angle. $\begin{align} & {\alpha }'=\frac{4\pi }{3}-\pi \\ & =\frac{4\pi -3\pi }{3} \\ & =\frac{\pi }{3} \end{align}$ Hence, $\tan \frac{\pi }{3}=\sqrt{3}$ Tangent is positive in quadrant III. Therefore, $\tan \left( -\frac{8\pi }{3} \right)=\tan \frac{\pi }{3}$ Substitute $\sqrt{3}$ for $\tan \frac{\pi }{3}$. $\tan \left( -\frac{8\pi }{3} \right)=\sqrt{3}$ Add $\pi $ to $\frac{-5\pi }{6}$, to find the reference angle. $\begin{align} & {{\alpha }'}'=\left( -\frac{5\pi }{6} \right)+\pi \\ & =\frac{-5\pi +6\pi }{6} \\ & =\frac{\pi }{6} \end{align}$ Hence, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ Cosecant is negative in quadrant III. Therefore, $\cos \frac{-5\pi }{6}=-\cos \frac{\pi }{6}$ Substitute $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$. $\cos \frac{-5\pi }{6}=-\frac{\sqrt{3}}{2}$ Substitute $\left( -1 \right)$ for $\sin \frac{3\pi }{2}$ , $\sqrt{3}$ for $\tan \left( -\frac{8\pi }{3} \right)$ and $\frac{\sqrt{3}}{2}$ for $\cos \frac{-5\pi }{6}$ in equation (1). $\begin{align} & x=\sin \frac{3\pi }{2}\tan \left( -\frac{8\pi }{3} \right)+\cos \left( -\frac{5\pi }{6} \right) \\ & =\left( -1 \right)\left( \sqrt{3} \right)+\left( -\frac{\sqrt{3}}{2} \right) \\ & =\frac{-2\sqrt{3}-\sqrt{3}}{2} \\ & =-\frac{3\sqrt{3}}{2} \end{align}$
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