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: 95


The exact value of the expression is $1$.

Work Step by Step

Let us consider the following equation. $y=\left( h\circ g \right)\left( \frac{17\pi }{3} \right)$ And the condition is $\begin{align} & g\left( x \right)=\cos x \\ & h\left( x \right)=2x \\ \end{align}$ Apply the condition in equation $y$. $y=h\left( g\left( \frac{17\pi }{3} \right) \right)$ Further solve, $y=2\left( \cos \left( \frac{17\pi }{3} \right) \right)$ …… (1) The angle $\frac{17\pi }{3}$ lies between $6\pi $ and $\frac{3\pi }{2}$, in quadrant III. And, calculate the reference angle as follows $\begin{align} & {\theta }'=6\pi -\frac{17\pi }{3} \\ & =\frac{18\pi -17\pi }{3} \\ & =\frac{\pi }{3} \end{align}$ Hence, $\cos \frac{\pi }{3}=\frac{1}{2}$ Here, the cosine is positive in quadrant III. So, $\cos \frac{17\pi }{3}=\cos \frac{\pi }{3}$ Put $\frac{1}{2}$ for $\cos \frac{\pi }{3}$. $\cos \frac{17\pi }{3}=\frac{1}{2}$ Put $\frac{1}{2}$ for $\cos \left( \frac{17\pi }{3} \right)$ in equation (1). $\begin{align} & y=2\left( \frac{1}{2} \right) \\ & =1 \end{align}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.