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: 96


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

Work Step by Step

Let us consider the following equation. $y=\left( h\circ f \right)\left( \frac{11\pi }{4} \right)$ The condition is $\begin{align} & f\left( x \right)=\sin x \\ & h\left( x \right)=2x \\ \end{align}$ Now, apply the condition in equation $y$. $y=h\left( f\left( \frac{11\pi }{4} \right) \right)$ Further solve, $y=2\left( \sin \left( \frac{11\pi }{4} \right) \right)$ …… (1) The angle $\frac{11\pi }{4}$ lies between $\frac{5\pi }{2}$ and $3\pi $, in quadrant II. Calculate the reference angle as follows $\begin{align} & {\theta }'=3\pi -\frac{11\pi }{4} \\ & =\frac{12\pi -11\pi }{4} \\ & =\frac{\pi }{4} \end{align}$ Hence, $\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Here, the sine is positive in quadrant II. So, $\sin \frac{11\pi }{4}=\sin \frac{\pi }{4}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{\pi }{4}$. $\sin \frac{11\pi }{4}=\frac{\sqrt{2}}{2}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{11\pi }{4}$ in equation (1). $\begin{align} & y=2\left( \frac{\sqrt{2}}{2} \right) \\ & =\sqrt{2} \end{align}$
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