## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 576: 89

#### Answer

The exact value of the expression is $-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)$.

#### Work Step by Step

Let us consider the following equation. $x=\sin \frac{11\pi }{4}\cos \frac{5\pi }{6}+\cos \frac{11\pi }{4}\sin \frac{5\pi }{6}$ …… (1) So, in quadrant I, all the trigonometric functions are positive. In quadrant II, sine and cosecant are positive and all other trigonometric functions are negative. In quadrant III, tangent and cotangent are positive and all other trigonometric functions are negative. In quadrant IV, cosine and secant are positive and all other trigonometric functions are negative. The angle $\frac{11\pi }{4}$ lies between $\frac{5\pi }{2}$ and $3\pi$, in quadrant II. Thus, 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 $\sin \frac{\pi }{4}$. $\sin \frac{11\pi }{4}=\frac{\sqrt{3}}{2}$ The angle $\frac{5\pi }{6}$ lies between $\frac{\pi }{2}$ and $\pi$ , in quadrant II. And, calculate the reference angle as follows \begin{align} & {\theta }'=\pi -\frac{5\pi }{6} \\ & =\frac{6\pi -5\pi }{6} \\ & =\frac{\pi }{6} \end{align} So, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ Here, the cosine is positive in quadrant II. Therefore, $\cos \frac{5\pi }{6}=\cos \frac{\pi }{6}$ Put $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$. $\cos \frac{5\pi }{6}=\frac{\sqrt{3}}{2}$ The angle $\frac{11\pi }{4}$ lies between $\frac{5\pi }{2}$ and $3\pi$ , in quadrant II. Also, 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} So, $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Here, the cosine is positive in quadrant II. Therefore, $\cos \frac{11\pi }{4}=\cos \frac{\pi }{4}$ Put $\frac{\sqrt{2}}{2}$ for $\cos \frac{\pi }{4}$. $\cos \frac{11\pi }{4}=\frac{\sqrt{2}}{2}$ The angle $\frac{5\pi }{6}$ lies between $\frac{\pi }{2}$ and $\pi$, in quadrant II. And, calculate the reference angle as follows \begin{align} & {\theta }'=\pi -\frac{5\pi }{6} \\ & =\frac{6\pi -5\pi }{6} \\ & =\frac{\pi }{6} \end{align} Hence, $\sin \frac{\pi }{6}=\frac{1}{2}$ Here, the sine is positive in quadrant II. Therefore, $\sin \frac{5\pi }{6}=\sin \frac{\pi }{6}$ Put $\frac{1}{2}$ for $\sin \frac{\pi }{6}$. $\sin \frac{5\pi }{6}=\frac{1}{2}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{11\pi }{4}$ , $-\frac{\sqrt{3}}{2}$ for $\cos \frac{5\pi }{6}$ , $\frac{1}{2}$ for $\sin \frac{5\pi }{6}$ and $\frac{\sqrt{2}}{2}$ for $\cos \frac{11\pi }{4}$ in equation (1). \begin{align} & x=\frac{\sqrt{2}}{2}\left( -\frac{\sqrt{3}}{2} \right)+\left( -\frac{\sqrt{2}}{2} \right)\left( \frac{1}{2} \right) \\ & =-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ & =-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right) \end{align}

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