Precalculus (6th Edition) Blitzer

Published by Pearson

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 81

Answer

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

Work Step by Step

$\cos \frac{23\pi }{4}$ lies in quadrant IV. Subtract $4\pi$ from $\frac{23\pi }{4}$ to find the positive co-terminal angles less than $2\pi$. \begin{align} & \theta =\frac{23\pi }{4}-4\pi \\ & =\frac{23\pi -16\pi }{4} \\ & =\frac{7\pi }{4} \end{align} The positive acute angle formed by the terminal side of $\theta$ and the x-axis is the reference angle ${\theta }'$. Since $\frac{7\pi }{4}$ angle lies in quadrant IV, subtract $\frac{7\pi }{4}$ from $2\pi$ to find the reference angle. \begin{align} & {\theta }'=2\pi -\frac{7\pi }{4} \\ & =\frac{8\pi -7\pi }{4} \\ & =\frac{\pi }{4} \end{align} Hence, $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ In quadrant I all the trigonometric function 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. Here, the cosine is positive in quadrant IV. Therefore, $\cos \frac{23\pi }{4}=\cos \frac{\pi }{4}$ Substitute $\frac{\sqrt{2}}{2}$ for $\cos \frac{\pi }{4}$. $\cos \frac{23\pi }{4}=\frac{\sqrt{2}}{2}$ The exact value of the expression is $\frac{\sqrt{2}}{2}$.

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