Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 41

Answer

The required value of $\cos 157.5{}^\circ $ is $-\frac{\sqrt{2+\sqrt{2}}}{2}$.

Work Step by Step

We have to find the value of $\cos 157.5{}^\circ $; the formula is used which represents the cos identity. The term $\cos 157.5{}^\circ $ comes under the second quadrant where the value of sine and cosec functions is positive and the rest of the functions are negative. And the value of the second quadrant is between $90{}^\circ \,\text{to}\,18\text{0}{}^\circ $. $\begin{align} & \cos 157.5{}^\circ =\cos \frac{315{}^\circ }{2} \\ & =-\sqrt{\frac{1+\cos 315{}^\circ }{2}} \\ & =-\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} \\ & =-\sqrt{\frac{2+\sqrt{2}}{4}} \end{align}$ Now, taking the square root of 4, that is 2, we get: $\cos 157.5{}^\circ =-\frac{\sqrt{2+\sqrt{2}}}{2}$ Thus, the value of $\cos 157.5{}^\circ $ is $-\frac{\sqrt{2+\sqrt{2}}}{2}$.
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