Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.4 - Half-Angle Formulas - 5.4 Problem Set - Page 304: 27

Answer

$-\sqrt {\frac{3-2\sqrt 2}{6}}$

Work Step by Step

As $B$ is in QIII, $\cos B$ is negative. $\cos B=-\sqrt {1-\sin^{2}B}$ $=-\sqrt {1-(-\frac{1}{3})^{2}}=-\sqrt {\frac{8}{9}}$ $=-\frac{2\sqrt 2}{3}$ $\cos \frac{B}{2}=\pm\sqrt {\frac{1+\cos B}{2}}$ $=\pm \sqrt {\frac{1-\frac{2\sqrt 2}{3}}{2}}=\pm\sqrt {\frac{3-2\sqrt 2}{6}}$ We need to know where $\frac{A}{2}$ terminates to determine the sign of $\cos \frac{B}{2}$. Given that $B$ is in QIII: Or $180^{\circ}\lt B\lt270^{\circ}$ Then $\frac{180^{\circ}}{2}\lt\frac{B}{2}\lt\frac{270^{\circ}}{2}$ That is, $90^{\circ}\lt\frac{B}{2}\lt135^{\circ}$ Therefore, $\cos \frac{B}{2}=-\sqrt {\frac{3-2\sqrt 2}{6}}$
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