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: 34

Answer

$\frac{3\sqrt {10}}{10}$

Work Step by Step

As $B$ is in QI, $\cos B$ is positive. $\cos B=\sqrt {1-\sin^{2}B}$ $=\sqrt {1-(\frac{3}{5})^{2}}=\sqrt {\frac{16}{25}}$ $=\frac{4}{5}$ $\cos \frac{B}{2}=\pm\sqrt {\frac{1+\cos B}{2}}$ $=\pm \sqrt {\frac{1+\frac{4}{5}}{2}}=\pm \sqrt {\frac{9}{10}}=\pm\frac{3}{\sqrt {10}}=\pm\frac{3\sqrt {10}}{10}$ We need to know where $\frac{B}{2}$ terminates to determine the sign of $\cos \frac{B}{2}$. Given that $B$ is in QI: Or $0^{\circ}\lt B\lt90^{\circ}$ Then $\frac{0^{\circ}}{2}\lt\frac{B}{2}\lt\frac{90^{\circ}}{2}$ That is, $0^{\circ}\lt\frac{B}{2}\lt45^{\circ}$ So $\frac{B}{2}$ terminates in the first quadrant. In the first quadrant, cosine is positive. Therefore, $\cos \frac{B}{2}=\frac{3\sqrt {10}}{ 10}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.