Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 249: 68

Answer

$\frac{1}{2}~cot~\frac{x}{2}-\frac{1}{2}~tan~\frac{x}{2} = cot~x$

Work Step by Step

We can verify that the equation is an identity: $\frac{1}{2}~cot~\frac{x}{2}-\frac{1}{2}~tan~\frac{x}{2}$ $=\frac{1}{2}\cdot \frac{cos~\frac{x}{2}}{sin~\frac{x}{2}}-\frac{1}{2}\cdot \frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}$ $=\frac{1}{2}~\Big(\frac{cos~\frac{x}{2}}{sin~\frac{x}{2}}-\frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}\Big)$ $=\frac{1}{2}~\Big(\frac{cos^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}-\frac{sin^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}\Big)$ $=\frac{1}{2}~\Big(\frac{cos^2~\frac{x}{2}-sin^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}\Big)$ $=\frac{cos^2~\frac{x}{2}-sin^2~\frac{x}{2}}{2~sin~\frac{x}{2}~cos~\frac{x}{2}}$ $= \frac{cos~x}{sin~x}$ $= cot~x$
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