Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 243: 55


$\frac{sin~x}{1+cos~x} = tan~\frac{x}{2}$

Work Step by Step

$\frac{sin~x}{1+cos~x}$ When we graph this function, we can see that it looks like the graph of $~~tan~\frac{x}{2}$ We can verify this algebraically: $\frac{sin~x}{1+cos~x} = \frac{sin~(\frac{x}{2}+\frac{x}{2})}{1+cos~(\frac{x}{2}+\frac{x}{2})}$ $\frac{sin~x}{1+cos~x} = \frac{sin~\frac{x}{2}~cos~\frac{x}{2}+cos~\frac{x}{2}~sin~\frac{x}{2}}{1+cos~\frac{x}{2}~cos~\frac{x}{2}-sin~\frac{x}{2}~sin~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{2~sin~\frac{x}{2}~cos~\frac{x}{2}}{1-sin^2~\frac{x}{2}+cos^2~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{2~sin~\frac{x}{2}~cos~\frac{x}{2}}{cos^2~\frac{x}{2}+cos^2~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{2~sin~\frac{x}{2}~cos~\frac{x}{2}}{2~cos^2~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = \frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}$ $\frac{sin~x}{1+cos~x} = tan~\frac{x}{2}$
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