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: 56


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

Work Step by Step

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