Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Review Exercises - Page 243: 43

Answer

$-\frac{sin~2x+sin~x}{cos~2x-cos~x} = cot~\frac{x}{2}$

Work Step by Step

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