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: 69


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

Work Step by Step

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