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: 47


$\frac{2~(sin~x-sin^3~x)}{cos~x} = sin~2x$

Work Step by Step

$\frac{2~(sin~x-sin^3~x)}{cos~x}$ When we graph this function, it looks like the graph of $~~sin~2x$ We can verify this algebraically: $\frac{2~(sin~x-sin^3~x)}{cos~x}$ $= \frac{2~sin~x(1-sin^2~x)}{cos~x}$ $= \frac{2~sin~x(cos^2~x)}{cos~x}$ $= 2~sin~x~cos~x$ $=sin~2x$
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