Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.5 - Multiple-Angle and Product-to-Sum Formulas - 7.5 Exercises - Page 548: 26

Answer

$tan~3x=\frac{-tan^3x+3~tan~x}{-3~tan^2x+1}=\frac{tan^3x-3~tan~x}{3~tan^2x-1}$

Work Step by Step

$tan~3x=tan~(x+2x)=\frac{tan~x+tan~2x}{1-tan~x~tan~2x}=\frac{tan~x+\frac{2~tan~x}{1-tan^2x}}{1-tan~x~\frac{2~tan~x}{1-tan^2x}}=\frac{\frac{tan~x-tan^3x+2~tan~x}{1-tan^2x}}{\frac{1-tan^2x-2~tan^2x}{1-tan^2x}}=\frac{-tan^3x+3~tan~x}{-3~tan^2x+1}=\frac{tan^3x-3~tan~x}{3~tan^2x-1}$
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