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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 65 (page 496)

In Problems $47 - 68$ , establish each identity.
$\tan (3 胃) = \frac{3 \tan 胃 - \tan^{3} 胃}{1 - 3 \tan^{2} 胃}$

Short Answer

Expert verified

The identity $\tan (3 胃) = \frac{3 \tan 胃 - \tan^{3} 胃}{1 - 3 \tan^{2} 胃}$ is established.

Step by step solution

01

Step 1 Given identity is,

$\tan (3 胃) = \frac{3 \tan 胃 - \tan^{3} 胃}{1 - 3 \tan^{2} 胃}$

The formulae used in this problem are,

$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \tan (2 胃) = \frac{2 \tan 胃}{1 - \tan^{2} 胃}$

02

Step 2 Use the formula in the required places and prove that the expressions in the both sides are equal.

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

Hence, proved.

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Most popular questions from this chapter

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