Chapter 13.9, Problem 35WE

To determine

To proof: The identity $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$ .

Answer to Problem 35WE

The identity $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$ is proved.

Explanation of Solution

The identity is

  $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

Calculation:

Split the angle into the sum of two angles whose trigonometric functions have exact values.

  $\tan 3\theta =\tan (2\theta +\theta )$

An addition formula for the tangent is $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ .

Apply this formula to the expression.

  $\tan (2\theta +\theta )=\frac{\tan 2\theta +\tan \theta }{1-\tan 2\theta \tan \theta }$ .

The double angle formula for tangent is $\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$ .

Replace $\tan 2\theta$ with $\frac{2\tan \theta }{1-{\tan }^2\theta }$ in the expression.

  $\tan (2\theta +\theta )=\frac{\frac{2\tan \theta }{1-{\tan }^2\theta }+\tan \theta }{1-\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)\tan \theta }$.

Simplify.

  $\begin{array}{c}\frac{\frac{2\tan \theta }{1-{\tan }^2\theta }+\tan \theta }{1-\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)\tan \theta }=\frac{\frac{2\tan \theta }{1-{\tan }^2\theta }+\frac{\tan \theta (1-{\tan }^2\theta )}{1-{\tan }^2\theta }}{1-\left(\frac{2{\tan }^2\theta }{1-{\tan }^2\theta }\right)}\\ =\frac{\frac{2\tan \theta +\tan \theta -{\tan }^3\theta }{1-{\tan }^2\theta }}{\frac{1-{\tan }^2\theta -2{\tan }^2\theta }{1-{\tan }^2\theta }}\\ =\frac{2\tan \theta +\tan \theta -{\tan }^3\theta }{1-{\tan }^2\theta }.\frac{1-{\tan }^2\theta }{1-{\tan }^2\theta -2{\tan }^2\theta }\\ =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\end{array}$

Therefore, the identity is proved.