Question

What is the formula for $\tan 3x$?

Answer 1

Write the formula for $\tan 3x$.

In trigonometric formula $\mathit{t}\mathit{a}\mathit{n}\left(C+D\right)\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{C}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{D}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{C}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{D}}$.

Put $C=x,D=2x$:

$$\begin{gathered} \tan \left(x+2x\right)=\frac{\tan x+\tan 2x}{1-\tan x\tan 2x} \\ \begin{array}{rcl}& \Rightarrow & \tan \left(x+2x\right)=\frac{\tan x+\left({\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}\right)}{1-\tan x\left({\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}\right)}\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{2}\mathit{x}\mathbf{=}\frac{\mathbf{2}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{x}}\right]\\ & \Rightarrow & \tan \left(3x\right)=\frac{{\displaystyle \frac{\tan x-{\tan }^{3}x+2\tan x}{1-{\tan }^{2}x}}}{{\displaystyle \frac{1-{\tan }^{2}x-2{\tan }^{2}x}{1-{\tan }^{2}x}}}\\ & \Rightarrow & \tan \left(3x\right)=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\end{array} \end{gathered}$$

.Hence, the formula is $\tan \left(3x\right)=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}$