Question

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

Answer 1

Find the formula for $\tan 3A$.

In the 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 $C=A,D=2A$:

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

Hence, the formula is $\begin{array}{rcl}\mathit{t}\mathit{a}\mathit{n}\left(3A\right)& \mathbf{=}& \frac{\mathbf{3}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{A}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{3}}\mathbf{A}}{\mathbf{1}\mathbf{-}\mathbf{3}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{A}}\end{array}$.