Prove that \(\frac{\tan^3x}{1-3\tan^2x}=\frac18(\tan3x-3\tan x)\).

Solution

a. We know that \(\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}\).

\[ \begin{aligned} \tan3x&=\frac{3\tan x-\tan^3x}{1-3\tan^2x}\\ &=\frac{3\tan x-9\tan^3x+8\tan^3x}{1-3\tan^2x}\\ &=\frac{8\tan^3x+3\tan x(1-3\tan^2x)}{1-3\tan^2x}\\ &=\frac{8\tan^3x}{1-3\tan^2x}+3\tan x. \end{aligned} \]

Therefore, \(\tan3x-3\tan x=\frac{8\tan^3x}{1-3\tan^2x}\), hence \(\frac{\tan^3x}{1-3\tan^2x}=\frac18(\tan3x-3\tan x)\).

b. Let \(S_n=\sum_{k=1}^{n}\frac{3^k\tan^3(a/3^k)}{1-3\tan^2(a/3^k)}\).

Using the result above,

\[ \frac{3^k\tan^3(a/3^k)}{1-3\tan^2(a/3^k)} =\frac{3^k}{8}\left(\tan\frac{a}{3^{k-1}}-3\tan\frac{a}{3^k}\right). \]

Therefore,

\[ \begin{aligned} S_n&=\frac38\left(\tan a-3\tan\frac a3\right)+\frac{3^2}{8}\left(\tan\frac a3-3\tan\frac a{3^2}\right)+\cdots+\frac{3^n}{8}\left(\tan\frac a{3^{n-1}}-3\tan\frac a{3^n}\right)\\ &=\frac38\tan a-\frac{3^{n+1}}8\tan\frac a{3^n}\\ &=\frac38\left(\tan a-3^n\tan\frac a{3^n}\right). \end{aligned} \]