Question

How do you simplify $$\tan 4\theta$$ to trigonometric functions of a unit $$\theta$$?

Answer 1

Hint: First of all assume $$4\theta$$ equal to $$2\times 2\theta$$ and consider $$2\theta$$ as x. Now, apply the half angle formula of the tangent function given as: - $$\tan 2x={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}$$ and simplify the expression by substituting $$x=2\theta$$. Now, again apply the half angle formula given as: - $$\tan 2\theta ={\displaystyle \frac{2\tan \theta }{1-{\tan }^2\theta }}$$ and simplify the expression further to get the answer.

Complete step by step answer:
Here, we have been provided with the trigonometric expression $$\tan 4\theta$$ and we are asked to simplify it in an expression in which we have the terms $$\tan \theta$$ only.
Now, the angle $$4\theta$$ means we have a $$4^{th}$$ multiple of angle $$\theta$$. We can write $$4\theta$$ as $$2\times 2\theta$$. Now, let us assume $$2\theta$$ equal to x, so the angle becomes 2x. Therefore, the expression can be written as: -
$$\Rightarrow \tan 4\theta =\tan 2x$$
Applying the half angle formula given as: - $$\tan 2x={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}$$, we get,
$$\Rightarrow \tan 4\theta ={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}$$
Substituting $$x=2\theta$$, we get,
$$\Rightarrow \tan 4\theta ={\displaystyle \frac{2\tan 2\theta }{1-{\tan }^{2}2\theta }}$$
Now, again applying the half angle formula given as: - $$\tan 2\theta ={\displaystyle \frac{2\tan \theta }{1-{\tan }^2\theta }}$$, we get,
$$\begin{array}{rl}& \Rightarrow \tan 4\theta ={\displaystyle \frac{2\left({\displaystyle \frac{2\tan \theta }{1-{\tan }^2\theta }}\right)}{1-\tan {\left({\displaystyle \frac{2\tan \theta }{1-{\tan }^2\theta }}\right)}^2}}\\ & \Rightarrow \tan 4\theta ={\displaystyle \frac{\left({\displaystyle \frac{4\tan \theta }{1-{\tan }^2\theta }}\right)}{{\displaystyle \frac{{(1-{\tan }^2\theta )}^2-4{\tan }^2\theta }{{(1-{\tan }^2\theta )}^2}}}}\end{array}$$
On simplification we get,
$$\Rightarrow \tan 4\theta ={\displaystyle \frac{4\tan \theta }{{(1-{\tan }^2\theta )}^2-4{\tan }^2\theta }}\times {\displaystyle \frac{{(1-{\tan }^2\theta )}^2}{(1-{\tan }^2\theta )}}$$
Cancelling the common factors, we get,
$$\begin{array}{rl}& \Rightarrow \tan 4\theta ={\displaystyle \frac{4\tan \theta (1-{\tan }^2\theta )}{1+{\tan }^4\theta -2{\tan }^2\theta -4{\tan }^2\theta }}\\ & \Rightarrow \tan 4\theta ={\displaystyle \frac{4\tan \theta -4{\tan }^3\theta }{1+{\tan }^4\theta -6{\tan }^2\theta }}\end{array}$$
Hence, the above expression is our answer.

Note:
One may note that the half angle formula is derived by using the basic tangent formula given as: - $$\tan (x+y)={\displaystyle \frac{\tan x+\tan y}{1-\tan x\tan y}}$$ where both x and y are considered equal. Note that you can also use the relation $$\tan 4\theta =\tan (\theta +3\theta )$$ to solve the question where $$\tan 3\theta$$ is given as: - $$\tan 3\theta ={\displaystyle \frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }}$$. It is important for you to remember all the half angle relations of sine, cosine and tangent functions.