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: - \[\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}\] and simplify the expression by substituting \[x=2\theta \]. Now, again apply the half angle formula given as: - \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\] 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: - \[\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}\], we get,
\[\Rightarrow \tan 4\theta =\dfrac{2\tan x}{1-{{\tan }^{2}}x}\]
Substituting \[x=2\theta \], we get,
\[\Rightarrow \tan 4\theta =\dfrac{2\tan 2\theta }{1-{{\tan }^{2}}2\theta }\]
Now, again applying the half angle formula given as: - \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\], we get,
\[\begin{align}
  & \Rightarrow \tan 4\theta =\dfrac{2\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)}{1-\tan {{\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)}^{2}}} \\
 & \Rightarrow \tan 4\theta =\dfrac{\left( \dfrac{4\tan \theta }{1-{{\tan }^{2}}\theta } \right)}{\dfrac{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta }{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}} \\
\end{align}\]
On simplification we get,
\[\Rightarrow \tan 4\theta =\dfrac{4\tan \theta }{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta }\times \dfrac{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}{\left( 1-{{\tan }^{2}}\theta \right)}\]
Cancelling the common factors, we get,
\[\begin{align}
  & \Rightarrow \tan 4\theta =\dfrac{4\tan \theta \left( 1-{{\tan }^{2}}\theta \right)}{1+{{\tan }^{4}}\theta -2{{\tan }^{2}}\theta -4{{\tan }^{2}}\theta } \\
 & \Rightarrow \tan 4\theta =\dfrac{4\tan \theta -4{{\tan }^{3}}\theta }{1+{{\tan }^{4}}\theta -6{{\tan }^{2}}\theta } \\
\end{align}\]
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: - \[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\] where both x and y are considered equal. Note that you can also use the relation \[\tan 4\theta =\tan \left( \theta +3\theta \right)\] to solve the question where \[\tan 3\theta \] is given as: - \[\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }\]. It is important for you to remember all the half angle relations of sine, cosine and tangent functions.