Question

If we have a variable as $x=\tan \dfrac{\pi }{18}$ then find the value of $3{{x}^{6}}-27{{x}^{4}}+33{{x}^{2}}$.
A. 1
B. 2
C. $3\sqrt{3}$
D. $\dfrac{1}{3}$

Answer 1

We first use the multiple angle formula of trigonometry for $\tan \left( 3\alpha \right)=\dfrac{3\tan \alpha -{{\tan }^{3}}\alpha }{1-3{{\tan }^{2}}\alpha }$. We put the value of $\alpha =\dfrac{\pi }{18}$ in the equation to find an equation of $x=\tan \dfrac{\pi }{18}$. We solve the equation and then take the square of that. We expand the square form to find the value of the problem $3{{x}^{6}}-27{{x}^{4}}+33{{x}^{2}}$.

Complete step-by-step solution:
We have the trigonometric multiple angle formula of $\tan \left( 3\alpha \right)=\dfrac{3\tan \alpha -{{\tan }^{3}}\alpha }{1-3{{\tan }^{2}}\alpha }$.
We put the value $\alpha =\dfrac{\pi }{18}$ in the equation. Also, we use the given identity $x=\tan \dfrac{\pi }{18}$. We also have the identity value of $\dfrac{1}{\sqrt{3}}=\tan \dfrac{\pi }{6}$.
$\begin{aligned} & \tan \left( \dfrac{3\pi }{18} \right)=\dfrac{3\tan \left( \dfrac{\pi }{18} \right)-{{\tan }^{3}}\left( \dfrac{\pi }{18} \right)}{1-3{{\tan }^{2}}\left( \dfrac{\pi }{18} \right)} \\\ & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \\\ \end{aligned}$
We now solve the equation and get $3{{x}^{2}}-1=\sqrt{3}x\left( {{x}^{2}}-3 \right)$.
Now we take square both sides of the equation and get ${{\left( 3{{x}^{2}}-1 \right)}^{2}}=3{{x}^{2}}{{\left( {{x}^{2}}-3 \right)}^{2}}$.
We break the squares using the formula of ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
$\begin{aligned} & {{\left( 3{{x}^{2}}-1 \right)}^{2}}=3{{x}^{2}}{{\left( {{x}^{2}}-3 \right)}^{2}} \\\ & \Rightarrow 9{{x}^{4}}+1-6{{x}^{2}}=3{{x}^{2}}\left( {{x}^{4}}+9-6{{x}^{2}} \right) \\\ \end{aligned}$
Now we solve the rest of the equation and get
$\begin{aligned} & 9{{x}^{4}}+1-6{{x}^{2}}=3{{x}^{2}}\left( {{x}^{4}}+9-6{{x}^{2}} \right) \\\ & \Rightarrow 9{{x}^{4}}+18{{x}^{4}}+1-6{{x}^{2}}-27{{x}^{2}}-3{{x}^{6}}=0 \\\ & \Rightarrow 3{{x}^{6}}-27{{x}^{4}}+33{{x}^{2}}=1 \\\ \end{aligned}$
Therefore, the value of $3{{x}^{6}}-27{{x}^{4}}+33{{x}^{2}}$ is 1. The correct option is A.

Note: Another way of solving would have been to try and find the solution going from $a=\tan \dfrac{\pi }{9}$ and using the formula of $\tan \left( 2\alpha \right)$. The process would have been more difficult as we need to use $\tan \dfrac{\pi }{3}$ and $\tan \dfrac{\pi }{6}$ to find the value of $a=\tan \dfrac{\pi }{9}$.