Question

If $x = \tan \dfrac{\pi }{{18}}$ then $3{x^6} - 27{x^4} + 33{x^2}$ is equal to:
A. 1
B. 2
C. $3\sqrt 3 $
D. $\dfrac{1}{3}$

Answer 1

Hint: This problem can be easily solved if we know the value of given trigonometric formula. The rest of the work is to substitute the obtained value of x in the given equation to obtain the final solution.
Thus what we need to know at first is the basic trigonometric values of $\tan \theta $ . As we know $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ . Thus when $\theta = 0,\tan \theta = 0$ and when $\theta = \dfrac{\pi }{2},\tan \theta $ is not defined. Similarly when$\theta = \dfrac{\pi }{{18}},\tan \theta = 0.176$ . This value can be substituted in the given formula on x to find the solution.

Complete step-by-step solution:
Step 1: Given a trigonometric relation of x which is $x = \tan \dfrac{\pi }{{18}}$ . If we see the numerical value of this given formula we get, $\tan \dfrac{\pi }{{18}} = 0.176$ .
Thus the value of x=0.176.
Now our work gets simpler by substituting the obtained value in the given equation.
Step 2: The given equation on x is $3{x^6} - 27{x^4} + 33{x^2}$. As we have the value of x, substituting in this equation we get,
$3{x^6} - 27{x^4} + 33{x^2}$=$3 \times {(0.176)^6} - 27 \times {(0.176)^4} + 33 \times {(0.176)^2}$
$ = 3 \times (0.00003) - 27 \times (0.00096) + 33 \times (0.031)$
$ \approx 1$
Step 3: Thus we got the value of the given equation by substitution as 1 which is option A.
The value of $3{x^6} - 27{x^4} + 33{x^2}$ with $x = \tan \dfrac{\pi }{{18}}$ is 1.

Option A is the correct answer.

Note: The common error which can occur is while finding the value of trigonometric formula and calculation errors while substituting the numerical value in the given equation. Basic trigonometric values of $\sin \theta $ , $\cos \theta $ and $\tan \theta $ functions are to be learned. The basic $\theta $ values which must be known is for $\theta = 0,\dfrac{\pi }{2},\dfrac{\pi }{3},\dfrac{\pi }{6},\pi $ .
For $\theta = 0:\sin \theta = 0,\cos \theta = 1,\tan \theta = 0$ and
For $\theta = \dfrac{\pi }{2}:\sin \theta = 1,\cos \theta = 0,\tan \theta $ is not defined.
For $\theta = \dfrac{\pi }{3}:\sin \theta = \dfrac{{\sqrt 3 }}{2},\cos \theta = \dfrac{1}{2},\tan \theta = \sqrt 3 $
For $\theta = \dfrac{\pi }{6}:\sin \theta = \dfrac{1}{2},\cos \theta = \dfrac{{\sqrt 3 }}{2},\tan \theta = \dfrac{1}{{\sqrt 3 }}$
For $\theta = \pi :\sin \theta = 0,\cos \theta = - 1,\tan \theta = 0$