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

If x=tan⁡π18x = \tan \dfrac{\pi }{{18}} then 3x6−27x4+33x23{x^6} - 27{x^4} + 33{x^2} is equal to:
A. 1
B. 2
C. 333\sqrt 3
D. 13\dfrac{1}{3}

Explanation

Solution

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⁡θ\tan \theta . As we know tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} . Thus when θ=0,tan⁡θ=0\theta = 0,\tan \theta = 0 and when θ=π2,tan⁡θ\theta = \dfrac{\pi }{2},\tan \theta is not defined. Similarly whenθ=π18,tan⁡θ=0.176\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⁡π18x = \tan \dfrac{\pi }{{18}} . If we see the numerical value of this given formula we get, tan⁡π18=0.176\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 3x6−27x4+33x23{x^6} - 27{x^4} + 33{x^2}. As we have the value of x, substituting in this equation we get,
3x6−27x4+33x23{x^6} - 27{x^4} + 33{x^2}=3×(0.176)6−27×(0.176)4+33×(0.176)23 \times {(0.176)^6} - 27 \times {(0.176)^4} + 33 \times {(0.176)^2}
=3×(0.00003)−27×(0.00096)+33×(0.031)= 3 \times (0.00003) - 27 \times (0.00096) + 33 \times (0.031)
≈1\approx 1
Step 3: Thus we got the value of the given equation by substitution as 1 which is option A.
The value of 3x6−27x4+33x23{x^6} - 27{x^4} + 33{x^2} with x=tan⁡π18x = \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⁡θ\sin \theta , cos⁡θ\cos \theta and tan⁡θ\tan \theta functions are to be learned. The basic θ\theta values which must be known is for θ=0,π2,π3,π6,π\theta = 0,\dfrac{\pi }{2},\dfrac{\pi }{3},\dfrac{\pi }{6},\pi .
For θ=0:sin⁡θ=0,cos⁡θ=1,tan⁡θ=0\theta = 0:\sin \theta = 0,\cos \theta = 1,\tan \theta = 0 and
For θ=π2:sin⁡θ=1,cos⁡θ=0,tan⁡θ\theta = \dfrac{\pi }{2}:\sin \theta = 1,\cos \theta = 0,\tan \theta is not defined.
For θ=π3:sin⁡θ=32,cos⁡θ=12,tan⁡θ=3\theta = \dfrac{\pi }{3}:\sin \theta = \dfrac{{\sqrt 3 }}{2},\cos \theta = \dfrac{1}{2},\tan \theta = \sqrt 3
For θ=π6:sin⁡θ=12,cos⁡θ=32,tan⁡θ=13\theta = \dfrac{\pi }{6}:\sin \theta = \dfrac{1}{2},\cos \theta = \dfrac{{\sqrt 3 }}{2},\tan \theta = \dfrac{1}{{\sqrt 3 }}
For θ=π:sin⁡θ=0,cos⁡θ=−1,tan⁡θ=0\theta = \pi :\sin \theta = 0,\cos \theta = - 1,\tan \theta = 0