ISI2016-DCG-61

Question

The value of $\sin^{6}\frac{\pi}{81}+\cos^{6}\frac{\pi}{81}-1+3\sin^{2}\frac{\pi}{81}\:\cos^{2}\frac{\pi}{81}$ is

• A. $\tan^{6}\frac{\pi}{81}$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

Answer: $\mathbf B$

Solution:

Given:$$\sin^6\frac{\pi}{81}+\cos^6\frac{\pi}{81} -1 + 3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81} \tag{i}$$

Now,

$\sin^6\frac{\pi}{81} = \left(\sin^2\frac{\pi}{81}\right)^3,\; \text{and}\;\cos^6\frac{\pi}{81} = \left (\cos^2\frac{\pi}{81}\right)^3 \tag{ii}$

Using identity: $a^3 + b^3 = (a+b)^3 -3ab(a+b)\tag{iii}$

and, $\sin^2\theta + \cos^2\theta = 1\tag{iv}$

Using equations $\;\mathrm{(ii)\;, (iii)\;and\;(iv)}$, we get:

$\sin^6\frac{\pi}{81}+\cos^6\frac{\pi}{81}=\left(\sin^2\frac{\pi}{81}\right)^3 + \left(\cos^2\frac{\pi}{81}\right)^3$

$ = \underbrace{\sin^2\frac{\pi}{81} + \cos^2\frac{\pi}{81}}_\text{=1} -3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81}\left(\underbrace{\sin^2\frac{\pi}{81} + \cos^2\frac{\pi}{81}}_\text{=1}\right) $

$= 1 - 3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81} \tag{v}$

From equations $\mathrm {(i)\;and\;(v)}$, we get:

$ \require{cancel} = \cancel {1} - \cancel{3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81}} -\cancel{1} + \cancel{3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81}}\\ = \mathbf 0$

$\therefore \mathbf B$ is the correct answer.