Question

If $\sin\theta + \cos\theta = x,$ then $\sin^{6}\theta + \cos^{6}\theta = \frac{1}{4}\lbrack 4 - 3(x^{2} - 1)^{2}\rbrack$ for

• A. All real x
• B. $x^{2} \leq 2$
• C. $x^{2} \geq 2$
• D. None of these

Answer 1

Answer: (B)

$x^{2} \leq 2$

Answer 2

On squaring the given relation

$\sin 2\theta = x^{2} - 1 \leq 1 \Rightarrow x^{2} \leq 2$

or $- \sqrt{2} \leq x \leq \sqrt{2}$ $\lbrack\because\sin 2\theta \leq 1\rbrack$

Now $\sin^{6}\theta + \cos^{6}\theta$

$= (\sin^{2}\theta + \cos^{2}\theta)^{3} - 3\sin^{2}\theta\cos^{2}\theta(\sin^{2}\theta + \cos^{2}\theta)$

$= 1 - 3\sin^{2}\theta\cos^{2}\theta = 1 - \frac{3}{4}\sin^{2}2\theta$

$= 1 - \frac{3}{4}(x^{2} - 1)^{2} = \frac{1}{4}\{ 4 - 3(x^{2} - 1)^{2}\}$

Thus the given result will hold true only when $x^{2} \leq 2$and not for all real values of x.