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3.Trigonometrical Ratios, Functions and Identities
medium
If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for
A
All real $x$
B
${x^2} \le 2$
C
${x^2} \ge 2$
D
None of these
Solution
(b) On squaring the given relation
$\sin 2\theta = {x^2} – 1 \le 1 \Rightarrow {x^2} \le 2$
or $ – \sqrt 2 \le x \le \sqrt 2 $ $[\because \,\,\sin 2\theta \le 1]$
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 ${x^2} \le 2$ and not for all real values of $x.$
Standard 11
Mathematics
