જો sin theta + cos theta = x, તો {sin ^6}thet | vedclass

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Question

(b) On squaring the given relation

$\sin 2	heta = {x^2} - 1 \le 1 \Rightarrow {x^2} \le 2$

or $ - \sqrt 2 \le x \le \sqrt 2 $    &nbs

Vedclass · Accepted Answer

${x^2} \le 2$

Vedclass · Answer

(b) On squaring the given relation

$\sin 2	heta = {x^2} - 1 \le 1 \Rightarrow {x^2} \le 2$

or $ - \sqrt 2 \le x \le \sqrt 2 $          $[\because \,\,\sin 2	heta  \le 1]$

Now ${\sin ^6}	heta + {\cos ^6}	heta $

$ = {({\sin ^2}	heta + {\cos ^2}	heta )^3} - 3{\sin ^2}	heta {\cos ^2}	heta ({\sin ^2}	heta + {\cos ^2}	heta )$

$ = 1 - 3{\sin ^2}	heta {\cos ^2}	heta = 1 - \frac{3}{4}{\sin ^2}2	heta $

$ = 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 $x.$

જો $\sin \theta + \cos \theta = x,$ તો ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ એ . . .. માટે શક્ય બને.

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