{sin ^6}theta + {cos ^6}theta + 3{sin ^2}thet | vedclass

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Question

(c) ${\sin ^6}	heta + {\cos ^6}	heta + 3\,{\sin ^2}	heta \,{\cos ^2}	heta $

$ = {({\sin ^2}	heta + {\cos ^2}	heta )^3} - 3{\sin ^2}	heta {\c

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(c) ${\sin ^6}	heta + {\cos ^6}	heta + 3\,{\sin ^2}	heta \,{\cos ^2}	heta $

$ = {({\sin ^2}	heta + {\cos ^2}	heta )^3} - 3{\sin ^2}	heta {\cos ^2}	heta + 3{\sin ^2}	heta {\cos ^2}	heta = 1.$ 

Trick : Put $	heta = {0^o},$

we get the value of expression equal to $1. $

Again put $	heta = {45^o},$ the value remains $1,$ it means that the expression is independent of $	heta$ and is equal to $1.$

${\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = $

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