For any theta , in ,left( {frac{pi }{4}, | vedclass

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Question

$3\,{(1 - \sin 2	heta )^2}\, + \,6(1 + \sin 2	heta )\, + \,4\,{\sin ^6}	heta $

$ = 3\,(1 - 2\sin \,2	heta  + {\sin ^2}2	heta ) + \,6 + 6\

Vedclass · Accepted Answer

$13 - 4\,{\cos ^6}\,	heta \,$

Vedclass · Answer

$3\,{(1 - \sin 2	heta )^2}\, + \,6(1 + \sin 2	heta )\, + \,4\,{\sin ^6}	heta $

$ = 3\,(1 - 2\sin \,2	heta  + {\sin ^2}2	heta ) + \,6 + 6\sin 2	heta  + \,4\,{\sin ^6}	heta $

$ = \,9 + 3{\sin ^2}2	heta  + 4\,{\sin ^6}	heta $

$ = \,9 + 12{\sin ^2}	heta {\cos ^2}	heta  + 4\,{(1 - {\cos ^2}	heta )^3}$

$ = \,9 + 12(1 - {\cos ^2}	heta ){\cos ^2}	heta  + 4\,(1 - 3{\cos ^2}	heta  + 3{\cos ^4}	heta  - {\cos ^6}	heta )$

$ = \,13 + 12{\cos ^2}	heta  - 12{\cos ^4}	heta  - 12{\cos ^2}	heta  + 12{\cos ^4}	heta  - 4{\cos ^6}	heta $

$ = \,13 - 4{\cos ^6}	heta $

For any $\theta \, \in \,\left( {\frac{\pi }{4},\frac{\pi }{2}} \right)$, the expression $3\,{\left( {\sin \,\theta - \cos \,\theta } \right)^4} + 6{\left( {\sin \,\theta + \cos \,\theta } \right)^2} + 4\,{\sin ^6}\,\theta $ equals

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