3,left[ {{{sin }^4},left( {frac{{3pi }}{2} - | vedclass

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(b) $3\left\{ {{{\sin }^4}\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}(3\pi + \alpha )} \right\}$
$ - 2\left\{ {{{\sin }^6}\left( {\fra

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$1$

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(b) $3\left\{ {{{\sin }^4}\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}(3\pi + \alpha )} \right\}$
$ - 2\left\{ {{{\sin }^6}\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right\}$
$ = 3\,\{ {( - \cos \alpha )^4} + {( - \sin \alpha )^4}\} - 2\,\{ {\cos ^6}\alpha + {\sin ^6}\alpha \} $
=$3\,\,\{ {({\cos ^2}\alpha + {\sin ^2}\alpha )^2} - 2{\sin ^2}\alpha {\cos ^2}\alpha \} $
$ - 2\,\{ {({\cos ^2}\alpha + {\sin ^2}\alpha )^3} - 3{\cos ^2}\alpha {\sin ^2}\alpha ({\cos ^2}\alpha + {\sin ^2}\alpha )\} $
$ = 3 - 6{\sin ^2}\alpha {\cos ^2}\alpha - 2 + 6{\sin ^2}\alpha {\cos ^2}\alpha = 3 - 2 = 1$
ट्रिक : $\alpha = 0,\frac{\pi }{2}$ रखने पर व्यंजक का मान $1$ आता है अर्थात् यह $\alpha $ से स्वतंत्र है।

$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $

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