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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(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\left\{ {\,{{( - \cos \alpha )}^4} + {{( - \sin \alpha )}^4}} \right\} - 2\,\left\{ {{{\cos }^6}\alpha + {{\sin }^6}\alpha } \right\}$
$ = 3\left\{ {{{\cos }^2}\alpha + {{\sin }^2}\alpha {)^2} - 2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right\}$
$ - 2\left\{ {{{({{\cos }^2}\alpha + {{\sin }^2}\alpha )}^3} - 3{{\cos }^2}\alpha {{\sin }^2}\alpha ({{\cos }^2}\alpha + {{\sin }^2}\alpha )} \right\}$
$ = 3 - 6{\sin ^2}\alpha {\cos ^2}\alpha - 2 + 6{\sin ^2}\alpha {\cos ^2}\alpha = 3 - 2 = 1$
Trick : Put $\alpha = 0,\frac{\pi }{2},$ the value of expression remains $1$ i.e., it is independent of $\alpha $.

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

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