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

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(B) दी गई व्यंजक: $3\left[ {{{\sin }^4}\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}(3\pi + \alpha )} \right] - 2\left[ {{{\sin }^6}\left

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$1$

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(B) दी गई व्यंजक: $3\left[ {{{\sin }^4}\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}(3\pi + \alpha )} \right] - 2\left[ {{{\sin }^6}\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right]$त्रिकोणमितीय न्यूनीकरण सूत्रों का उपयोग करते हुए:$\sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha$,$\sin(3\pi + \alpha) = -\sin \alpha$$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$,$\sin(5\pi - \alpha) = \sin \alpha$इन मानों को प्रतिस्थापित करने पर:$= 3\left[ {(-\cos \alpha)^4 + (-\sin \alpha)^4} \right] - 2\left[ {(\cos \alpha)^6 + (\sin \alpha)^6} \right]$$= 3(\cos^4 \alpha + \sin^4 \alpha) - 2(\cos^6 \alpha + \sin^6 \alpha)$सर्वसमिकाओं $a^2+b^2 = (a+b)^2 - 2ab$ और $a^3+b^3 = (a+b)^3 - 3ab(a+b)$ का उपयोग करते हुए:$= 3[(\cos^2 \alpha + \sin^2 \alpha)^2 - 2\sin^2 \alpha \cos^2 \alpha] - 2[(\cos^2 \alpha + \sin^2 \alpha)^3 - 3\sin^2 \alpha \cos^2 \alpha(\cos^2 \alpha + \sin^2 \alpha)]$चूंकि $\sin^2 \alpha + \cos^2 \alpha = 1$:$= 3[1 - 2\sin^2 \alpha \cos^2 \alpha] - 2[1 - 3\sin^2 \alpha \cos^2 \alpha]$$= 3 - 6\sin^2 \alpha \cos^2 \alpha - 2 + 6\sin^2 \alpha \cos^2 \alpha$$= 3 - 2 = 1$

$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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