3left[ sin^4left( frac{3pi}{2} - alpha right) | vedclass

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(B) Given expression: $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} + \al

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(B) Given expression: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right]$Using trigonometric reduction formulas:$\sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha$,$\sin(3\pi + \alpha) = -\sin \alpha$,$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$,$\sin(5\pi - \alpha) = \sin \alpha$Substituting these values:$= 3\left[ (-\cos \alpha)^4 + (-\sin \alpha)^4 \right] - 2\left[ (\cos \alpha)^6 + (\sin \alpha)^6 \right]$$= 3(\cos^4 \alpha + \sin^4 \alpha) - 2(\cos^6 \alpha + \sin^6 \alpha)$Using identities $\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$ and $\sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha$:$= 3(1 - 2\sin^2 \alpha \cos^2 \alpha) - 2(1 - 3\sin^2 \alpha \cos^2 \alpha)$$= 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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