Question

Please help me in this question........

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Answer 1

HELLO DEAR,

we know,
sin(3π/2 - a) = cosa.
sin(3π + a) = sina.
sin(π/2 + a) = cosa.
sin(5π - a) = sina.

now,

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

$\bold{3[cos^4\alpha + sin^4\alpha] - 2[cos^6\alpha + sin^6\alpha]}$

$\bold{3[(cos^2\alpha)^2+( sin^2\alpha)^2] - 2[(cos^2\alpha)^3 + (sin^2\alpha)^3]}$

$\bold{3[(cos^2\alpha + sin^2\alpha)^2 - 2sin^2\alpha*cos^2\alpha] - 2[(cos^2\alpha + sin^2\alpha)^3 - 3sin^2\alpha*cos^2\alpha]}$

$\bold{3[1 - 2sin^2\alpha*cos^2\alpha] - 2[1 - 3sin^2\alpha*cos^2\alpha]}$

$\bold{[3 - 2 - 6sin^2\alpha*cos^2\alpha + 6sin^2\alpha*cos^2\alpha]}$

$\bold{3 - 2 = 1}$

hence, $\bold{3[sin^4(\frac{3\pi}{2} - \alpha) + sin^4(3\pi + \alpha)] - 2[sin^6(\frac{\pi}{2} + \alpha) + sin^6(5\pi - \alpha)] = 1}$

I HOPE ITS HELP YOU DEAR,
THANKS