The expression [1 - sin(3pi - alpha) + cos(3p | vedclass

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Question

(B) First,simplify each trigonometric term using reduction formulas:$ \sin(3\pi - \alpha) = \sin(\pi - \alpha) = \sin \alpha $$ \cos(3\pi + \alpha) =

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$ - \sin 2\alpha $

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(B) First,simplify each trigonometric term using reduction formulas:$ \sin(3\pi - \alpha) = \sin(\pi - \alpha) = \sin \alpha $$ \cos(3\pi + \alpha) = -\cos \alpha $$ \sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha $$ \cos(\frac{5\pi}{2} - \alpha) = \cos(\frac{\pi}{2} - \alpha) = \sin \alpha $Substituting these into the expression:$ [1 - \sin \alpha - \cos \alpha] [1 - (-\cos \alpha) + \sin \alpha ] $$ = [1 - (\sin \alpha + \cos \alpha)] [1 + (\sin \alpha + \cos \alpha)] $Using the identity $(a - b)(a + b) = a^2 - b^2$:$ = 1^2 - (\sin \alpha + \cos \alpha)^2 $$ = 1 - (\sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha) $$ = 1 - (1 + \sin 2\alpha) $$ = - \sin 2\alpha $

The expression $[1 - \sin(3\pi - \alpha) + \cos(3\pi + \alpha)] [1 - \sin(\frac{3\pi}{2} - \alpha) + \cos(\frac{5\pi}{2} - \alpha)]$ when simplified reduces to:

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