If sin A + sin 2A = x and cos A + cos 2A = y, | vedclass

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(A) Given: $x = \sin A + \sin 2A$ and $y = \cos A + \cos 2A$.Calculate $x^2 + y^2$:$x^2 + y^2 = (\sin A + \sin 2A)^2 + (\cos A + \cos 2A)^2$$= \sin^2

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$2y$

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(A) Given: $x = \sin A + \sin 2A$ and $y = \cos A + \cos 2A$.Calculate $x^2 + y^2$:$x^2 + y^2 = (\sin A + \sin 2A)^2 + (\cos A + \cos 2A)^2$$= \sin^2 A + \sin^2 2A + 2 \sin A \sin 2A + \cos^2 A + \cos^2 2A + 2 \cos A \cos 2A$$= (\sin^2 A + \cos^2 A) + (\sin^2 2A + \cos^2 2A) + 2(\cos 2A \cos A + \sin 2A \sin A)$$= 1 + 1 + 2 \cos(2A - A)$$= 2 + 2 \cos A = 2(1 + \cos A)$.Now,substitute this into the expression $({x^2} + {y^2})({x^2} + {y^2} - 3)$:$= [2(1 + \cos A)][2(1 + \cos A) - 3]$$= [2(1 + \cos A)][2 + 2 \cos A - 3]$$= [2(1 + \cos A)][2 \cos A - 1]$$= 2(2 \cos^2 A + \cos A - 1)$$= 2(2 \cos^2 A - 1 + \cos A)$$= 2(\cos 2A + \cos A) = 2y$.

If $\sin A + \sin 2A = x$ and $\cos A + \cos 2A = y,$ then $({x^2} + {y^2})({x^2} + {y^2} - 3) = $

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