If y = sin^2 alpha + cos^2(alpha + beta) + 2 | vedclass

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(C) Given $y = \sin^2 \alpha + \cos^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$.Using the identity $\cos(\alpha + \beta) + 2 \s

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(C) Given $y = \sin^2 \alpha + \cos^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$.Using the identity $\cos(\alpha + \beta) + 2 \sin \alpha \sin \beta = \cos(\alpha - \beta)$,we can rewrite the expression:$y = \sin^2 \alpha + \cos(\alpha + \beta) [\cos(\alpha + \beta) + 2 \sin \alpha \sin \beta]$$y = \sin^2 \alpha + \cos(\alpha + \beta) \cos(\alpha - \beta)$Using the product-to-sum formula $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$:$y = \sin^2 \alpha + \frac{1}{2} [\cos(2\alpha) + \cos(2\beta)]$Using the identity $\cos(2\alpha) = 1 - 2 \sin^2 \alpha$:$y = \sin^2 \alpha + \frac{1}{2} [1 - 2 \sin^2 \alpha + \cos(2\beta)]$$y = \sin^2 \alpha + \frac{1}{2} - \sin^2 \alpha + \frac{1}{2} \cos(2\beta)$$y = \frac{1}{2} + \frac{1}{2} \cos(2\beta)$Since $\beta$ is a constant,$y$ is a constant value.Therefore,the third derivative $\frac{d^3y}{d\alpha^3} = 0$.

If $y = \sin^2 \alpha + \cos^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$,then find $\frac{d^3y}{d\alpha^3}$ (keeping $\beta$ as a constant).

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