2 sin^2 beta + 4 cos(alpha + beta) sin alpha | vedclass

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Question

(C) આપેલ પદાવલિ: $E = 2 \sin^2 \beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + \cos 2(\alpha + \beta)$નિત્યસમ $2 \sin^2 \beta = 1 - \cos 2\bet

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$\cos 2\alpha$

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(C) આપેલ પદાવલિ: $E = 2 \sin^2 \beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + \cos 2(\alpha + \beta)$નિત્યસમ $2 \sin^2 \beta = 1 - \cos 2\beta$ અને $\cos 2(\alpha + \beta) = 2 \cos^2(\alpha + \beta) - 1$ નો ઉપયોગ કરતા:$E = (1 - \cos 2\beta) + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + (2 \cos^2(\alpha + \beta) - 1)$$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta$$2 \sin \alpha \sin \beta = \cos(\alpha - \beta) - \cos(\alpha + \beta)$ નો ઉપયોગ કરતા:$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 2 \cos(\alpha + \beta) [\cos(\alpha - \beta) - \cos(\alpha + \beta)]$$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 2 \cos(\alpha + \beta) \cos(\alpha - \beta) - 2 \cos^2(\alpha + \beta)$$E = 2 \cos(\alpha + \beta) \cos(\alpha - \beta) - \cos 2\beta$$2 \cos A \cos B = \cos(A + B) + \cos(A - B)$ નો ઉપયોગ કરતા:$E = (\cos 2\alpha + \cos 2\beta) - \cos 2\beta$$E = \cos 2\alpha$

$2 \sin^2 \beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + \cos 2(\alpha + \beta) = $

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