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

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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 $2\sin^2 \beta = 1 - \cos 2\beta$

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$\cos 2\alpha$

Vedclass · Answer

(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)$$E = 2\cos^2(\alpha + \beta) - \cos 2\beta + 4\cos(\alpha + \beta)\sin \alpha \sin \beta$$E = 2\cos(\alpha + \beta) [\cos(\alpha + \beta) + 2\sin \alpha \sin \beta] - \cos 2\beta$$2\sin \alpha \sin \beta = \cos(\alpha - \beta) - \cos(\alpha + \beta)$ નો ઉપયોગ કરતા:$E = 2\cos(\alpha + \beta) [\cos(\alpha + \beta) + \cos(\alpha - \beta) - \cos(\alpha + \beta)] - \cos 2\beta$$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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