જો sin 2theta + sin 2phi = frac{1}{2} અને cos | vedclass

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(B) આપેલ સમીકરણો: $\sin 2	heta + \sin 2\phi = \frac{1}{2}$ $(i)$ $\cos 2	heta + \cos 2\phi = \frac{3}{2}$ $(ii)$ બંને સમીકરણોનો વર્ગ કરતા: $(\sin 2\

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$\frac{5}{8}$

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(B) આપેલ સમીકરણો: $\sin 2	heta + \sin 2\phi = \frac{1}{2}$ $(i)$ $\cos 2	heta + \cos 2\phi = \frac{3}{2}$ $(ii)$ બંને સમીકરણોનો વર્ગ કરતા: $(\sin 2	heta + \sin 2\phi)^2 = \frac{1}{4}$ $\sin^2 2	heta + \sin^2 2\phi + 2\sin 2	heta \sin 2\phi = \frac{1}{4}$ $(iii)$ $(\cos 2	heta + \cos 2\phi)^2 = \frac{9}{4}$ $\cos^2 2	heta + \cos^2 2\phi + 2\cos 2	heta \cos 2\phi = \frac{9}{4}$ $(iv)$ $(iii)$ અને $(iv)$ નો સરવાળો કરતા: $(\sin^2 2	heta + \cos^2 2	heta) + (\sin^2 2\phi + \cos^2 2\phi) + 2(\cos 2	heta \cos 2\phi + \sin 2	heta \sin 2\phi) = \frac{1}{4} + \frac{9}{4}$ $1 + 1 + 2\cos(2	heta - 2\phi) = \frac{10}{4}$ $2 + 2\cos 2(	heta - \phi) = \frac{5}{2}$ $2\cos 2(	heta - \phi) = \frac{5}{2} - 2 = \frac{1}{2}$ $\cos 2(	heta - \phi) = \frac{1}{4}$ નિત્યસમ $\cos 2A = 2\cos^2 A - 1$ નો ઉપયોગ કરતા: $2\cos^2(	heta - \phi) - 1 = \frac{1}{4}$ $2\cos^2(	heta - \phi) = 1 + \frac{1}{4} = \frac{5}{4}$ $\cos^2(	heta - \phi) = \frac{5}{8}$

જો $\sin 2\theta + \sin 2\phi = \frac{1}{2}$ અને $\cos 2\theta + \cos 2\phi = \frac{3}{2}$ હોય,તો $\cos^2(\theta - \phi) =$

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