If cos alpha + cos beta = frac{3}{2} and sin | vedclass

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(B) Given $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$.Using sum-to-product formulas:$2 \cos \frac{\alpha+\be

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$\frac{7}{5}$

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(B) Given $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$.Using sum-to-product formulas:$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{3}{2}$ $(i)$$2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{2}$ $(ii)$Dividing $(ii)$ by $(i)$ gives $	an \frac{\alpha+\beta}{2} = \frac{1}{3}$.Since $	heta = \frac{\alpha+\beta}{2}$,we have $	an 	heta = \frac{1}{3}$.We need to find $\sin 2	heta + \cos 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta} + \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$.Substituting $	an 	heta = \frac{1}{3}$:$= \frac{2(1/3)}{1 + 1/9} + \frac{1 - 1/9}{1 + 1/9} = \frac{2/3}{10/9} + \frac{8/9}{10/9} = \frac{6}{10} + \frac{8}{10} = \frac{14}{10} = \frac{7}{5}$.

If $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$ and $\theta$ is the arithmetic mean of $\alpha$ and $\beta$,then $\sin 2\theta + \cos 2\theta$ is equal to

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