If cos alpha + cos beta = frac{1}{3} and sin | vedclass

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(A) Given $\cos \alpha + \cos \beta = \frac{1}{3} \dots (i)$ and $\sin \alpha + \sin \beta = \frac{1}{4} \dots (ii)$.Squaring and adding $(i)$ and $(i

Vedclass · Accepted Answer

$\frac{7}{25}$

Vedclass · Answer

(A) Given $\cos \alpha + \cos \beta = \frac{1}{3} \dots (i)$ and $\sin \alpha + \sin \beta = \frac{1}{4} \dots (ii)$.Squaring and adding $(i)$ and $(ii)$:$(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = (\frac{1}{3})^2 + (\frac{1}{4})^2$$(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{1}{9} + \frac{1}{16}$$1 + 1 + 2 \cos (\alpha - \beta) = \frac{16 + 9}{144} = \frac{25}{144}$$2 + 2 \cos (\alpha - \beta) = \frac{25}{144} \implies 2 \cos (\alpha - \beta) = \frac{25}{144} - 2 = -\frac{263}{144}$.Alternatively,using sum-to-product formulas:$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{3} \dots (iii)$$2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{4} \dots (iv)$Dividing $(iv)$ by $(iii)$ gives $	an \frac{\alpha+\beta}{2} = \frac{3}{4}$.Then $\cos (\alpha+\beta) = \frac{1 - 	an^2 \frac{\alpha+\beta}{2}}{1 + 	an^2 \frac{\alpha+\beta}{2}} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.

If $\cos \alpha + \cos \beta = \frac{1}{3}$ and $\sin \alpha + \sin \beta = \frac{1}{4}$,then $\cos (\alpha + \beta) = $

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