If cos alpha + cos beta = a,sin alpha + sin b | vedclass

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(B) Given $\cos \alpha + \cos \beta = a$ and $\sin \alpha + \sin \beta = b$. Squaring and adding these equations: $(\cos \alpha + \cos \beta)^2 + (\si

Vedclass · Accepted Answer

$a^2 + b^2 - 3$

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(B) Given $\cos \alpha + \cos \beta = a$ and $\sin \alpha + \sin \beta = b$. Squaring and adding these equations: $(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = a^2 + b^2$ $(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = a^2 + b^2$ $1 + 1 + 2 \cos(\alpha - \beta) = a^2 + b^2$ $2 + 2 \cos(2 	heta) = a^2 + b^2$ $2(1 + \cos 2 	heta) = a^2 + b^2$ $2(2 \cos^2 	heta) = a^2 + b^2$ $4 \cos^2 	heta = a^2 + b^2$ $\cos^2 	heta = \frac{a^2 + b^2}{4}$. Now,$\frac{\cos 3 	heta}{\cos 	heta} = \frac{4 \cos^3 	heta - 3 \cos 	heta}{\cos 	heta} = 4 \cos^2 	heta - 3$. Substituting $\cos^2 	heta = \frac{a^2 + b^2}{4}$: $4 \left( \frac{a^2 + b^2}{4} ight) - 3 = a^2 + b^2 - 3$.

If $\cos \alpha + \cos \beta = a$,$\sin \alpha + \sin \beta = b$ and $\alpha - \beta = 2 \theta$,then $\frac{\cos 3 \theta}{\cos \theta} = $

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