If sin (theta-alpha), sin theta and sin (thet | vedclass

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(B) Given that $\sin (	heta-\alpha), \sin 	heta, \sin (	heta+\alpha)$ are in $H.P.$$\Rightarrow \frac{1}{\sin (	heta-\alpha)}, \frac{1}{\sin 	het

Vedclass · Accepted Answer

$1-4 \cos ^2 \frac{\alpha}{2}$

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(B) Given that $\sin (	heta-\alpha), \sin 	heta, \sin (	heta+\alpha)$ are in $H.P.$$\Rightarrow \frac{1}{\sin (	heta-\alpha)}, \frac{1}{\sin 	heta}, \frac{1}{\sin (	heta+\alpha)}$ are in $A.P.$$	herefore \frac{2}{\sin 	heta} = \frac{1}{\sin (	heta-\alpha)} + \frac{1}{\sin (	heta+\alpha)}$$\Rightarrow \frac{2}{\sin 	heta} = \frac{\sin (	heta+\alpha) + \sin (	heta-\alpha)}{\sin (	heta-\alpha) \sin (	heta+\alpha)}$Using the formula $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$ and $\sin(A+B) \sin(A-B) = \sin^2 A - \sin^2 B$:$\Rightarrow \frac{2}{\sin 	heta} = \frac{2 \sin 	heta \cos \alpha}{\sin^2 	heta - \sin^2 \alpha}$$\Rightarrow \sin^2 	heta - \sin^2 \alpha = \sin^2 	heta \cos \alpha$$\Rightarrow \sin^2 	heta (1 - \cos \alpha) = \sin^2 \alpha$Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:$\Rightarrow \sin^2 	heta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$$\Rightarrow \sin^2 	heta = 2 \cos^2 \frac{\alpha}{2}$$\Rightarrow 1 - \cos^2 	heta = 2 \cos^2 \frac{\alpha}{2}$$\Rightarrow \cos^2 	heta = 1 - 2 \cos^2 \frac{\alpha}{2}$Multiplying by $2$ and subtracting $1$ to get $\cos 2 	heta = 2 \cos^2 	heta - 1$:$\Rightarrow 2 \cos^2 	heta - 1 = 2(1 - 2 \cos^2 \frac{\alpha}{2}) - 1$$\Rightarrow \cos 2 	heta = 2 - 4 \cos^2 \frac{\alpha}{2} - 1 = 1 - 4 \cos^2 \frac{\alpha}{2}$

If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in $H.P.$,then the value of $\cos 2 \theta$ is

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