If sin theta + sin 2theta + sin 3theta = sin | vedclass

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Question

(A) Given equations are:$(1) \sin 	heta + \sin 2	heta + \sin 3	heta = \sin \alpha$$(2) \cos 	heta + \cos 2	heta + \cos 3	heta = \cos \alpha$Usin

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$\alpha / 2$

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(A) Given equations are:$(1) \sin 	heta + \sin 2	heta + \sin 3	heta = \sin \alpha$$(2) \cos 	heta + \cos 2	heta + \cos 3	heta = \cos \alpha$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ for the first equation:$(\sin 3	heta + \sin 	heta) + \sin 2	heta = \sin \alpha$$2 \sin 2	heta \cos 	heta + \sin 2	heta = \sin \alpha$$\sin 2	heta (2 \cos 	heta + 1) = \sin \alpha$ ... $(i)$Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ for the second equation:$(\cos 3	heta + \cos 	heta) + \cos 2	heta = \cos \alpha$$2 \cos 2	heta \cos 	heta + \cos 2	heta = \cos \alpha$$\cos 2	heta (2 \cos 	heta + 1) = \cos \alpha$ ... $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{\sin 2	heta (2 \cos 	heta + 1)}{\cos 2	heta (2 \cos 	heta + 1)} = \frac{\sin \alpha}{\cos \alpha}$$	an 2	heta = 	an \alpha$Therefore,$2	heta = \alpha$,which implies $	heta = \alpha / 2$.

If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha$ and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha$,then $\theta$ is equal to

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