If sin theta = sin alpha,then sin frac{theta} | vedclass

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(D) Given $\sin 	heta = \sin \alpha$,the general solution is $	heta = n\pi + (-1)^n \alpha$ for $n \in \mathbb{Z}$.For $n = 0$,$	heta = \alpha$,so

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All of the above

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(D) Given $\sin 	heta = \sin \alpha$,the general solution is $	heta = n\pi + (-1)^n \alpha$ for $n \in \mathbb{Z}$.For $n = 0$,$	heta = \alpha$,so $\sin \frac{	heta}{3} = \sin \frac{\alpha}{3}$.For $n = 1$,$	heta = \pi - \alpha$,so $\sin \frac{	heta}{3} = \sin \left( \frac{\pi}{3} - \frac{\alpha}{3} ight)$.For $n = -1$,$	heta = -\pi - \alpha$,so $\sin \frac{	heta}{3} = \sin \left( -\frac{\pi}{3} - \frac{\alpha}{3} ight) = -\sin \left( \frac{\pi}{3} + \frac{\alpha}{3} ight)$.Thus,all the given options are possible values.

If $\sin \theta = \sin \alpha$,then $\sin \frac{\theta}{3}$ is equal to:

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