If sin theta = sin alpha,then which of the fo | vedclass

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(A) Given $\sin 	heta = \sin \alpha$. Subtracting $\sin \alpha$ from both sides,we get $\sin 	heta - \sin \alpha = 0$. Using the trigonometric ident

Vedclass · Accepted Answer

$\frac{	heta+\alpha}{2}$ is any odd multiple of $\frac{\pi}{2}$ or $\frac{	heta-\alpha}{2}$ is any multiple of $\pi$

Vedclass · Answer

(A) Given $\sin 	heta = \sin \alpha$. Subtracting $\sin \alpha$ from both sides,we get $\sin 	heta - \sin \alpha = 0$. Using the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}ight) \sin \left(\frac{C-D}{2}ight)$,we have: $2 \cos \left(\frac{	heta+\alpha}{2}ight) \sin \left(\frac{	heta-\alpha}{2}ight) = 0$. This product is zero if either $\cos \left(\frac{	heta+\alpha}{2}ight) = 0$ or $\sin \left(\frac{	heta-\alpha}{2}ight) = 0$. If $\cos \left(\frac{	heta+\alpha}{2}ight) = 0$,then $\frac{	heta+\alpha}{2} = (2n+1) \frac{\pi}{2}$,which means $\frac{	heta+\alpha}{2}$ is an odd multiple of $\frac{\pi}{2}$. If $\sin \left(\frac{	heta-\alpha}{2}ight) = 0$,then $\frac{	heta-\alpha}{2} = n\pi$,which means $\frac{	heta-\alpha}{2}$ is a multiple of $\pi$. Thus,the condition is that $\frac{	heta+\alpha}{2}$ is an odd multiple of $\frac{\pi}{2}$ $OR$ $\frac{	heta-\alpha}{2}$ is a multiple of $\pi$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

If $\sin \theta = \sin \alpha$,then which of the following is true?

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