If sin theta = -frac{3}{4},then sin 2 theta = | vedclass

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(B) Given $\sin 	heta = -\frac{3}{4}$.We know that $\cos^2 	heta = 1 - \sin^2 	heta$.$\cos^2 	heta = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \fr

Vedclass · Accepted Answer

$-\frac{3 \sqrt{7}}{8}$

Vedclass · Answer

(B) Given $\sin 	heta = -\frac{3}{4}$.We know that $\cos^2 	heta = 1 - \sin^2 	heta$.$\cos^2 	heta = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$.Thus,$\cos 	heta = \pm \frac{\sqrt{7}}{4}$.Using the formula $\sin 2 	heta = 2 \sin 	heta \cos 	heta$:If $\cos 	heta = \frac{\sqrt{7}}{4}$,then $\sin 2 	heta = 2 	imes (-\frac{3}{4}) 	imes (\frac{\sqrt{7}}{4}) = -\frac{3 \sqrt{7}}{8}$.If $\cos 	heta = -\frac{\sqrt{7}}{4}$,then $\sin 2 	heta = 2 	imes (-\frac{3}{4}) 	imes (-\frac{\sqrt{7}}{4}) = \frac{3 \sqrt{7}}{8}$.Since the options provided include $-\frac{3 \sqrt{7}}{8}$,the correct option is $B$.

If $\sin \theta = -\frac{3}{4}$,then $\sin 2 \theta = $

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