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If $\sin \theta=-\frac{3}{4}$, then $\sin 2 \theta=$
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The correct answer is:
$-\frac{3 \sqrt{7}}{8}$
If $\sin \theta=\frac{-3}{4}$
$\begin{aligned} & \cos \theta=\sqrt{1-\sin ^2 \theta} \\ & \cos \theta=\sqrt{1-\left(\frac{-3}{4}\right)^2} \\ & \cos \theta=\sqrt{\frac{16-9}{4}}=\sqrt{\frac{7}{16}}\end{aligned}$
$\therefore \sin 2 \theta=2 \sin \theta \cdot \cos \theta$
$=2 \times\left(\frac{-3}{4}\right)\left(\frac{\sqrt{7}}{4}\right)=\frac{-3 \sqrt{7}}{8}$
$\begin{aligned} & \cos \theta=\sqrt{1-\sin ^2 \theta} \\ & \cos \theta=\sqrt{1-\left(\frac{-3}{4}\right)^2} \\ & \cos \theta=\sqrt{\frac{16-9}{4}}=\sqrt{\frac{7}{16}}\end{aligned}$
$\therefore \sin 2 \theta=2 \sin \theta \cdot \cos \theta$
$=2 \times\left(\frac{-3}{4}\right)\left(\frac{\sqrt{7}}{4}\right)=\frac{-3 \sqrt{7}}{8}$
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