Find $\sin 2 x$ , given that $\sin x = \frac { 3 } { 5 }$and ($x)$is in quadrant II
A) $- \frac { 24 } { 25 }$

B) $\frac { 24 } { 7 }$

C) $ \frac{4 \sqrt{6}}{7} $

D) $\frac { \sqrt { 6 } } { 25 }$

E) $- \frac { 4 } { 7 }$

Question

Find $\sin 2 x$ , given that $\sin x = \frac { 3 } { 5 }$and ($x)$is in quadrant II
A) $- \frac { 24 } { 25 }$

B) $\frac { 24 } { 7 }$

C) $ \frac{4 \sqrt{6}}{7} $

D) $\frac { \sqrt { 6 } } { 25 }$

E) $- \frac { 4 } { 7 }$

Quizplus · Accepted Answer

The formula for $\sin 2x$ is $2\sin x \cos x$. Given $\sin x = \frac{3}{5}$ and $x$ is in quadrant II, where sine is positive and cosine is negative. To find $\cos x$, we use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, which gives $\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$. Therefore, $\sin 2x = 2 \cdot \frac{3}{5} \cdot -\frac{4}{5} = -\frac{24}{25}$.

Find sin⁡2x\sin 2 xsin2x , Given That sin⁡x=35\sin x = \frac { 3 } { 5 }sinx=53​

Find $\sin 2 x$ , Given That $\sin x = \frac { 3 } { 5 }$