Solve the problem.
-If $\sin \theta = - \frac { 4 } { 5 }$, and $\theta$ terminates in quadrant IV, then find $\sin 2 \theta$.
A) $- \frac { 24 } { 25 }$
B) $\frac { 24 } { 25 }$
C) $- \frac { 7 } { 25 }$
D) $\frac { 7 } { 25 }$

Question

Quizplus · Accepted Answer

The formula for $\sin 2\theta$ is $2\sin\theta\cos\theta$. Given $\sin\theta = -\frac{4}{5}$, and knowing that in the fourth quadrant, sine is negative and cosine is positive, we find $\cos\theta$ using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. Thus, $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. Therefore, $\sin 2\theta = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{24}{25}$.

Solve the Problem sin⁡θ=−45\sin \theta = - \frac { 4 } { 5 }sinθ=−54​

Solve the Problem $\sin \theta = - \frac { 4 } { 5 }$