Question

find the exact circular function value.

\\( \sin \left( \frac { 4 \pi } { 3 } \
ight) \\)

\\( \bigcirc \\) a. \\( - 1 \\)

\\( \bigcirc \\) b. \\( - \frac { 1 } { 2 } \\)

\\( \bigcirc \\) c. \\( \frac { \sqrt { 3 } } { 2 } \\)

\\( \bigcirc \\) d. \\( - \frac { \sqrt { 3 } } { 2 } \\)

Explanation:

Step1: Determine the reference angle

The angle \(\frac{4\pi}{3}\) is in the third - quadrant. The formula for the reference angle \(\theta'\) of an angle \(\theta\) in the third - quadrant is \(\theta'=\theta-\pi\).
For \(\theta = \frac{4\pi}{3}\), we have \(\theta'=\frac{4\pi}{3}-\pi=\frac{4\pi - 3\pi}{3}=\frac{\pi}{3}\).

Step2: Use the sign of the sine function in the third - quadrant

In the third - quadrant (\(\pi<\theta<\frac{3\pi}{2}\)), the sine function is negative.
We know that \(\sin(A)=\sin(\pi + B)=-\sin(B)\) when \(A=\pi + B\) (here \(A = \frac{4\pi}{3}\) and \(B=\frac{\pi}{3}\)).
Since \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\), then \(\sin(\frac{4\pi}{3})=-\sin(\frac{\pi}{3})\).

Answer:

D. \(-\frac{\sqrt{3}}{2}\)