find the exact value of $ \sin 60 ^ { \circ } + \sin 120 ^ { \circ } - \sin 240 ^ { \circ } + \sin 300 ^ { \circ } $.
select the correct choice below and fill in any answer boxes in your choice.
$ \bigcirc $ a. $ \sin 60 ^ { \circ } + \sin 120 ^ { \circ } - \sin 240 ^ { \circ } + \sin 300 ^ { \circ } = $
(type an exact answer, using radicals as needed.)
$ \bigcirc $ b. the answer is undefined.

Question

Accepted Answer

# Explanation:
## Step1: Find the value of each trigonometric function
- $\sin60^{\circ}=\frac{\sqrt{3}}{2}$
- $\sin120^{\circ}=\sin(180^{\circ} - 60^{\circ})=\sin60^{\circ}=\frac{\sqrt{3}}{2}$
- $\sin240^{\circ}=\sin(180^{\circ}+60^{\circ})=-\sin60^{\circ}=-\frac{\sqrt{3}}{2}$
- $\sin300^{\circ}=\sin(360^{\circ}-60^{\circ})=-\sin60^{\circ}=-\frac{\sqrt{3}}{2}$

## Step2: Substitute the values into the expression
$$
\begin{align*}
&\sin60^{\circ}+\sin120^{\circ}-\sin240^{\circ}+\sin300^{\circ}\
=&\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}-(-\frac{\sqrt{3}}{2})+(-\frac{\sqrt{3}}{2})\
=&\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\
=&\sqrt{3}
\end{align*}
$$

# Answer:
A. $\sin60^{\circ}+\sin120^{\circ}-\sin240^{\circ}+\sin300^{\circ}=\sqrt{3}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Find the value of each trigonometric function

Step2: Substitute the values into the expression

Answer: