Problem 22 - Precalculus Essentials

Question

a. Use the unit circle shown for Exercises \(5-18\) on the previous page to find the value of the trigonometric function. b. Use even and odd properties of trigonometric functions and your answer from part (a) to find the value of the same trigonometric function at the indicated real number. a. \(\sin \frac{2 \pi}{3}\) b. \(\sin \left(-\frac{2 \pi}{3}\right)\)

Step-by-Step Solution

Answer

a. \(\sin \frac{2 \pi}{3} = \frac{\sqrt{3}}{2}\) b. \(\sin \left(-\frac{2 \pi}{3}\right) = -\frac{\sqrt{3}}{2}\)

Step 1: Find the sin value for \( \frac{2 \pi}{3} \)

On the unit circle, \( \frac{2 \pi}{3} \) is in the second quadrant. The points on the unit circle in the second quadrant have positive y-values and that’s why the sin values - which correspond to the y-values in the unit circle - are also positive. For \( \frac{2 \pi}{3} \), the corresponding sin value is \( \frac{\sqrt{3}}{2} \).

Step 2: Use the even and odd properties of trigonometric functions to find the sin value for \( -\frac{2 \pi}{3} \)

Since sine is an odd function, we have the identity: \(\sin(-x) = -\sin(x)\). Therefore, \(\sin \left(-\frac{2 \pi}{3}\right) = - \sin \left(\frac{2 \pi}{3}\right) = -\frac{\sqrt{3}}{2} \).