Question

in exercises 67 - 70, find the value of the unique real number \\( \theta \\) between 0 and \\( 2 \pi \\) that satisfies the two given conditions. 67. \\( \sin \theta = \frac { 1 } { 2 } \\) and \\( \tan \theta < 0 \\).

Explanation:

Step1: Determine the quadrants

Since \(\sin\theta=\frac{1}{2}>0\), \(\theta\) is in Quadrant I or II.
Since \(\tan\theta < 0\), \(\theta\) is in Quadrant II or IV.
So, \(\theta\) is in Quadrant II.

Step2: Find the reference angle

We know that \(\sin\theta=\frac{1}{2}\), and the reference angle \(\theta_{r}\) satisfies \(\sin\theta_{r}=\frac{1}{2}\). So, \(\theta_{r}=\frac{\pi}{6}\).

Step3: Calculate \(\theta\) in Quadrant II

For an angle \(\theta\) in Quadrant II, \(\theta=\pi-\theta_{r}\).
Substitute \(\theta_{r}=\frac{\pi}{6}\) into the formula: \(\theta=\pi - \frac{\pi}{6}=\frac{5\pi}{6}\).

Answer:

\(\frac{5\pi}{6}\)