Question

1. determine the exact answer of tan 150°. include a quick sketch of the angle. 3
2. find the two values of angle θ on the interval 0° ≤ θ ≤ 360°. give exact answers where possible. 3

\\( \sin \theta = - \frac { 1 } { 2 } \\)

Explanation:

Step1: Determine the reference angle

Since \(150^{\circ}=180^{\circ} - 30^{\circ}\), the reference angle is \(30^{\circ}\).

Step2: Determine the sign of \(\tan150^{\circ}\)

\(150^{\circ}\) is in the second quadrant. In the second quadrant, \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) is negative (because \(\sin\theta> 0\) and \(\cos\theta<0\) in the second quadrant).

Step3: Calculate \(\tan150^{\circ}\)

We know that \(\tan\alpha=-\tan(180^{\circ}-\alpha)\) for \(\alpha = 30^{\circ}\). And \(\tan30^{\circ}=\frac{\sqrt{3}}{3}\), so \(\tan150^{\circ}=-\tan30^{\circ}=-\frac{\sqrt{3}}{3}\).

For the sketch:

• Draw the coordinate axes.
• Start from the positive \(x\) - axis and rotate counter - clockwise by \(150^{\circ}\). The terminal side of the \(150^{\circ}\) angle lies in the second quadrant.

For \(\sin\theta=-\frac{1}{2}\):

Step1: Find the reference angle

We know that \(\sin\alpha=\frac{1}{2}\) when \(\alpha = 30^{\circ}\) (since \(\sin30^{\circ}=\frac{1}{2}\)).

Step2: Determine the quadrants

Since \(\sin\theta<0\), \(\theta\) is in the third or fourth quadrant.

Step3: Calculate the values of \(\theta\)

In the third quadrant: \(\theta=180^{\circ}+30^{\circ}=210^{\circ}\)
In the fourth quadrant: \(\theta = 360^{\circ}-30^{\circ}=330^{\circ}\)

Answer:

\(\tan150^{\circ}=-\frac{\sqrt{3}}{3}\); The values of \(\theta\) for \(\sin\theta=-\frac{1}{2}\) (\(0^{\circ}\leq\theta\leq360^{\circ}\)) are \(210^{\circ}\) and \(330^{\circ}\)