Question

1. no calculator. find each angle $\theta$ with the given trigonometric value. use the domain $0^{circ} leq \theta<360^{circ}$. support with ordered pair, reference angle, and sketch. hint: there may be 2 values for $\theta$. a. $cos \theta=\frac{sqrt{3}}{2}$ b. $sin \theta=\frac{-sqrt{2}}{2}$ c. $sin \theta=1$

Explanation:

Step1: Recall the unit - circle values

We know that on the unit circle, \(\cos\theta=x\) and \(\sin\theta = y\). The reference angles and their trigonometric values: \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), \(\sin90^{\circ} = 1\)

Step2: Solve for \(\theta\) in \(\cos\theta=\frac{\sqrt{3}}{2}\)

Since \(\cos\theta=\frac{\sqrt{3}}{2}\), and \(\cos\theta\) is positive in the first and fourth quadrants.
The reference angle \(\theta_{r}=30^{\circ}\)
For the first - quadrant: \(\theta = 30^{\circ}\)
For the fourth - quadrant: \(\theta=360^{\circ}-30^{\circ}=330^{\circ}\)

Step3: Solve for \(\theta\) in \(\sin\theta=\frac{-\sqrt{2}}{2}\)

Since \(\sin\theta\) is negative, it is in the third and fourth quadrants. The reference angle \(\theta_{r} = 45^{\circ}\)
For the third - quadrant: \(\theta=180^{\circ}+45^{\circ}=225^{\circ}\)
For the fourth - quadrant: \(\theta=360^{\circ}-45^{\circ}=315^{\circ}\)

Step4: Solve for \(\theta\) in \(\sin\theta = 1\)

We know that \(\sin90^{\circ}=1\). In the domain \(0^{\circ}\leq\theta<360^{\circ}\), \(\theta = 90^{\circ}\)

Answer:

a. \(\theta = 30^{\circ}\) or \(\theta=330^{\circ}\)
b. \(\theta = 225^{\circ}\) or \(\theta=315^{\circ}\)
c. \(\theta = 90^{\circ}\)