Question

1. find all possible angles for θ using the domain ( 0 leq \theta < 2pi ).

a) ( sin \theta = \frac{sqrt{3}}{2} )
b) ( cos \theta = -\frac{1}{2} )
c) ( cos \theta = 1 )
d) ( cos \theta = \frac{sqrt{2}}{2} )

Explanation:

Step1: Recall the unit - circle values

For \(y = \sin\theta\), \(\sin\theta=\frac{\sqrt{3}}{2}\). The reference angle \(\theta_{r}=\frac{\pi}{3}\). Since \(\sin\theta>0\), \(\theta\) is in the first and second quadrants.
\(\theta=\frac{\pi}{3}\) (first - quadrant) and \(\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}\) (second - quadrant)

Step2: For \(x = \cos\theta\), \(\cos\theta=-\frac{1}{2}\)

The reference angle \(\theta_{r}=\frac{\pi}{3}\). Since \(\cos\theta<0\), \(\theta\) is in the second and third quadrants.
\(\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}\) (second - quadrant) and \(\theta=\pi+\frac{\pi}{3}=\frac{4\pi}{3}\) (third - quadrant)

Step3: For \(\cos\theta = 1\)

We know that \(\cos\theta=x\) on the unit circle. When \(x = 1\), \(\theta = 0\) (since \(\cos(0)=1\) and \(0\leq\theta<2\pi\))

Step4: For \(\cos\theta=\frac{\sqrt{2}}{2}\)

The reference angle \(\theta_{r}=\frac{\pi}{4}\). Since \(\cos\theta>0\), \(\theta\) is in the first and fourth quadrants.
\(\theta=\frac{\pi}{4}\) (first - quadrant) and \(\theta = 2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\) (fourth - quadrant)

Answer:

a) \(\theta=\frac{\pi}{3},\frac{2\pi}{3}\)
b) \(\theta=\frac{2\pi}{3},\frac{4\pi}{3}\)
c) \(\theta = 0\)
d) \(\theta=\frac{\pi}{4},\frac{7\pi}{4}\)