Question

the movement of the progress bar may be uneven because questions can be worth
which points are on the graph of $y = \cos \theta$?
\bigcirc \\ (3\pi, 1) and \left(\frac{\pi}{2}, 0\
ight)
\bigcirc \\ (4\pi, 1) and \left(\frac{\pi}{3}, \frac{1}{2}\
ight)
\bigcirc \\ (3\pi, - 1) and \left(\frac{\pi}{2}, 1\
ight)
\bigcirc \\ (4\pi, 0) and \left(\frac{\pi}{3}, \frac{\sqrt{3}}{2}\
ight)

Explanation:

Step1: Recall the cosine function properties

The cosine function \(y = \cos\theta\) has a period of \(2\pi\), i.e., \(\cos(\theta + 2k\pi)=\cos\theta\) for \(k\in\mathbb{Z}\), and \(\cos(2k\pi)=1\), \(\cos((2k + 1)\pi)=- 1\), \(\cos(\frac{\pi}{2}+k\pi)=0\), \(\cos(\frac{\pi}{3})=\frac{1}{2}\), \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\).

Step2: Check each option

• Option 1:
• For \(\theta = 3\pi\), \(\cos(3\pi)=\cos(\pi + 2\pi)=\cos(\pi)=-1

eq1\). So \((3\pi,1)\) is not on the graph of \(y = \cos\theta\).

• Option 2:
• For \(\theta = 4\pi\), \(\cos(4\pi)=\cos(0 + 4\pi)=\cos(0)=1\).
• For \(\theta=\frac{\pi}{3}\), \(\cos(\frac{\pi}{3})=\frac{1}{2}\).
• Option 3:
• For \(\theta=\frac{\pi}{2}\), \(\cos(\frac{\pi}{2}) = 0

eq1\). So \((\frac{\pi}{2},1)\) is not on the graph of \(y=\cos\theta\).

• Option 4:
• For \(\theta = 4\pi\), \(\cos(4\pi)=1

eq0\). So \((4\pi,0)\) is not on the graph of \(y=\cos\theta\).

Answer:

\((4\pi,1)\) and \((\frac{\pi}{3},\frac{1}{2})\) (the second - option).