Question

using the partial view of the unit circle shown below, which supporting statement best supports the fact that the period of \\( \cos (x) \\) is \\( 2 \pi \\)?
\\( \cos (\frac{\pi}{3})=\frac{1}{2} \\)
\\( \cos (\frac{\pi}{4})=\frac{\sqrt{2}}{2} \\)
\\( \cos (\frac{4 \pi}{3})=\frac{1}{2} \\)
\\( \cos (\frac{4 \pi}{3})=\frac{\sqrt{2}}{2} \\)

Explanation:

Step1: Recall the period formula

The period of \(y = \cos(x)\) is \(T = 2\pi\). By the definition of period, \(\cos(x)=\cos(x + 2k\pi)\), \(k\in\mathbb{Z}\).

Step2: Analyze each option

• For \(\cos(\frac{\pi}{3})=\frac{1}{2}\), there is no indication of the period property here.
• For \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), no indication of the period property.
• For \(\cos(\frac{13\pi}{6})\), we can rewrite \(\frac{13\pi}{6}=2\pi+\frac{\pi}{6}\). Since \(\cos(x + 2\pi)=\cos(x)\), and from the unit - circle \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\), so \(\cos(\frac{13\pi}{6})=\cos(2\pi+\frac{\pi}{6})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\). This shows the period property \(T = 2\pi\).
• For \(\cos(\frac{11\pi}{6})\), \(\frac{11\pi}{6}=2\pi-\frac{\pi}{6}\), \(\cos(2\pi - x)=\cos(x)\), but it is more about the cosine's even - like property (\(\cos(A - B)\) formula) rather than the most straightforward period demonstration compared to \(\cos(x+2\pi)=\cos(x)\)

Answer:

\(\cos(\frac{13\pi}{6})=\frac{\sqrt{3}}{2}\)