Question

1. \\( \cos \frac { 13 \pi } { 2 } \\)

a) - 1
b) 1
c) 0
d) undefined

Explanation:

Step1: Use the cosine periodicity formula

The cosine function has a period of \(2\pi\), so \(\cos(x + 2k\pi)=\cos(x)\) for any integer \(k\). We can rewrite \(\frac{13\pi}{2}\) as \(\frac{12\pi + \pi}{2}=6\pi+\frac{\pi}{2}\). Here \(x = \frac{\pi}{2}\) and \(k = 3\). Then \(\cos(\frac{13\pi}{2})=\cos(6\pi+\frac{\pi}{2})\). According to the formula \(\cos(x + 2k\pi)=\cos(x)\), we have \(\cos(6\pi+\frac{\pi}{2})=\cos(\frac{\pi}{2})\).

Step2: Evaluate \(\cos(\frac{\pi}{2})\)

We know from the unit - circle definition of the cosine function. For an angle \(\theta\) in standard position \((x,y)\) on the unit circle \(x^{2}+y^{2}=1\), \(\cos\theta=x\). When \(\theta=\frac{\pi}{2}\), the point on the unit circle is \((0,1)\), so \(\cos(\frac{\pi}{2}) = 0\).

Answer:

C. 0