7) \\( \\cos \\frac { 13 \\pi } { 2 } \\)\na) - 1\nb) 1\nc) 0\nd) undefined
7) \\( \\cos \\frac { 13 \\pi } { 2 } \\)\na) - 1\nb) 1\nc) 0\nd) undefined
Answer
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