QUESTION IMAGE
Question
use the unit circle to find the value of \\(\cos \frac{\pi}{2}\\) and periodic properties of trigonometric functions to find the value of \\(\cos \frac{9\pi}{2}\\).
image of unit circle with points \\(\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\
ight)\\), \\(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\
ight)\\), \\((1,0)\\), \\(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\
ight)\\), \\((0,-1)\\), \\(\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\
ight)\\), \\((-1,0)\\), \\((0,1)\\), and angles \\(\frac{\pi}{4}\\), \\(\frac{3\pi}{4}\\)
select the correct choice below and fill in any answer boxes in your choice.
a. \\(\cos \frac{\pi}{2} = 0\\)
(type an exact answer, using radicals as needed. simplify your answer. rationalize the denominator.)
\\(\bigcirc\\) b. the solution is undefined.
select the correct choice below and fill in any answer boxes in your choice.
\\(\bigcirc\\) a. \\(\cos \frac{9\pi}{2} = \square\\)
(type an exact answer, using radicals as needed. simplify your answer. rationalize the denominator.)
\\(\bigcirc\\) b. the solution is undefined.
Step1: Recall cosine on unit circle
On the unit circle, for an angle \(\theta\), \(\cos\theta\) is the x - coordinate of the point \((x,y)\) corresponding to \(\theta\). For \(\theta=\frac{\pi}{2}\), the point is \((0,1)\), so \(\cos\frac{\pi}{2} = 0\) (already given in option A).
Step2: Use periodicity of cosine
The cosine function \(y = \cos x\) has a period of \(2\pi\), i.e., \(\cos(x + 2k\pi)=\cos x\) for any integer \(k\). We can rewrite \(\frac{9\pi}{2}\) as \(\frac{\pi}{2}+4\pi\) (since \(4\pi = 2\times2\pi\), and \(k = 2\) here). So \(\cos\frac{9\pi}{2}=\cos(\frac{\pi}{2}+4\pi)\).
Step3: Apply periodic property
Using the periodic property \(\cos(x + 2k\pi)=\cos x\) with \(x=\frac{\pi}{2}\) and \(k = 2\), we get \(\cos(\frac{\pi}{2}+4\pi)=\cos\frac{\pi}{2}\). Since \(\cos\frac{\pi}{2}=0\), then \(\cos\frac{9\pi}{2}=0\).
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For \(\cos\frac{9\pi}{2}\), the correct choice is A and \(\cos\frac{9\pi}{2}=\boldsymbol{0}\)