# How do you find the exact value of cos((11pi)/3)

How do you find the exact value of $\mathrm{cos}\left(\frac{11\pi }{3}\right)$?
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Sanai Douglas
Given: $\mathrm{cos}\left(\frac{11\pi }{3}\right)$
The cosine function has a period of $2\pi$ . This means the values of the cosine repeat every period.
See if $\frac{11\pi }{3}$ is $>2\pi$
$\frac{2\pi }{1}=\frac{2\pi }{1}\ast \frac{3}{3}=\frac{6\pi }{3}$
$\frac{11\pi }{3}=\frac{6\pi }{3}+\frac{5\pi }{3}=2\pi +\frac{5\pi }{3}$
Yes $\frac{11\pi }{3}$ is $>2\pi$
The equivalent cosine value of $\frac{11\pi }{3}$ is $\frac{5\pi }{3}$
On a trigonometric circle, the $\mathrm{cos}\left(\frac{5\pi }{3}\right)=\frac{1}{2}$

Maghrabimh
Explanation:
I'm not sure why we have an entire subject dedicated to just two right triangles (30,60,90 and 45,45,90) but we sure seem to.
$\mathrm{cos}\left(\frac{11\pi }{3}\right)$
$=\mathrm{cos}\left(\frac{11\pi }{3}-4\pi \right)$
$=\mathrm{cos}\left(-\frac{\pi }{3}\right)$
$=\mathrm{cos}\left(\frac{\pi }{3}\right)$
$=\mathrm{cos}{60}^{\circ }$
$=\frac{1}{2}$