# Trigonometric relations

Determine the angle $v$ between $\pi/2$ and $\pi$ that meet $\cos v = \cos(23\pi/18)$. The answer should be able to be written like $v=a\pi/b$ where $a/b$ is a abbreviated fraction.

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Please explain what you've done so far and where you're having trouble. –  Andrew Uzzell Jan 4 '13 at 17:06

Recall that $$\cos(\pi+ \theta) = \cos(\pi - \theta)$$ Note that $$\dfrac{23}{18} \pi = \pi + \dfrac5{18} \pi$$ Now finish it off.

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$$\cos \frac{23}{18}\pi=\cos\left(2n\pi\pm \frac{23}{18}\pi\right)$$ where $n$ is any integer.

So we need $$\frac\pi2< 2n\pi\pm \frac{23}{18}\pi <\pi$$

For '+', $$9< 36n+23< 18\implies -\frac{14}{36}<n<-\frac5{36}$$ so there can be no integral $n$

For '-', $$9< 36n-23< 18\implies \frac{32}{36}<n<\frac{41}{36}\implies n=1$$

Putting $n=1,$ we get, $2\pi-\frac{23}{18}\pi=\frac{13}{18}\pi$

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