# Suppose $v\in [\pi/2, \pi]$ with $\tan v=-17$, find the exact expression. Find $\sin(v-3\pi)$

I need help solving the problem thank you!

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Use the formula $\sin(A-B) =\sin A \cos B - \sin B \cos A$ to simplify $\sin(v-3\pi)$ keeping in mind that $\sin(3\pi) = 0$ and $\cos(3\pi) = -1$.
$\tan v = -17$ implies that $\sin v = \cfrac {17}{\sqrt{290}}$ and $\cos v = \cfrac {-1}{\sqrt{290}}$. Use thes values and you will get your answer. I hope you understand.
In the interval given $\forall \theta \in [\pi/2, \pi]$, $$\sin \theta \ge 0$$ $$\cos \theta \le 0$$ $$\tan\theta \le 0$$
no but $-\cfrac {17}{\sqrt{290}}$ –  user31280 Oct 29 '12 at 3:58