Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need help solving the problem thank you!

share|cite|improve this question
up vote 0 down vote accepted

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 $\forall \theta \in [\pi/2, \pi]$, $$\sin \theta \ge 0$$ $$\cos \theta \le 0$$ $$\tan\theta \le 0$$

share|cite|improve this answer
no but $-\cfrac {17}{\sqrt{290}}$ – user31280 Oct 29 '12 at 3:58
i got it just got it thank you!!! – Maximiliano Oct 29 '12 at 4:00
what formula would you use for cos v= – Maximiliano Oct 29 '12 at 4:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.