Take the 2-minute tour ×
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.

How do I get from 2nd last to last step? How did they simplify cos & sin $99\times \frac{5\pi}{6}$?

share|improve this question

3 Answers 3

Well, $99\cdot \frac{5\pi}{6}=33\cdot \frac{5\pi}{2}=\frac{165\pi}{2}$, so

$\sin\left(\frac{165\pi}{2}\right)=\sin\left(82\pi + \frac{\pi}{2}\right)=\sin(\frac{\pi}{2})=1$. Similarly, $\cos\left(\frac{165\pi}{2}\right)=0$.

share|improve this answer

$99\cdot \frac {5 \pi}{6}=82.5\pi $ and you can ignore the multiples of $2 \pi$

share|improve this answer

$\cos \left(99 \times \frac{5 \pi}{6} \right) = \cos \left(33 \times \frac{5 \pi}{2} \right) = \cos \left(\frac{165 \pi}{2} \right) = 0$ since $\cos(\frac{n \pi}{2}) = 0$ whenever $n$ is odd.

$\sin \left(99 \times \frac{5 \pi}{6} \right) = \sin \left(33 \times \frac{5 \pi}{2} \right) = \sin \left(\frac{165 \pi}{2} \right) = \sin \left(82 \pi + \frac{\pi}{2} \right) = 1$ since $\sin(2n \pi + \frac{\pi}{2}) = 1$.

share|improve this answer

Your Answer

 
discard

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.