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.

I am to calculate $\cos(v+\frac{\pi}{6})$ when $\cos v = -\frac{2}{3}$ and $0 \lt v \lt \pi$.

I know I can change $\cos(v+\frac{\pi}{6})$ into

$$\cos v \times \cos \frac{\pi}{6} - \sin v \times \sin \frac{\pi}{6}$$

but I fail to see how this helps me, or for that matter any other step I could take. Am I completely off?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

No, you’re doing fine. You now have $\cos\left(v+\frac{\pi}6\right)$ expressed in terms of four quantities: $\cos v$, $\cos\frac{\pi}6$, $\sin\frac{\pi}6$, and $\sin v$. You know (or should know) the numerical values of the first three, so all that’s left is to determine $\sin v$. Now $\cos v$ is negative, so $v$ is in the second or third quadrant. But you also know that $0<v<\pi$, so in fact $v$ is in the second quadrant. Now use the fact that $\sin^2 v+\cos^2v=1$ to get $\sin^2v$, and decide whether you want the positive or negative square root of this for an angle in the second quadrant.

share|improve this answer
    
This helped a lot, I found the answer and it seems to be correct ($-\frac{2\sqrt{3}+\sqrt{5}}{6}$). Thank you! –  Quispiam Sep 12 '12 at 10:07

Hint:

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

and

$\sin x=\sqrt {1-\cos^2 x}$ (as $0<x<\pi$ and $\sin x$ is positive in this interval)

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.