# Calculate $\cos(v+\frac{\pi}{6})$ when $\cos v = -\frac{2}{3}$

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?

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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.

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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)

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