Solving for Cos Exactly

Question

How to solve $\cos(\dfrac{5\pi}{6})$ and $\cos(\dfrac{7\pi}{6})$ exactly?

I couldn't use special triangles to solve this either.

Answer 1

Remember these 2 identities: $$\begin{align} &\cos \left( \pi - x \right) = - \cos(x) \\ & \cos(\pi + x ) = -\cos (x) \end{align}$$

Or you can simply write: $$\cos (\pi - x) = \cos (\pi + x) =-\cos (x) $$

You need to find $\cos \left(\cfrac{5\pi}{6}\right)$. Try to break this down in order to use the above identities.

You can write $\cos \left(\cfrac{5\pi}{6}\right)$ as $\cos \left( \cfrac{6\pi - \pi}{6}\right)$

Or: $$\cos \left(\cfrac{6\pi}{6} - \cfrac{\pi}{6} \right) = \cos \left(\pi - \cfrac{\pi}{6} \right) $$

Now, you may use $\cos \left( \pi - x \right) = -\cos x$ identity here.

For the second part: $\cos \left(\cfrac{7\pi}{6}\right) = ? $

Break $\cfrac{7\pi}{6}$ into: $\color{blue}{\cfrac{6\pi + \pi}{6}}$

So, you get:

$$\cos \left( \pi + \cfrac{\pi}{6} \right)$$

Thus, you can use the identity of $\cos (\pi + x) = -\cos x$

Answer 2

HINT: $$\cos(\dfrac{\pi}{6})=\dfrac{\sqrt{3}}{2}.$$ $$\dfrac{5\pi}{6}=\pi-\dfrac{\pi}{6}.$$ $$\dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}.$$ $$\cos(\pi+x)=-\cos x.$$ $$\cos(-x)=\cos x.$$