# Solving for Cos Exactly

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.

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 $\cos (\pi + x) = -\cos x$

• ${}{}{}{}{}{}{}+1$ – Jika Jun 27 '14 at 16:16
• Thank You Jika for the +1 :) – Kushashwa Ravi Shrimali Jun 27 '14 at 16:16
• I would just use the unit circle ^.^ – Cruncher Jun 27 '14 at 17:12
• Do those identities also work for sin and tan? – A_for_ Abacus Jun 27 '14 at 18:00
• @A_for_Abacus: No, but there are analogous versions. The ones for sine (and cosine) can be seen by drawing out generic triangles. The ones for tangent are better done algebraically once you know the sine and cosine ones. – Eric Stucky Jun 27 '14 at 23:36

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

• cos(-x) = cos x .. I think you have done a typo there. – Kushashwa Ravi Shrimali Jun 27 '14 at 16:02
• Yes. Thank you. I fixed it. – Jika Jun 27 '14 at 16:03
• Looks fine now. :) – Kushashwa Ravi Shrimali Jun 27 '14 at 16:03
• LaTeX tip : Instead of using $\$ \ at each line, you may just use the code like this : \begin{align} & \cos (\pi + x) = -\cos x \\ & \cos(-x) = \cos x \end{align} – Kushashwa Ravi Shrimali Jun 27 '14 at 16:14
• ${}{}{}{}$. Thanks. – Jika Jun 27 '14 at 16:16