2

This is homework so if there is another example that can illustrate the technique I would happily accept that as guidance. The only thing I have been able to find is a question asking about $\cos(2\pi/7)$, which I think is a much harder problem.

I dont have the faintest idea how to solve this and the textbook (Hungerford) doesn't have any examples at all. Ive tried looking for resources online but havent found any that I was able to understand.

So can anyone show me how to solve these types of problems? Thanks a bunch.

Falker
  • 21
  • 1
  • 3

2 Answers2

2

If $\theta={2\pi \over 5}$, then $ \cos 2\theta=\cos 3\theta$. Now use trigonometric identities.

Per Erik Manne
  • 11,106
1

Note that $2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$ therefore $\cos(2⋅\dfrac{2\pi}{5})=\cos(3⋅\dfrac{2\pi}{5})$. Put $\cos(\dfrac{2\pi}{5})=x$. Using the formulas $\cos2x=2\cos2x−1,\cos 3x=4\cos 3x−3\cos x$, we have $4x^3−2x^2−3x+1=0⇔(x−1)(4x^2+2x−1)=0$. Because $\cos(\dfrac{2\pi}{5})≠1$, we get $4x^2+2x−1=0.$ Another way, $\cos(\dfrac{2\pi}{5})>0$, then $\cos \dfrac{2\pi}{5} = \dfrac{-1 + \sqrt{5}}{2}$.

fenyvesr
  • 11