Originally Posted by

**s_ingram** Hi folks,

By considering the roots of cos 5t = 0 prove that

$\cos 5t = 16 \cos t (\cos t - \cos \frac{\pi}{10}) (\cos t - \cos \frac{3\pi}{10}) (\cos t - \cos \frac{7\pi}{10}) (\cos t - \cos \frac{9\pi}{10})$

in the first part of the problem we had to prove using de Moivre's theorem, that $\cos 5t = 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$ I don't know if this is relevant for the second part.

The roots of $\cos 5t = 0$ are $5t = \cos^{-1} (0)$ i.e. $5t = \dfrac{\pi}{2} + n\pi$ which gives

t = $\dfrac{\pi}{10}$ , $\dfrac{3\pi}{10}$, $\dfrac{\pi}{2}$, $\dfrac{7\pi}{10}$, $\dfrac{9\pi}{10}$

The form of the answer suggests that I can build a 5th order polynomial using these roots in the same way that for a quadratic equation in x, I could build it from the 2 roots $\alpha$ and $\beta$ as

$(x - \alpha)(x - \beta)$

but this would give

$ (\cos t - \dfrac{\pi}{10}) $ ... etc not

$(\cos t - \cos \dfrac{\pi}{10}) $ ....

I am not sure whether I am on track or not. Can anyone supply a hint?

thanks