1. ## building complex polynomials

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

2. ## Re: building complex polynomials

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
If $t=\dfrac{\pi}{10}$, then $\cos t = \cos \dfrac{\pi}{10}$.

Yes, you are on the right track. You will wind up with:

$$\cos (5t) = k\cos t \left( \cos t - \cos \dfrac{\pi}{10} \right) \left( \cos t - \cos \dfrac{3\pi}{10} \right) \left( \cos t - \cos \dfrac{7\pi}{10} \right) \left( \cos t - \cos \dfrac{9\pi}{10} \right)$$

Then, just solve for $k$ when you plug in some known value for $t$. For example, $t = 0$.

$$1 = k\left(1-\cos \dfrac{\pi}{10} \right)\left(1- \cos \dfrac{3\pi}{10} \right) \left( 1 - \cos \dfrac{7\pi}{10} \right) \left(1 - \cos \dfrac{9\pi}{10} \right)$$

Solving for $k$ gives:

$$k = \dfrac{1}{\left(1-\cos \dfrac{\pi}{10} \right)\left(1- \cos \dfrac{3\pi}{10} \right) \left( 1 - \cos \dfrac{7\pi}{10} \right) \left(1 - \cos \dfrac{9\pi}{10} \right) }$$

But I don't see how to justify putting $(cos - cos (root)) instead of (cos - root). I am confused by trigonometric equations. They don't seem to work like algebraic equations. 4. ## Re: building complex polynomials It works exactly like algebraic expressions. You can use $$\cos(5t)=K\prod_{n\in \mathbb{Z}} \left( t-\dfrac{(2n+1)\pi}{10} \right)$$ But this is an infinite product.$\cos t $has the advantage of being$2\pi $cyclic. 5. ## Re: building complex polynomials Not sure why it's an infinite product. There are only 5 terms in this case. But building up (cos - root) makes sense. It is analogous to algebraic equations. I don't see how you can just change the root to the cos of the root even if cos is 2 pi cyclic. 6. ## Re: building complex polynomials Originally Posted by s_ingram Not sure why it's an infinite product. There are only 5 terms in this case. But building up (cos - root) makes sense. It is analogous to algebraic equations. I don't see how you can just change the root to the cos of the root even if cos is 2 pi cyclic. The term must be zero when$t$equals the root. At$t = \dfrac{\pi}{10}$, you do not have$\cos t = \dfrac{\pi}{10}$. You have$\cos t = \cos \dfrac{\pi}{10}$. Term by term: $$\cos t - \cos \dfrac{\pi}{10} = 0\text{ when }t = \dfrac{\pi}{10} + n\pi$$ $$\cos t - \cos \dfrac{3\pi}{10} = 0\text{ when }t = \dfrac{3\pi}{10} + n\pi$$ $$\cos t = 0\text{ when }t = \dfrac{\pi}{2} + n\pi$$ $$\cos t - \cos \dfrac{7\pi}{10} = 0\text{ when }t = \dfrac{7\pi}{10} + n\pi$$ $$\cos t - \cos \dfrac{9\pi}{10} = 0\text{ when }t = \dfrac{9\pi}{10} + n\pi$$ Assume you want to use$\sin t$instead of$\cos t$. $$\sin \left( \dfrac{\pi}{10} + n\pi \right) = \sin \dfrac{\pi}{10} \cos (n\pi) + \sin (n\pi) \cos \dfrac{\pi}{10} = (-1)^n \sin \dfrac{\pi}{10}$$ So, this will not generate all of the roots over$\cos 5t$. 7. ## Re: building complex polynomials Let$\displaystyle p(x)=16x^5-20x^3+5x$where$\displaystyle p(\cos t)=\cos 5t$now$\displaystyle p\left(\cos \frac{(2k-1)\pi }{10}\right)=\cos \text{ }\frac{(2k-1)\pi }{2}=0$so$\displaystyle \cos \frac{(2k-1)\pi }{10}$,$\displaystyle k=1,2,3,4,5$are all roots of$\displaystyle p(x)$therefore we can factor$\displaystyle p(x)=16\prod _{k=1}^5 \left(x-\cos \frac{(2k-1)\pi }{10}\right)$now replace$\displaystyle x=cos t\displaystyle \cos 5t=p(\cos t)=16\prod _{k=1}^5 \left(\cos t-\cos \frac{(2k-1)\pi }{10}\right)$8. ## Re: building complex polynomials I have been plodding throughout this, trying to figure out what you are trying to drill into me, but it isn't working! I am perfectly happy (now!) that$\cos 5t = \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})\$

but where is the k springing from?