# De Moivre's theorem

• September 30th 2010, 03:11 AM
arze
De Moivre's theorem
By comparing expressions for $(\cos\theta+i\sin\theta)^5$ given by De Moivre's theorem and by the binomial theorem, prove that
$\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$.
By considering the equation $\cos 5\theta=0$, prove that
$\cos (\frac{\pi}{10}).\cos (\frac{3\pi}{10})=\frac{1}{4}\sqrt{5}$

My problem is in the second part.
$\cos 5\theta=0$ means that $16\cos^5\theta-20\cos^3\theta+5\cos\theta=0$, but after this I don't see how $\cos (\frac{\pi}{10}).\cos (\frac{3\pi}{10})=\frac{1}{2}\sqrt{5}$ relates to it.
Thanks!
• September 30th 2010, 03:59 AM
HallsofIvy
$5(\pi/10)= \pi/2$ and $5(3\pi/10)= 3\pi/2$, both of which have cosine 0 so both $cos(\pi/10)$ and $cos(3\pi/10)$ satisfy that equation.
• September 30th 2010, 05:01 AM
Lukybear
I think you sovle equation and get your 5 roots:
which are:
cos (2k(pi)), where k = 0,1,2,3,4,

Then you manipulate it trigonoemtrically. But i cant see how currently :(
• September 30th 2010, 05:20 AM
TheCoffeeMachine
$\displaystyle \cos\frac{\pi}{10}\cdot\cos\frac{3\pi}{10} = \frac{1}{2}\cos{\frac{\pi}{5}+\frac{1}{2}\cos{\fra c{2\pi}{5} = \frac{1}{4}\sqrt{5} \ne \frac{1}{2}\sqrt{5}.$

The book needs a reality check. :)
• September 30th 2010, 05:29 AM
arze
Quote:

Originally Posted by HallsofIvy
$5(\pi/10)= \pi/2$ and $5(3\pi/10)= 3\pi/2$, both of which have cosine 0 so both $cos(\pi/10)$ and $cos(3\pi/10)$ satisfy that equation.

Umm, I understand that part. My problem is with the proof. Do you mean find the roots of the equation $16\cos 5\theta-20\cos 3\theta+5\cos\theta=0$?
• September 30th 2010, 05:30 AM
arze
Quote:

Originally Posted by TheCoffeeMachine
$\displaystyle \cos\frac{\pi}{10}\cdot\cos\frac{3\pi}{10} = \frac{1}{2}\cos{\frac{\pi}{5}+\frac{1}{2}\cos{\fra c{2\pi}{5} = \frac{1}{4}\sqrt{5} \ne \frac{1}{2}\sqrt{5}.$

The book needs a reality check. :)

sorry, my bad, its a typo
• September 30th 2010, 07:53 AM
TheCoffeeMachine
Ok. Here we go - we have:

$\cos{5\theta} = 16\cos^5\theta-20\cos^3\theta+5\cos\theta$

$\displaystyle \Rightarrow \frac{\cos{5\theta}}{\cos{\theta}} = 16\cos^4\theta-20\cos^2\theta+5$.

So $\cos{5\theta} = 0 \Rightarrow 16\cos^4\theta-20\cos^2\theta+5 = 0$.

Now, $\displaystyle \cos{5\theta} = 0 \Rightarrow \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}, ...$

Obviously we cannot have $\displaystyle{\theta} = \frac{5\pi}{10}$ because it yields $\cos{\theta} = 0$.

Also $\displaystyle \cos\frac{9\pi}{10} = -\cos{\frac{\pi}{10}}$ and $\displaystyle \cos\frac{7\pi}{10} = -\cos{\frac{3\pi}{10}}$.

So [tex]\displaystyle \cos\frac{\pi}{10}[/Math], [tex]\displaystyle \cos\frac{3\pi}{10}[/Math], [tex]\displaystyle-\cos\frac{3\pi}{10}[/Math], and $\displaystyle-\cos\frac{\pi}{10}$

are the roots of $\displaystyle 16\cos^4\theta-20\cos^2\theta+5 = 0$.

Consider $\displaystyle f(\theta) = 16\cos^4\theta-20\cos^2\theta+5$.

By Vieta's relations if $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$, $\alpha_{4}$, are the roots of $f(\theta)$, then:

$\displaystyle \alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4} = \frac{5}{16}$.

But we already know that these roots are $\displaystyle \cos\frac{\pi}{10}, \cos\frac{3\pi}{10}, -\cos\frac{3\pi}{10}, -\cos\frac{\pi}{10}$,

so we have [tex]\displaystyle \cos\frac{\pi}{10}\cos\frac{3\pi}{10}\cos\frac{3\p i}{10}\cos\frac{\pi}{10} = \frac{5}{16} \Rightarrow \cos^2\frac{\pi}{10}\cos^2\frac{3\pi}{10} = \frac{5}{16}[/Math].

$\displaystyle \therefore \boxed{\cos\frac{\pi}{10}\cos\frac{3\pi}{10} = \frac{1}{4}\sqrt{5}}.$