By comparing expressions for given by De Moivre's theorem and by the binomial theorem, prove that

.

By considering the equation , prove that

My problem is in the second part.

means that , but after this I don't see how relates to it.

Thanks!

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- Sep 30th 2010, 04:11 AMarzeDe Moivre's theorem
By comparing expressions for given by De Moivre's theorem and by the binomial theorem, prove that

.

By considering the equation , prove that

My problem is in the second part.

means that , but after this I don't see how relates to it.

Thanks! - Sep 30th 2010, 04:59 AMHallsofIvy
and , both of which have cosine 0 so both and satisfy that equation.

- Sep 30th 2010, 06:01 AMLukybear
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 :( - Sep 30th 2010, 06:20 AMTheCoffeeMachine

The book needs a reality check. :) - Sep 30th 2010, 06:29 AMarze
- Sep 30th 2010, 06:30 AMarze
- Sep 30th 2010, 08:53 AMTheCoffeeMachine
Ok. Here we go - we have:

.

So .

Now,

Obviously we cannot have because it yields .

Also and .

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

are the roots of .

Consider .

By Vieta's relations if , , , , are the roots of , then:

.

But we already know that these roots are ,

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}[/tex].