and , both of which have cosine 0 so both and satisfy that equation.
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!
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].