Question

@ganeshie8 @Miracrown @dan815 @UsukiDoll

Answer 1

\[\cot(\pi/15) - 4\sin(\pi/15) = \sqrt{15}\]

Answer 2

My work so far.
This expression is equivalent to \(\cot(16\pi/15) + 4\sin(16\pi/15)\).
\[x := e^{2\pi i /11}\]Then:\[\cot(16\pi/15) = \frac{\cos(16\pi/15)}{\sin(16\pi/15) } = \dfrac{(x^4 + \overline{x^4})/2}{(x^4 - \overline{x^4})/2i}\]\[\Rightarrow i\cot(16\pi/15) = \dfrac{x^{11}+ x^4}{x^{11} - x^4} = \dfrac{x^7+1}{x^7 - 1}\]

Answer 3

Similarly,\[4i \sin(16\pi/15) = 2(x^4 - x^{11})\]

Answer 4

I quit... what is this?!

Answer 5

Complex numbers.

Answer 6

like complex analysis?

Answer 7

If we add those two, we get\[i \cot(16\pi/15) + 4i \sin(16\pi/15) = \frac{x^7+1}{x^7 - 1} + 2(x^4 - x^{11})\]Somehow, I want to make that fraction a polynomial using properties of roots of unity. Any ideas?

Answer 8

@UsukiDoll No, I'm just trying to prove a trigonometric identity using complex numbers. Specifically, I mean roots of unity. Is that complex analysis? I don't know.

Answer 9

I'm still a noob when it comes to proofs D:

Answer 10

Also, we have to show that the right-hand side is \(i\sqrt{15}\) and we'll be done.

Answer 11

I know I can write \(1\) as \(x^{15}\) or \(x^{30}\) or so on. Which exponent should I choose so as to make it divisible by \(x^7 + 1\)?

Answer 12

\(x^7 - 1\) actually, which means that we're also looking at a seventh root of unity.

Answer 13

@ganeshie8 Speak out. :P