Intense trigonometric question

[newchie]

Homework Statement

prove that for any integer n

tannθ = ((n choose 1)tanθ - (n choose 3)tan^3θ + (n choose 5)tan^5θ)-.../ (( n choose 0) -(n choose 2)tan^2θ + (n choose 4)tan^4θ-...)

The first 3 numbers on top and bottom look like these however the sum and differences keep going, and you have to prove they are equal

Homework Equations

Demoivres
Induction

The Attempt at a Solution

I tried to use demoivres but all the terms seem to messing up, and if i use induction i get very weird numbers, I just would like an approach on this problem

 

[A. Bahat]

This is a very interesting problem. You were right that de Moivre's formula is involved, so we should write e^{in\theta}=\cos(n\theta)+i\sin(n\theta).
But we also have e^{in\theta}=(\cos(\theta)+i\sin(\theta))^n. Just expand this out (using the binomial theorem) and set these expressions equal to each other.

 

[newchie]

This is a very interesting problem. You were right that de Moivre's formula is involved, so we should write e^{in\theta}=\cos(n\theta)+i\sin(n\theta).
But we also have e^{in\theta}=(\cos(\theta)+i\sin(\theta))^n. Just expand this out (using the binomial theorem) and set these expressions equal to each other.

Could you write it out for me I am still confused

 

[A. Bahat]

By the binomial theorem, (\cos\theta+i\sin\theta)^n=\sum_{k=0}^n {n\choose k}\,\cos^{n-k}(θ)\cdot i^k\sin^k(θ). So we know that \cos(nθ)+i\sin(nθ)=\sum_{k=0}^n {n\choose k}\,\cos^{n-k}(θ)\cdot i^k\sin^k(θ). The terms with even k are real, while the terms with odd k are imaginary. Now set the real part of the left-hand side (the cosine term) equal to the real part of the right-hand side (the sum going over even k). Likewise, doing this with the imaginary parts will involve the sine term. This will give expressions for \cos(nθ) and \sin(nθ). Then you can just take the quotient of these to get \tan(nθ).

 

[newchie]

I see! Thanks for all your help, if you had time would you mind putting up the full explained proof. I have done demoivres before for cos and sin, but its a bit confusing for tan, do i make it sin/cos then do something I am sorry i just don't get how this will be used. I get the expansion just not how we can use it

 

[A. Bahat]

Okay, since the real part of that sum on the right-hand side is \sum_{\substack{0\leq k\leq n\\k\text{ even}}}{n\choose k}\cos^{n-k}(θ)\cdot i^k\sin^k(θ) and the real part on the left-hand side is just \cos(nθ), we have \cos(nθ)=\sum_{\substack{0\leq k\leq n\\k\text{ even}}}{n\choose k}\cos^{n-k}(θ)\cdot i^k\sin^k(θ)={n\choose 0}\cos^n(θ)-{n\choose 2}\cos^{n-2}(θ)\sin^2(θ)+\cdots (this sum obviously ends either on k=n-1 or k=n, depending on whether n is even or odd).

The imaginary part is basically the same and gives you an expression for sine.

 

[SammyS]

Okay, since the real part of that sum on the right-hand side is \sum_{\substack{0\leq k\leq n\\k\text{ even}}}{n\choose k}\cos^{n-k}(θ)\cdot i^k\sin^k(θ) and the real part on the left-hand side is just \cos(nθ), we have \cos(nθ)=\sum_{\substack{0\leq k\leq n\\k\text{ even}}}{n\choose k}\cos^{n-k}(θ)\cdot i^k\sin^k(θ)={n\choose 0}\cos^n(θ)-{n\choose 2}\cos^{n-2}(θ)\sin^2(θ)+\cdots (this sum obviously ends either on k=n-1 or k=n, depending on whether n is even or odd).
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[A. Bahat]

Sorry, I got ahead of myself! It took out a big chunk of my previous response. Is it okay now?

 

[SammyS]

Sorry, I got ahead of myself! It took out a big chunk of my previous response. Is it okay now?

Hopefully, what remains will be OK.