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.