# Intense trigonometric question!

1. Jan 19, 2012

### newchie

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Demoivres
Induction

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 19, 2012

### 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.

3. Jan 19, 2012

### newchie

Could you write it out for me im still confused

4. Jan 19, 2012

### 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 $\tan(nθ)$.

5. Jan 19, 2012

### 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 im sorry i just dont get how this will be used. I get the expansion just not how we can use it

6. Jan 19, 2012

### 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.

Last edited: Jan 19, 2012
7. Jan 19, 2012

### SammyS

Staff Emeritus
Have you read the rules for these Forums?

Under Homework Help it says:
On helping with questions: ... Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.​

8. Jan 19, 2012

### A. Bahat

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

9. Jan 19, 2012

### SammyS

Staff Emeritus
Hopefully, what remains will be OK.