3
$\begingroup$

I've figured out the approach. Writing the expansion of $(1 + x)^n$, then replacing $x$ with $i \tan (A)$.

Then separating real and imaginary part and $\tan(nA)$ will be equal to Im/Real.

But, after reaching $(1 + i \tan(A))^n$, I'm unable to convert it into De-Moivre's form from which I could proceed further.

Prove that

$$\tan(nA) = \dfrac{\dbinom{n}1t - \dbinom{n}3 t^3 + \dbinom{n}5 t^5 \pm \cdots}{1 - \dbinom{n}2 t^2 + \dbinom{n}4 t^4 \pm \cdots}$$ where $t= \tan(A)$.

  • 0
    I see you've already "accepted" an answer, but the one I've posted has a different viewpoint that is also worth knowing about.2012-06-25
  • 1
    So far I'm the only one who's up-voted this even those Marvis has also posted an answer and several people have commented on it.2012-06-25

2 Answers 2

2

\begin{align} (1 + i \tan(A))^n & = \left( 1 + i \dfrac{\sin(A)}{\cos(A)}\right)^n\\ & = \left(\dfrac{\cos(A) + i \sin(A)}{\cos(A)} \right)^n\\ & = \dfrac{e^{inA}}{\cos^n(A)}\\ & = \dfrac{\cos(nA) + i \sin(nA)}{\cos^n(A)} \end{align} Hence, the real part is $\dfrac{\cos(nA)}{\cos^n(A)}$ and the imaginary part is $\dfrac{\sin(nA)}{\cos^n(A)}$.

Now you should be able to finish it off by dividing the imaginary part by the real part.

  • 0
    Not sure why this is necessary. If you expand (1+itan(A))^n and collect the real parts (= the even terms), you get the denominator; collect to the imaginary parts (the odd terms), you get the numerator. Done?2012-06-25
  • 1
    @anthonyquas Bazinga had problems with applying the De-Moivre's formula and not with finding the real and imaginary part of $(1+i \tan(A))^n$ by expanding the terms out.2012-06-25
  • 0
    OK. It looks like you understood the OP's question better than I did. Was the fact that Im(1+itan A)^n/Re(1+itan A)^n assumed known in the Q, or was this what he was asking for help with?2012-06-25
  • 0
    From what I understood, looks like OP is fine with expanding $(1+i \tan(A))^n$ as $$\left(1 - \dbinom{n}2 t^2 + \dbinom{n}4 t^4 \pm \cdots \right) + i \left(\dbinom{n}1t - \dbinom{n}3 t^3 + \dbinom{n}5 t^5 \pm \cdots\right).$$ But is unable to see why Im/Re gives $\tan(nA)$. I understood this, since the OP says he has problems with converting this using De Moivre's formula. But yes I do agree that the question is unclear and it needs to be written in a better way. Looks like OP was told by someone to consider $(1+i\tan(A))^n$ and find Im/Re to get tan(nA).2012-06-25
  • 0
    Actually I had done some questions before to find the expansion of cot(npi/4) where A = pi/4, so tanA = 1. Thus I knew the approach but I wasn't able to find the missing link.2012-06-25
2

Generally $$ \tan(\alpha+\beta+\gamma+\cdots) = \frac{e_1-e_3+e_5-e_7+\cdots}{e_0 -e_2+e_4-e_6+\cdots}\tag{1} $$ where $e_k$ is the sum of all products of $k$ of the tangents $\tan\alpha,\tan\beta,\tan\gamma,\ldots\ {}$. For example, if there are just four variables, $\alpha,\beta,\gamma,\delta$, then $$ e_2 = \tan\alpha\tan\beta + \tan\alpha\tan\gamma + \tan\alpha\tan\delta + \tan\beta\tan\gamma+\tan\beta\tan\delta +\tan\gamma\tan\delta $$ and $$ e_3 = \tan\alpha\tan\beta\tan\gamma+\tan\alpha\tan\beta\tan\delta+\tan\alpha\tan\gamma\tan\delta+\tan\beta\tan\gamma\tan\delta, $$ and so on. And of course $e_0=1$ (except when there are $0$ variables, in which case $e_0=0$). It's easy to prove $(1)$ by mathematical induction on the number of variables.

So if you want $\tan(n\alpha)$, it's just the case where all of the variables are the same variable, $\alpha$. So for example, if there are four variables, then $e_2 = \dbinom 4 2 \tan\alpha\tan\alpha = 6\tan^2\alpha$ and $e_3 = \dbinom 4 3 \tan^3\alpha$, etc.

  • 0
    Thank you for showing another approach.2012-06-26