# proof

• May 2nd 2009, 04:53 AM
beq!x
proof
prove this : $\displaystyle [(1+itg \alpha ) /(1-itg\alpha)]^n=(1+itg (n\alpha))/(1-itg(n\alpha))$
$\displaystyle i$ is for complex numbers
• May 2nd 2009, 06:01 AM
running-gag
Quote:

Originally Posted by beq!x
prove this : $\displaystyle [(1+itg \alpha ) /(1-itg\alpha)]^n=(1+itg (n\alpha))/(1-itg(n\alpha))$
$\displaystyle i$ is for complex numbers

Hi

$\displaystyle \left[\frac{1+itg \alpha}{1-itg\alpha}\right]^n=\left[\frac{1+i\frac{\sin\alpha}{\cos\alpha}}{1-i\frac{\sin\alpha}{\cos\alpha}}\right]^n = \left[\frac{\cos\alpha+i\sin\alpha}{\cos\alpha-i\sin\alpha}\right]^n = \left[\frac{e^{i\alpha}}{e^{-i\alpha}}\right]^n = \frac{e^{in\alpha}}{e^{-in\alpha}}$

$\displaystyle \left[\frac{1+itg \alpha}{1-itg\alpha}\right]^n=\frac{\cos n\alpha+i\sin n\alpha}{\cos n\alpha-i\sin n\alpha}=\frac{1+itg (n\alpha)}{1-itg (n\alpha)}$
• May 2nd 2009, 07:15 AM
HallsofIvy
Ahh! "tg" is "tan". I was wondering what $\displaystyle g(\alpha)$ was!
• May 2nd 2009, 08:36 AM
Soroban
Hello, beq!x!

Quote:

Prove: .$\displaystyle \left[\frac{1+i\tan z}{1-i\tan x}\right]^n \;=\;\frac{1+i\tan(nz)} {1-i\tan(nz)}$

Left side: .$\displaystyle \left[\frac{1 + i\dfrac{\sin z}{\cos z}}{1 - i\dfrac{\sin z}{\cos z}}\right]^n \;=\; \left[\frac{\dfrac{\cos z + i\sin z}{\cos z}} {\dfrac{\cos x - i\sin x}{\cos z}} \right]^n \;=\;$ .$\displaystyle \left[\frac{\cos z + i\sin z}{\cos z - i\sin z}\right]^n \;=\;\frac{(\cos z + i\sin z)^n}{(\cos z - i\sin z)^n}$

Applying DeMoivre's Theorem, we have: .$\displaystyle \frac{\cos(nz) + i\sin(nz)}{\cos(nz) - i\sin(nz)}$

Divide numerator and denominator by $\displaystyle \cos(nz)$

. . $\displaystyle =\;\frac{\;\dfrac{\cos(nz)}{\cos(nz)} + i\dfrac{\sin(nz)}{\cos(nz)}\;} {\dfrac{\cos(nz)}{\cos(nz)} - i\dfrac{\sin(nz)}{\cos(nz)}} \;=\;\frac{1 + i\tan(nz)}{1 - i\tan(nz)} \quad\hdots\quad \text{ta-}DAA!$