1. ## Complex numbers

Hi I have 2 really really short questions and one medium length one I need help with!

1.)Does
$\displaystyle 4isin(\frac{pi}{2}-2x) - 4cos(\frac{pi}{2}-2x)$ equal to $\displaystyle 4isin(\frac{pi}{2}+2x) - 4cos(\frac{pi}{2}+2x)$?

2.)How do I, using complex number methods, deduce that cos(3x) = 4cos^3(x) - 3cos(x)?

3.)Does |a-bi| =a+bi?

Thank you so much!

2. ## Re: Complex numbers

Hello, Tutu!

2) How do I, using complex number methods, deduce that: $\displaystyle \cos(3x) \:=\: 4\cos^3\!x - 3\cos x\,?$

You should be familiar with this identity: .$\displaystyle \left(\cos\theta + i\sin\theta)^n \:=\:\cos(n\theta) + i\sin(n\theta)$

For $\displaystyle n = 3$, we have: .$\displaystyle (\cos x + i\sin x)^3 \:=\:\cos(3x) + i\sin(3x)$ .[1]

Expand the left side:

. . $\displaystyle (\cos x + i\sin x)^3 \;=\;\cos^3\!x + 3i\cos^2x\sin x - 3\cos x\sin^2\!x - i\sin^3\!x$

n . . . . . . . . . . . . $\displaystyle =\;(\cos^3\!x - 3\cos x\sin^2x) + i(3\cos^2\!x\sin x - \sin^3\!x)$

Then [1] becomes: .$\displaystyle \cos(3x) + i\sin(3x) \;=\;(\cos^3\!x - 3\cos x\sin^2x) + i(3\cos^2\!x\sin x - \sin^3\!x)$

Equate real and imaginary components:

$\displaystyle \cos(3x) \:=\:\cos^3x - 3\cos x\sin^2\!x \:=\:\cos^3\!x - 3\cos x(1 - \cos^2\!x)$

. . Hence: .$\displaystyle \boxed{\cos(3x) \;=\;4\cos^3\!x - 3\cos x}$

$\displaystyle \sin(3x) \;=\;\3\cos^2\!x\sin x - \sin^3\!x \;=\;3(1-\sin^2\!x)\sin x - \sin^3\!x$

. . Hence: .$\displaystyle \boxed{\sin(3x) \;=\;3\sin x - 4\sin^3\!x}$

And we have the Triple-angle Identities for both sine and cosine.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle \tan(3x) \;=\;\frac{\sin(3x)}{\cos(3x)} \;=\;\frac{3\cos^2\!x\sin x - \sin^3\!x}{\cos^3\!x - 3\cos x\sin^2\!x}$

Divide top and bottom by $\displaystyle \cos^3\!x\!:$

$\displaystyle \tan(3x) \;=\;\dfrac{\frac{3\cos^2\!x\sin x}{\cos^3\!x} - \frac{\sin^3\!x}{\cos^3\!x}}{\frac{\cos^3\!x}{\cos ^3\!x} - \frac{3\cos x\sin^2\!x}{\cos^3\!x}} \;=\;\dfrac{3\left(\frac{\sin x}{\cos x}\right) - \left(\frac{\sin x}{\cos x}\right)^3}{1 - 3\left(\frac{\sin x}{\cos x}\right)^2}$

Therefore: .$\displaystyle \boxed{\tan(3x) \;=\;\frac{3\tan x - \tan^3\!x}{1 - 3\tan^2\!x}}$