calculating cos(nθ) and sin(nθ) with de Moivre

• Jun 19th 2010, 06:50 PM
Bart
calculating cos(nθ) and sin(nθ) with de Moivre
I'm not sure in which subforum I should post this, feel free to move it to another subforum.

From my course:

$(cos(\theta) + jsin(\theta))^3 = cos(3\theta) + jsin(3\theta)$
$cos^3(\theta) + 3jcos^2(\theta)sin(\theta) - 3cos(\theta)sin^2(\theta) - jsin^3(\theta) = cos(3\theta) + jsin(3\theta)$
<->
$cos(3\theta) = cos^3(\theta) - 3cos(\theta)sin^2(\theta)$ and $sin(3\theta) = 3cos^2(\theta)sin(\theta) - sin^3(\theta)$

I have no problem with the first two lines, I don't understand how you can logically deduce the last line from the first two lines.
After all, $cos(3\theta) = cos^3(\theta) + 3jcos^2(\theta)sin(\theta) - 3cos(\theta)sin^2(\theta) - jsin^3(\theta) - jsin(3\theta)$
• Jun 19th 2010, 07:09 PM
Ackbeet
In order to get the third line, they equated the real parts of both sides of the second line, and the imaginary parts of both sides of the second line.
• Jun 20th 2010, 12:09 PM
Bart
So that is comparable to when you would seperate X and Y in an ´ordinary` equation?
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