# Eulers formula trouble

• Sep 16th 2009, 05:54 AM
andreynr6
Eulers formula trouble
How do I write

2sin²5xcos3x

to a sum of cosine and/or sinusterms?
I should use Eulers formula, and control it by putting x=0 before and after I've done it.

Don't really know how to start, looked all the chapters in my book but still can't figure it out, think it's the ² term that makes me confused.

Any tip?
• Sep 16th 2009, 06:07 AM
songoku
Hi andreynr6

I'm not sure about the Euler formula, but I can use trigonometry manipulation to write 2sin²5xcos3x as a sum of cosine terms.

$\displaystyle 2 \sin^2 5x \cos 3x$

$\displaystyle =2\left(\frac{1}{2}-\frac{1}{2} \cos 10x\right) \cos 3x$

$\displaystyle =\cos 3x - \cos 3x \cos 10x$

$\displaystyle =\cos 3x - \frac{1}{2}(\cos 13x + \cos 7x)$
• Sep 16th 2009, 06:18 AM
andreynr6
ah, I did something like you also, but noticed my little mistake thanks to you.

Yeah, think I skip Euler for this time :)

Thanks!

/andrey, not so good in English.....yet
• Sep 17th 2009, 02:18 AM
pacman
I don't know what you are apt to, but this may help . . . .

2 sin^2 (5x) cos (3x) = 16sin^2 (x) cos (x) (cos 2x - 1/2)(cos 2x + cos 4x + 1/2)^2

graph below,
• Sep 17th 2009, 02:23 AM
pacman
or this one, because it depends on what you want to obtain . . . .

2 sin^2 5x cos 3x = 8[sin^2 (5x/2)][cos^2 (5x/2)][2cos^2 (3x/2) - 1]

factoring it will do . . .
• Sep 17th 2009, 03:18 AM
andreynr6
how did you do the first one? I wanna know, so I can do it myself..
• Sep 17th 2009, 07:36 AM
andreynr6
Solved it.

Here's the solution, thanks to all of you.
• Sep 17th 2009, 04:11 PM
pacman
ah, now i know what you wanted