I can't see how they did this operation.

http://img824.imageshack.us/img824/1111/simpr.png

thx for the help.

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- Jan 28th 2011, 02:01 PMkonvosI don't get this simplification
I can't see how they did this operation.

http://img824.imageshack.us/img824/1111/simpr.png

thx for the help. - Jan 28th 2011, 02:09 PMpickslides
Looks like simple trig identities to me i.e. $\displaystyle 2\sin x \cos x = \sin 2x$

- Jan 28th 2011, 02:09 PMArchie Meade
- Jan 28th 2011, 02:11 PMHoudini
Ok..1.sin2(x) + cos2(x) = 1=>sin2(x)=1+cos2(x)....but you have - in front of sin so you can see why -1+cos2(x)

2.sin(2x) = 2 sin x cos x

from 1 and 2 =>-sin2(x) + cos2(x)+2 sin x cos x=-1+cos2(x)+sin(2x) - Jan 28th 2011, 02:23 PMkonvos
thx for the help everybody

I'd no idea of this identity http://www.mathhelpforum.com/math-he...2177745298.png

I suppose I should review trig... - Jan 28th 2011, 02:48 PMDefunkt
$\displaystyle -sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =$

$\displaystyle = -sin^2x -cos^2x + cos^2x + cos^2x$

$\displaystyle = -(sin^2x + cos^2x) + 2cos^2x$

$\displaystyle = -1 + 2cos^2x$ - Jan 28th 2011, 04:44 PMArchie Meade
- Jan 28th 2011, 05:59 PMmr fantastic