I don't get this simplification

**konvos** · January 28th 2011, 02:01 PM

I can't see how they did this operation.

thx for the help.

**pickslides** · January 28th 2011, 02:09 PM

Looks like simple trig identities to me i.e. $2\sin x \cos x = \sin 2x$

**Archie Meade** · January 28th 2011, 02:09 PM

Originally Posted by konvos

I can't see how they did this operation.

thx for the help.

$sin(2x)=2sinxcosx$

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

and

$cos^2x+sin^2x=1\Rightarrow\ -\left(sin^2x+cos^2x\right)=-1$

**Houdini** · January 28th 2011, 02:11 PM

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)

**konvos** · January 28th 2011, 02:23 PM

thx for the help everybody

Originally Posted by Archie Meade

$sin(2x)=2sinxcosx$

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

and

$cos^2x+sin^2x=1\Rightarrow\ -\left(sin^2x+cos^2x\right)=-1$

I'd no idea of this identity

I suppose I should review trig...

**Defunkt** · January 28th 2011, 02:48 PM

$-sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =$
$= -sin^2x -cos^2x + cos^2x + cos^2x$
$= -(sin^2x + cos^2x) + 2cos^2x$
$= -1 + 2cos^2x$

**Archie Meade** · January 28th 2011, 04:44 PM

Originally Posted by konvos

thx for the help everybody

I'd no idea of this identity

I suppose I should review trig...

No, that's not an identity.
Just algebraic manipulation.
You could of course use identities to weave your way to the final line.

**mr fantastic** · January 28th 2011, 05:59 PM

Originally Posted by Defunkt

$-sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =$
$= -sin^2x -cos^2x + cos^2x + cos^2x$
$= -(sin^2x + cos^2x) + 2cos^2x$
$= -1 + 2cos^2x$

It's also probably worth noting that $\cos^2(x) - \sin^2(x) = \cos(2x)$ is a standard double angle formula (because I just know that the next question asked by the OP will be how to find y ....)

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