# Thread: I don't get this simplification

1. ## I don't get this simplification

I can't see how they did this operation.

thx for the help.

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

3. Originally Posted by konvos
I can't see how they did this operation.

thx for the help.
$\displaystyle sin(2x)=2sinxcosx$

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

and

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

4. 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)

5. thx for the help everybody

$\displaystyle sin(2x)=2sinxcosx$

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

and

$\displaystyle cos^2x+sin^2x=1\Rightarrow\ -\left(sin^2x+cos^2x\right)=-1$
I'd no idea of this identity
I suppose I should review trig...

6. $\displaystyle -sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =$
$\displaystyle = -sin^2x -cos^2x + cos^2x + cos^2x$
$\displaystyle = -(sin^2x + cos^2x) + 2cos^2x$
$\displaystyle = -1 + 2cos^2x$

7. 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.

8. Originally Posted by Defunkt
$\displaystyle -sin^2x + cos^2x = -sin^2x + (cos^2x - cos^2x) + cos^2x =$
$\displaystyle = -sin^2x -cos^2x + cos^2x + cos^2x$
$\displaystyle = -(sin^2x + cos^2x) + 2cos^2x$
$\displaystyle = -1 + 2cos^2x$
It's also probably worth noting that $\displaystyle \cos^2(x) - \sin^2(x) = \cos(2x)$ is a standard double angle formula (because I just know that the next question asked by the OP will be how to find y ....)