# Thread: Trigo Identities - Proving

1. ## Trigo Identities - Proving

Can anyone help me with proving this:
$sin^2 2x - sin^2 x = sin x sin 3x$

2. $\sin^2 2x - \sin^2 x =\sin x \sin 3x$

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

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

$4\sin^2 x \cos^2 x - \sin^2 x =\sin x (\cos 2x \sin x + 2\cos^2 x \sin x)$

$\sin^2 x (4\cos^2 x - 1)=\sin^2 x (\cos 2x + 2\cos^2 x)$

$4\cos^2 x - \cos^2 x - \sin^2 x = \cos^2 x - \sin^2 x + 2\cos^2 x$

$x=x$

3. i'm confused on how you got the last three steps:

4. Line 5 is got by factoring sin^2 x from line 4. Then divide both sides by sin^2 x to get line 6. Two sides are equal in line 6, so x=x and this statement is true for all x values (except x=0 because we divided both sides by sin^2 x, so you should check this for x=0 manually)

5. thanks for the clarification. i see now, good job, that was some nice mathematical manipulation.

6. ## Help

Firstly, thanks for your quick response. However, we are taught not to do proving like this... We have to do it something like this:

LHS: xxxxxxxx
= xxxxxxxxxx
= xxxxxxxxxx
= xxxxxxxxxx
= xxxxxxxxxx = RHS

Would you mind doing it in this format so that i can understand better? Thanks!

7. Do you know what LHS and RHS mean?

LHS: Left Hand Side

RHS: Right Hand Side

So put all the the steps to the left of the equals sign.

LHS: $(\sin 2x)^2 - \sin^2 x$

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

$= 4\sin^2 x \cos^2 x - \sin^2 x$

$= \sin^2 x (4\cos^2 x - 1)$

$= 4\cos^2 x - \cos^2 x - \sin^2 x$

RHS: $\sin x \sin(x+2x)$

$= \sin x (\cos 2x \sin x + \cos x \sin 2x)$

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

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

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