# Thread: Trigo Identities - Proving

1. ## Trigo Identities - Proving

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

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

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

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

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

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

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

$\displaystyle x=x$

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

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

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

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

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

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

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

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

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

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