Can anyone help me with proving this:
$\displaystyle sin^2 2x - sin^2 x = sin x sin 3x$
$\displaystyle \sin^2 2x - \sin^2 x =\sin x \sin 3x$
$\displaystyle (\sin 2x)^2 - \sin^2 x =\sin x \sin(x+2x)$
$\displaystyle (2\sin x \cos x)^2 - \sin^2 x =\sin x (\cos 2x \sin x + \cos x \sin 2x)$
$\displaystyle 4\sin^2 x \cos^2 x - \sin^2 x =\sin x (\cos 2x \sin x + 2\cos^2 x \sin x)$
$\displaystyle \sin^2 x (4\cos^2 x - 1)=\sin^2 x (\cos 2x + 2\cos^2 x)$
$\displaystyle 4\cos^2 x - \cos^2 x - \sin^2 x = \cos^2 x - \sin^2 x + 2\cos^2 x$
$\displaystyle x=x$
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)
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!
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: $\displaystyle \sin x \sin(x+2x)$
$\displaystyle = \sin x (\cos 2x \sin x + \cos x \sin 2x)$
$\displaystyle = \sin x (\cos 2x \sin x + 2\cos^2 x \sin x)$
$\displaystyle = \sin^2 x (\cos 2x + 2\cos^2 x)$
$\displaystyle = \cos^2 x - \sin^2 x + 2\cos^2 x = LHS$