1. ## prove this identity (with add and subtract formulas)

sin (x+y) sin (x-y) = sin^2x - sin^2y

i tried doing it but i ended up expanding which turned out ugly..and the result was sin^2x cos^2x - cos^2x sin^2x which did not equal the right side.... the add and subtract formulas are the following:

sin (x+y) = sinx cosy + cosx siny
sin (x-y) = sinx cosy - cosx siny

2. $\displaystyle (\sin(x) \cos(y) + \cos(x)\sin(y))(\sin(x) \cos(y) - \cos(x)\sin(y))$

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

Now use $\displaystyle \cos^2(y) = 1 - \sin^2(y)$ and $\displaystyle \cos^2(x) = 1 - \sin^2(x)$...

3. Originally Posted by magmagod
sin (x+y) sin (x-y) = sin^2x - sin^2y

i tried doing it but i ended up expanding which turned out ugly..and the result was sin^2x cos^2x - cos^2x sin^2x which did not equal the right side.... the add and subtract formulas are the following:

sin (x+y) = sinx cosy + cosx siny
sin (x-y) = sinx cosy - cosx siny

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

$\displaystyle L.H.S. = sin (x+y) \times sin (x-y) = (sinx.cosy + cosx. siny) \times (sinx. cosy - cosx. siny)$

$\displaystyle = (sin^{2}x.cos^{2}y) - (sinx. cosy) . (cosx .siny) + (cosx. siny).(sinx. cosy) - (cos^{2}x.sin^{2}y)$

$\displaystyle = (sin^{2}x.cos^{2}y) - (cos^{2}x.sin^{2}y)$

$\displaystyle = [sin^{2}x.(1-sin^{2}y)] - [(1-sin^{2}x).sin^{2}y]$

$\displaystyle = sin^{2}x - (sin^{2}x.sin^{2}y) -sin^{2}y + (sin^{2}y.sin^{2}x)$

$\displaystyle = sin^{2}x -sin^{2}y$

= R.H.S