1. ## grade 12 trig identities

Hey everyone, just wondering if anyone can figure this identity out.

$\displaystyle \ sin4x - sin2x/sin2x = \cos3x/cosx$

I already worked with the left side and ended up with $\displaystyle 2cos2x-1$ by using the double angle formula -> sin2x=2sinxcosx
Im stuck on what to do next.

any help is appreciated, thanks in advance.

2. Originally Posted by aries
Hey everyone, just wondering if anyone can figure this identity out.

$\displaystyle \ sin4x - sin2x/sin2x = \cos3x/cosx$

I already worked with the left side and ended up with $\displaystyle 2cos2x-1$ by using the double angle formula -> sin2x=2sinxcosx
Im stuck on what to do next.

any help is appreciated, thanks in advance.
Note that:

RHS = $\displaystyle \frac{\cos(x + 2x)}{\cos x} = \frac{\cos x \cos(2x) - \sin x \sin(2x)}{\cos x}$

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

LHS = $\displaystyle 2 \cos(2x) - 1 = \cos(2x) + \cos(2x) - 1$.

Now do you see how to get from LHS to RHS?