grade 12 trig identities

**aries** · January 12th 2008, 07:28 PM

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

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

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

any help is appreciated, thanks in advance.

**mr fantastic** · January 12th 2008, 07:59 PM

Originally Posted by aries

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

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

I already worked with the left side and ended up with $2cos2x-1$ 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 = $\frac{\cos(x + 2x)}{\cos x} = \frac{\cos x \cos(2x) - \sin x \sin(2x)}{\cos x}$

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

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

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

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