Verify (cos(3x)-cos(x))/(sin(3x)-sin(x)) = -tan(2x)

[helpwithtrig]

I'm supposed to verify
(cos(3x)-cos(x))/(sin(3x)-sin(x)) = -tan(2x)
I get so far and can't get past it.
Working just the left side:

(cos(2x+x)-cosx) / (sin(2x+x)-sinx)

(cos2xcosx-sin2xsinx-cosx) / (sin2xcosx+cos2xsinx-sinx)

((cos^2x-sin^2x)cosx-(2sinxcosx)sinx-cosx) / ((2sinxcosx)cosx + (cos^2x - sin^2x)sinx - sinx)
factoring out (cosx/sinx)
cotx ( cos^2x - sin^2x - 2sin^2x - 1) / (2cos^2x + cos^2x - sin^2x -1)
I group stuff:
cotx ( (cos^2x - sin^2x) - ( 1 - 2sin^2x )) / ( (cos^2x - sin^2x) + (2cos^2x - 1 ) )
and that leads to:
cotx ( cos(2x) - cos(2x) ) / (cos(2x) + cos(2x))
which is 0.

I have tried other substitutions.
Do you think the problem is wrong?

 

[stapel]

I'm supposed to verify (cos(3x)-cos(x))/(sin(3x)-sin(x)) = -tan(2x)
I get so far and can't get past it. Working just the left side:

(cos(2x+x)-cosx) / (sin(2x+x)-sinx)

I think you were on the right track here; you just didn't go quite far enough.

Try it like this:

. . . . .\(\displaystyle \frac{\cos{(2x\, +\, x)}\, +\, \cos{(2x\, -\, x)}}{\sin{(2x\, +\, x)}\, +\, \sin{(2x\, -\, x)}}\)

Now apply those identities you were using, simplify, and see what you get! :wink:

Eliz.

 

[helpwithtrig]

Thanks, that makes it soooo easy!!!!