Thread: Triple angle trig function

Solve:
sin3xcosx = cos3xsinx


Here's what I've done so far:

sin3xcosx = cos3xsinx
sin3xcosx - cos3xsinx = 0
sin(2x+x)cosx - cos(2x+x)sinx = 0
(sin2xcosx + cos2xsinx)cosx - (cos2xcosx - sin2xsinx)sinx = 0

Here's where I'm stuck. Should I distribute the cosx and sinx on either side? Or plug in the sin2x/cos2x double angle identities?
Any help is appreciated!

Originally Posted by live_laugh_luv27

Solve:
sin3xcosx = cos3xsinx

I would write this as and use the formula .

So, what should I do after I simplify the equation to sin(3x - x) = 0?

Originally Posted by live_laugh_luv27

So, what should I do after I simplify the equation to sin(3x - x) = 0?

3x – x = 2x, so your equation is sin(2x) = 0.