cos 3x = 4 cos^3 x - 3 cos x

I need help on where to start.

cos(3x)=4cos^{3}(x)-3cos(x)

Use the sum formula for cos, cos(3x)=cos(2x)cos(x)-sin(2x)sin(x)

cos(2x)cos(x)-sin(2x)sin(x)=4cos^{3}(x)-3cos(x)

Work on the left and try to turn it into the right.

Use the identities cos(2x)=cos^{2}(x)-sin^{2}(x) and

sin(2x)=2sin(x)cos(x)

Sub them in and it should fall into place.

Thanks for your help.

You're welcome. Let me know if yo got it to work OK.

Oh yeah, if you don't mind, please click on 'rate this answer' for me.:)

So far I did what you said and arrived at:

cos^2 x - sin^2 x cosx - 2sinx cosx sinx

I know cos^2 x +sin^2 x=1, but I don't see how it falls into place.

Is there a way that I can work on the right side to equal the left as well?

Well, let's continue then from where we left off.

I will make the identity subs cos^{2}(x)-sin^{2}(x)=cos(2x)

and

sin(2x)=2sin(x)cos(x)

Then we get:

(cos^{2}(x)-sin^{2}(x))cos(x)-2sin^{2}(x)cos(x)

Expand:

cos^{3}(x)-sin^{2}(x)cos(x)-2sin^{2}(x)cos(x)

=cos^{3}(x)-3sin^{2}(x)cos(x)

Now, make the sub sin^{2}(x)=1-cos^{2}(x)

=cos^{3}(x)-3(1-cos^{2}(x))cos(x)

=cos^{3}(x)-3cos(x)+3cos^{3}(x)

=\fbox{4cos^{3}(x)-3cos(x)}

There it is... :)

Thanks man for all of your help. Lol, I couldn't rate your answer again.

That's OK. I know. I hope this learned you what you got to do to solve them. It is a matter of seeing the right identity substitutions. Mind you, there are certainly other ways to go about it.
Someone else may see a simpler way, but this worked well.