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chris1080p · Jul 7, 2008, 05:00 PM

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

I need help on where to start.

galactus · Jul 8, 2008, 06:49 AM

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

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.

chris1080p · Jul 8, 2008, 09:56 AM

Thanks for your help.

galactus · Jul 8, 2008, 11:54 AM

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.:)

chris1080p · Jul 8, 2008, 12:06 PM

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?

galactus · Jul 8, 2008, 12:36 PM

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... :)

chris1080p · Jul 8, 2008, 12:44 PM

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

galactus · Jul 8, 2008, 12:46 PM

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.