# Prove Cos3x=4(cos^3)x-3cosx

• Nov 15th 2010, 10:59 AM
mike789
Prove Cos3x=4(cos^3)x-3cosx

Use the expansion of cos(A+B) to show that
Cos3x=4(cos^3)x-3cosx

This is what i did:
Cos 3x = (cos^2)x-(sin^2)x
= (3cos^2)x-1

I think i am doing something wrong..
• Nov 15th 2010, 11:21 AM
Unknown008
Well, you actually did something wrong...

Use the expansion of cos(A+B) as you were told.

$\cos(A+B) = \cos A\cos B - \sin A\sin B$

Now, you have cos(3x), make it cos(2x + x) and use the expansion. What do you get?
• Nov 15th 2010, 11:34 AM
mike789
Quote:

Originally Posted by Unknown008
Well, you actually did something wrong...

Use the expansion of cos(A+B) as you were told.

$\cos(A+B) = \cos A\cos B - \sin A\sin B$

Now, you have cos(3x), make it cos(2x + x) and use the expansion. What do you get?

Don't you have to use additon formulae?
Can you split cos3x?
• Nov 15th 2010, 11:37 AM
Unknown008
I'm sorry, what do you mean by addition formulae?
• Nov 15th 2010, 11:48 AM
mike789
Quote:

Originally Posted by Unknown008
I'm sorry, what do you mean by addition formulae?

cos(A+b) = cosAcosB-sinAsinB
• Nov 15th 2010, 11:50 AM
Unknown008
Um... is that not what I told you to use?
• Nov 15th 2010, 12:02 PM
mike789
Quote:

Originally Posted by Unknown008
Um... is that not what I told you to use?

But, it doesn’t prove that Cos3x=4(cos^3)x-3cosx

it gives me:
cos2XcosX-sin2XsinX
• Nov 15th 2010, 12:04 PM
Unknown008
Yes, but how about using it for the time being to see where it leads us? (Smile)
• Nov 15th 2010, 12:57 PM
e^(i*pi)
Quote:

Originally Posted by mike789
But, it doesn’t prove that Cos3x=4(cos^3)x-3cosx

it gives me:
cos2XcosX-sin2XsinX

And, in turn, $\cos(2x) = \cos(x+x)$