1. 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..

2. Well, you actually did something wrong...

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

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

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

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

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

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

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?

4. I'm sorry, what do you mean by addition formulae?

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

cos(A+b) = cosAcosB-sinAsinB

6. Um... is that not what I told you to use?

7. 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

8. Yes, but how about using it for the time being to see where it leads us?

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

it gives me:
cos2XcosX-sin2XsinX
And, in turn, $\displaystyle \cos(2x) = \cos(x+x)$

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cos3x = 4cos^3 - 3cosx

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