# Thread: using the difference formula

1. ## using the difference formula

So Im supposed to use the difference formula to figure out cos(2x). This is what I have so far:

cos(2x)=cos(3x-x)
cos(3x-x)=(cos3x)(cosx)+(sin3x)(sinx)

Im just not sure how to finish solving! (my math homework is done online so it's supposed to be as simplified as possible).

2. Are you sure you can use only the difference formula? Most people derive it with the sum formula using cos(2x) = cos(x+x)

3. yep we're supposed to use the difference formula.

4. In that case you will need these identities:

$\sin(3x) = 3\sin(x) - 4\sin^3(x)$
$\cos(3x) = 4\cos^3(x) - 3\cos(x)$

5. Hmmm...tried plugging them in and I'm still still stuck . Would you mind showing me what you'd do after they've been plugged in?

6. $\cos(3x)\cos(x) + \sin(3x)\sin(x)$ = $[4\cos^3(x) - 3\cos(x)] \cos(x) + [3\sin(x) - 4\sin^3(x)] \sin(x)$

= $4\cos^4(x) - 3\cos^2(x) + 3\sin^2(x) - 4\sin^4(x)$
= $4[\cos^4(x) - \sin^4(x)] - 3[ \cos^2(x) - \sin^2(x)]$

Now notice that $\cos^4(x) - \sin^4(x)$ can be written as a difference of two squares to $(\cos^2(x) + \sin^2(x) ) (\cos^2(x) - \sin^2(x))$.

Do you see where to go from here?

7. yep got it!!! thanks sooo much for all the help!