# using the difference formula

• Nov 6th 2009, 10:28 PM
maryanna91
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).
• Nov 6th 2009, 10:33 PM
Gusbob
Are you sure you can use only the difference formula? Most people derive it with the sum formula using cos(2x) = cos(x+x)
• Nov 6th 2009, 10:36 PM
maryanna91
yep we're supposed to use the difference formula.
• Nov 6th 2009, 10:49 PM
Gusbob
In that case you will need these identities:

$\sin(3x) = 3\sin(x) - 4\sin^3(x)$
$\cos(3x) = 4\cos^3(x) - 3\cos(x)$
• Nov 6th 2009, 11:05 PM
maryanna91
Hmmm...tried plugging them in and I'm still still stuck (Headbang). Would you mind showing me what you'd do after they've been plugged in?
• Nov 6th 2009, 11:56 PM
Gusbob
$\cos(3x)\cos(x) + \sin(3x)\sin(x)$ = $[4\cos^3(x) - 3\cos(x)] \cos(x) + [3\sin(x) - 4\sin^3(x)] \sin(x)$

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

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?
• Nov 7th 2009, 12:16 AM
maryanna91
yep got it!!! thanks sooo much for all the help!