using the difference formula

**maryanna91** · November 6th 2009, 09:28 PM

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).

**Gusbob** · November 6th 2009, 09:33 PM

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

**maryanna91** · November 6th 2009, 09:36 PM

yep we're supposed to use the difference formula.

**Gusbob** · November 6th 2009, 09:49 PM

In that case you will need these identities:

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

**maryanna91** · November 6th 2009, 10:05 PM

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?

**Gusbob** · November 6th 2009, 10:56 PM

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

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

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?

**maryanna91** · November 6th 2009, 11:16 PM

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

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