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View Full Version : Trig help!!! sin3xsinx=1/2(cos2x-cos4x)

clementine
06-22-2006, 12:57 AM
sin3xsinx=1/2(cos2x-cos4x)

tkhunny
06-22-2006, 01:00 AM
Unfortunately, you have failed to provide a problem statement. Are we finding solutions? Proving identities? Fishing?

Perhaps you could notice this:

cos(4x) = cos(3x+x) = cos(3x)cos(x) - sin(3x)sin(x)

sin(3x)sin(x) = cos(3x)cos(x) - cos(4x)

Are we getting anywhere?

clementine
06-22-2006, 01:12 AM
Hmm... We're supposed to find out both of them. Is there any way you could help me? (:

tkhunny
06-22-2006, 01:19 AM
If you are trying to prove an identity, I gave you a big hint.

Since it is an identity, the same process should produce ALL the solutions.

You still haven't provided an actual problem statement. Where did you get the problem? What were you asked to do? Use the words in the book.

clementine
06-22-2006, 01:22 AM
well, all my friend said was: "here it is. can you help me?"

I thought maybe multiply by 2 on both sides or somethin'...but that didn't look too right.

Seriously, I have no clue what the heck I'm doin'... Heh...

maybe it's something like

sin2x/4 - sin4x/8...??

soroban
06-22-2006, 01:23 AM
Hello, clementine!

I assume that we're supposed to prove this identity.
[Especially since it's a special case of a Product-to-sum Indentity.]

\sin(3x)\sin(x)\:=\:\frac{1}{2}\left[\cos(2x)\,-\,\cos(4x)\right]
On the right side: \,\cos(2x)\:=\:\cos(3x\,-\,x)\:=\:\cos(3x)\cos(x)\,+\,\sin(3x)\sin(x)

\;\;\;And we have: \,\cos(4x)\:=\:\cos(3x\,+\,x)\:=\:\cos(3x)\cos(x)\ ,-\,\sin(3x)\sin(x)

Subtract: \,\cos(2x)\,-\,\cos(4x)\:=\:2\sin(3x)\sin(x)

Divide by 2: \,\frac{1}{2}[\cos(2x)\,-\,\cos(4x)] \;= \;\sin(3x)\sin(x)

clementine
06-22-2006, 01:37 AM
Hmm... Thank you, Soroban! I think you're right. But if it weren't proving the identity but making the left and right alike...what would it be? Is that even possible with this equation?