clementine

06-22-2006, 12:57 AM

sin3xsinx=1/2(cos2x-cos4x)

someone please help!! PLEASE!!!

someone please help!! PLEASE!!!

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

clementine

06-22-2006, 12:57 AM

sin3xsinx=1/2(cos2x-cos4x)

someone please help!! PLEASE!!!

someone please help!! PLEASE!!!

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)

This leads to

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

Are we getting anywhere?

Perhaps you could notice this:

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

This leads to

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.

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...??

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)

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?

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