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#1 2008-03-16 18:36:57
- Jack M
- Guest
Trig Proof - I am having trouble taking this further
cosx -cos3x/sin3x-sinx=tan2x
I have taken it in several directions. so far I have:
cosxcos3x+sinxsin3x/sin3xcosx-cos3xsinx=tan2x
OR
cosx-cos2xcosx-sin2xsinx/sin2xcosx=tan2x
It's driving me a bit 
Any help would be greatly appreciated
#2 2008-03-16 21:32:47
- JaneFairfax
- Member
- Registered: 2007-02-23
- Posts: 6,868
Re: Trig Proof - I am having trouble taking this further
You can also use the sum formulae:
where A = x and B = 3x.
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#3 2008-03-17 02:54:10
- Kurre
- Member
- Registered: 2006-07-18
- Posts: 280
Re: Trig Proof - I am having trouble taking this further
I just plugged that into a calculator and it didnt match.. 
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#4 2008-03-17 03:46:35
- Daniel123
- Member
- Registered: 2007-05-23
- Posts: 663
Re: Trig Proof - I am having trouble taking this further
Yes it does.
The identity Jack is trying to prove is:
Last edited by Daniel123 (2008-03-17 08:01:52)
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#5 2008-03-17 06:51:57
- Kurre
- Member
- Registered: 2006-07-18
- Posts: 280
Re: Trig Proof - I am having trouble taking this further
aha, I thought he meant:
Hint: Use addition formula (sin(x+y)=sinxcosy+cosxsiny) and then sin2x=2sinxcosx (for cos, cos(x+y)=cosxcosy-sinxsiny, cos2x=cos^2x-sin^2x). And I think you have some errors in the denominator in the second approach.
Last edited by Kurre (2008-03-17 07:05:32)
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#6 2008-03-17 20:45:50
- Jack M
- Guest
Re: Trig Proof - I am having trouble taking this further
Proven!
Here's how:
(cosx-cos3x)/(sin3x-sinx)=tan2x
LHS=(-2sin(x+3x/2)sin(x-3x/2))/(2cos(3x+x/2)cos(3x-x/2))
-2sin2xsin-2x)/(2cos2xsinx)
-2sin2x-sinx)/(2cos2xsinx)
=(2sin2x)/(2cos2x)
=sin2x/cos2x
=RHS YAY!
Thank you all 
