# Confused about double angle identities

1. Dec 4, 2007

### Jatt

1. The problem statement, all variables and given/known data
sin4x-sin2x/sin2x=cos3x/cosx

2. Relevant equations
sin2x = 2sinxcosx

3. The attempt at a solution

LS = sin(2x + 2x) - sin2x/sin2x
= sin2xcos2x + cos2xsin2x - sin2x/sin2x
= 2sin2xcos2x - sin2x/sin2x
This is where i get stuck...
I don't know what happens if you try to: 2sinx2x - sin2x or 2sin2x/sin2x, is that possible or not? Can you guys help me solve this question? Thanks.

2. Dec 4, 2007

### rock.freak667

Quick question you are to prove that
$$\frac{sin4x-sin2x}{sin2x}=\frac{cos3x}{cosx}$$

?

3. Dec 4, 2007

yup.

4. Dec 4, 2007

### dlgoff

I believe there are two more Relevant equations you need.

5. Dec 4, 2007

### Jatt

dosen't matter i was able to solve it, but here's another problem which I'm now stuck with. $$cosx+cos2x+cos3x=cos2x(1+2cosx)$$.
I've tried many things with this problem, but always seem to get lost.
The only given identites which I'm given to use:
$$cos2x=cos^2x-sin^2x$$
$$cos2x=2cos^2x-1$$
$$cos2x=1-2sin^2x$$
$$sin2x=2sinxcosx$$

Last edited: Dec 4, 2007
6. Dec 4, 2007

### rock.freak667

OK, since you have to SOLVE and not prove....
if you expand the RHS you would see that the cos2x cancels out and you are left with

cos(x)+cos(3x)=2cos$^2$(x)

then expand out cos(3x) and see if anything gets simpler

7. Dec 4, 2007

### Jatt

lol, sorry for not stating this, but i have to prove not solve.

8. Dec 4, 2007

### rock.freak667

well then expand out the LHS
Recall that cos(3x)=cos(2x+x)

9. Dec 4, 2007

### Jatt

yea then i get: cos2xcosx + sin2xsinx + cos2x + cosx

10. Dec 4, 2007

### rock.freak667

cos2x=2cos^2(x)-1
and sin2x=2sinxcosx
cos^2(x)+sin^2(x)=1

expand out and find all in terms of cosx and hopefully it will work