How Do Double Angle Identities Simplify Trigonometric Equations?

[Jatt]

Homework Statement

sin4x-sin2x/sin2x=cos3x/cosx

Homework Equations

sin2x = 2sinxcosx

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.

 

[rock.freak667]

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

?

 

[Jatt]

yup.

 

[dlgoff]

I believe there are two more Relevant equations you need.

 

[Jatt]

doesn'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

 

[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

 

[Jatt]

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

 

[rock.freak667]

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

 

[Jatt]

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

 

[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