Trig Identities for Refraction

[GrahamCU]

Homework Statement

The problem is essentially (I've rephrased it, but this is what it is asking)
show that [ (2n₁cosa) / (n₂cosa + n₁cosb) ]²

is equal to (sin2a sin2b) / (sin²(a+b)cos²(a-b))

where these are for refraction through materials and satisfy n₁sina=n₂sinb



2. The attempt at a solution
I just took pictures of my work because it would be easier than typing it all out. As far as I can tell I didn't do anything wrong up till then
first:
http://img526.imageshack.us/img526/6325/0221091218.jpg
second:
http://img10.imageshack.us/img10/7136/0221091220.jpg

the lines are numbered so you can refer to them easily if you need to


tl;dr:
show how [2cosasinb]² = sin2a sin2b


Thanks

 

[Mark44]

Just looking at your bottom line, I think you have made a mistake somewhere.
Your last line is to show that 2cosa*sinb = sin2a*sin2b, which is not an identity.

Starting from the right side:
sin2a*sin2b = 2sina*cosa*2sinb*cosb = 4sina*cosa*sinb*cosb, which is different from the left side.

You don't show how you got your first expression to [ (2cosasinb) / (sin2a / 2 + sin2b / 2) ]^2, so it's possible you made a mistake getting to that point.

 

[GrahamCU]

ok thanks, I'll double check my work again, and if it doesn't work out, post my complete work to see

 

[GrahamCU]

Ok, I actually just made a mistake typing the question in here. I fixed it now, and posted pictures of my work so you can see how I got there. The last line (now correct in first post) should have been

show how
[2cosasinb]² = sin2a sin2b

Thanks for your help :)

 

[GrahamCU]

bump for help. is this an identity?

[2cosasinb]² = sin2a sin2b

 

[Mark44]

I think it might be. The equation is true for a = pi/6 and b = pi/3. I don't see how to prove that it's true for arbitrary a and b, but let me look into it...

 

[Mark44]

Not an identity. In my earlier post I neglected to square a number, so it's not true for a = pi/6 and b = pi/3.