# Trig Identities for Refraction

• GrahamCU
Sorry for any confusion.In summary, the conversation is discussing a problem involving refraction through materials and the equation [ (2n1cosa) / (n2cosa + n1cosb) ]2. The attempt at a solution involves showing that this equation is equal to (sin2a sin2b) / (sin2(a+b)cos2(a-b)), but the person notices a mistake in the last line and corrects it in a later post. They ask for help in determining if this is an identity, but it is determined that it is not.
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

Last edited by a moderator:
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.

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

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

bump for help. is this an identity?

[2cosasinb]2 = sin2a sin2b

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...

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.

## What are trig identities for refraction?

Trig identities for refraction are mathematical equations that describe the relationship between angles and the refraction of light as it passes through different media.

## Why are trig identities important for understanding refraction?

Trig identities are important because they allow us to calculate and predict the angles at which light will bend when passing through different materials, such as water or glass. This is crucial in fields such as optics and lens design.

## What is Snell's Law and how does it relate to trig identities for refraction?

Snell's Law is a fundamental principle in optics that describes the relationship between the angles of incidence and refraction of light passing through a boundary between two materials. It is closely related to trig identities for refraction, as they both involve the use of trigonometric functions to calculate the angle of refraction.

## How are trig identities for refraction used in real-world applications?

Trig identities for refraction have numerous real-world applications, including in the design of eyeglasses, cameras, and other optical devices. They are also used in meteorology to understand how light is refracted in the atmosphere, and in seismology to study how seismic waves bend as they travel through different layers of the earth.

## Are there any limitations to using trig identities for refraction?

While trig identities are useful for understanding the basics of refraction, they do not account for more complex phenomena, such as the effects of dispersion or multiple reflections. In these cases, more advanced mathematical models and simulations may be necessary.

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