- #1

rogeralms

- 19

- 0

## Homework Statement

Using the results of Problems 4.70, that is EQs. (4.98) and (4.99), show that

R

_{parallel}+ T

_{paralllel}= 1

## Homework Equations

R

_{parallel}= ( tan^2 ( theta

_{i}- theta

_{t}) ) / (tan^2 (theta

_{i}+ theta

_{t}) )

T

_{parallel}= (sin (2*theta

_{i}) * sin (2*theta

_{t}))/ sin^2 (theta

_{i}+ theta

_{t})

## The Attempt at a Solution

After getting this far (shown below) I took it to the math help center at my university and they couldn't solve it any further than what I had done:

First put both in the same denominator

sin^2 (theta

_{i}- theta

_{t)}) / cos^2(theta

_{i}- theta

_{t}) * cos^2(theta

_{i}+ theta

_{t}/sin^2(theta

_{i}+ theta

_{t}which gives a common denominator of cos^2(theta

_{i}-theta

_{t})* sin^2(theta

_{i}+ theta

_{t})

For brevity I will call theta

_{i}= i and theta

_{t}= t

Now we have sin^2(i-t)*cos^2(i+t) + sin (2*i)*sin(2*t)/ cos^2(i-t)*sin^2(i+t)

I tried (1 - cos^2(i-t)*(1-sin^2(i+t) + sin(2*i)*sin(2*t)/ cos^2(i-t)*sin^2(i+t)

which puts the minus on cos and plus angle on sin which matches the denominator but that is as far as I got which was further than the help desk at my university.

Can someone give me a hint as to which identities I should use to work this out?

You have my undying gratitude and about a million photons of positive energy sent to you for your help!