Verifying Rparallel + Tparalllel = 1

[rogeralms]

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!

 

[scurty]

Here's two identities that might help:

$\sin^2(x-y) = \sin^2(x+y) - \sin(2x)\sin(2y)$
$\cos^2(x-y) = \cos^2(x+y) + \sin(2x)\sin(2y)$