Theorem: n_1 sin(theta_B) - n_2 cos(theta_B) = 0 at tan(theta_B) = n_2/n_1

Theorem (Brewster-zero-residual canonical): at theta_B = arctan(n_2/n_1), sin theta_B = n_2/sqrt(n_1^2 + n_2^2) and cos theta_B = n_1/sqrt(n_1^2 + n_2^2), so n_1 sin theta_B - n_2 cos theta_B = (n_1 n_2 - n_2 n_1)/sqrt(n_1^2 + n_2^2) = 0…