Question

What is `tan^2theta` in terms of non-exponential trigonometric functions?

Answer 1

`tan^2(theta) = (1-cos(2theta))/(1+cos(2theta))`

Explanation:

You first need to remember that `cos(2theta) = 2cos^2(theta) - 1 = 1-2sin^2(theta)`. Those equalities give you a "linear" formula for `cos^2(theta)` and `sin^2(theta)`.

We now know that `cos^2(theta) = (1+cos(2theta))/2` and `sin^2(theta) = (1-cos(2theta))/2` because `cos(2theta) = 2cos^2(theta) - 1 iff 2cos^2(theta) = 1 + cos(2theta) iff cos^2(theta) = (1+cos(2theta))/2`. Same for `sin^2(theta)`.

`tan^2(theta) = sin^2(theta)/cos^2(theta) = (1-cos(2theta))/2 * 2/(1+cos(2theta)) = (1-cos(2theta))/(1+cos(2theta))`