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

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Topscooter
Jan 1, 2016

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))

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