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

1 Answer
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))#

Related questions
See all questions in Products, Sums, Linear Combinations, and Applications
Impact of this question
1614 views around the world
You can reuse this answer
Creative Commons License