If sin theta=3/5 and theta has its terminal side in quadrant II, how do you find the exact value of tan 2 theta?

1 Answer
Nghi N.
Apr 17, 2016

#- 24/7 #

Explanation:

#sin t = 3/5#. Find cos t.
#cos^2 t = 1 - sin^2 t = 1 - 9/25 = 16/25 #
#cos t = +- 4/5.#
Since t is located in Quadrant II, cos t is negative.
#cos t = - 4/5# --> #tan t = (sin t)/(cos t) = (3/5)(-4/5) = -3/4#
#tan 2t = (2tan t)/(1 - tan^2 t) = (-3/2)/(1 - 9/16) =#
#= -(3/2)/(7/16) = - (3/2)(16/7) = - 24/7#
Check by calculator
#tan t = -3/4# -> #t = - 36^@87# --> #tan 2t = tan (-73.74) = -3.43#
tan 2t = - 24/7 = 3.43. OK

Related questions
See all questions in Relating Trigonometric Functions
Impact of this question
16845 views around the world
You can reuse this answer
Creative Commons License