How do you determine all trigonometric ratios given sinθ=2/3 ?

1 Answer
Nghi N.
Nov 2, 2015

Find the six trig functions ratios, given #sin x = 2/3#

Explanation:

#sin x = 2/3# --> #x1 = 41^@81# and #x2 = 180 - 41.81 = 138^@19#
#cos^2 x = 1 - sin^2 x = 1 - 4/9 = 5/9# --> #cos x = +- sqrt5/3#
#tan x1 = tan 41.81 = (2/3)(3/sqrt5) = 2/sqrt5 = (2sqrt5)/5#
#tan x2 = tan 138.19 = - (2sqrt5)/5#
#cot 41.81 = sqrt5/2#
#cot 138.19 = - sqrt5/2#
#sec 41.81 = 3/sqrt5 = (3sqrt5)/5#
#sec 138.19 = - (3sqrt5)/5#
#csc 41.81 = csc 138.19 = 3/2#

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