Given #sin theta = 2/3# and #pi/2<theta<pi#, how do you find the value of the 5 other trigonometric functions?

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Answer 1 · 2015-05-06T00:48:01

Since sin x > 0 and #pi/2 < x < pi#, then x is in Quadrant II.

sin x = 2/3 -> sin^2 x = 4/9

#cos^2 x = 1 - 4/9 = 5/9 -> cos x = -(sqr5)/3 #(Quadrant II)

#tan x = (2/3):((-sqr5)/3) = -2/(sqr5)# = #-((2sqr5)/5)#

#cot x = -sqr5/2#

#sec = 1/cos x = -3/(sqr5)#

#csc = 1/sin x = 3/2#