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

Question

20834 views around the world

You can reuse this answer
Creative Commons License

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)

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

$cot x = -sqr5/2$

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

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

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