can any one explain how to solve this cos(t) = −1/5 and π<t<(3π/2), what is csc(t)?
so far I've gotten cos(t) = +/- ((2 square root of 6) / 5)
but i get lost in the rest
and i know "t" is in the 3rd quadrant, right?
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cos(t) = −1/5
0 < t < (3π/2)
cosec t =?
cos (t) = -1/5
sin (t) = ±√(24)/5
1/sin (t) = cosec t
= ± 5/ √(24)
AC² = AB² + BC²
(5)² = 1 + BC²
24 = BC²
±√(24) = BC
csc(t) = 1/(sin(t)
Given cos(t) = -1/5
sin²(t) = 1 - cos²(t)
sin²(t) = 1 - (-1/5)²
sin²(t) = 1 - 1/25
sin²(t) = 24/25
sin(t) = ±2√6 / 5
csc(t) = ±5/(2√6)
you are right t is in 3rd quadrant. so csc will be negative
csc)t) = - 5/(2√6)