How do you find the five remaining trigonometric function satisfying #csctheta=-3/2#, #costheta<0#?

Question

How do you find the five remaining trigonometric function satisfying #csctheta=-3/2#, #costheta<0#?

1 Answer

Impact of this question

2754 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T13:17:41

#csc t = 1/(sin t) = - 3/2# --> #sin t = - 2/3#
#sin t = - 2/3#
#cos^2 t = 1 - sin^2 t = 1 - 4/9 = 5/9# --> #cos t = +- sqrt5/3#
#cos t = - sqrt5/3# (cos t < 0)
#tan t = sin/(cos) = (- 2/3)(- 3/sqrt5) = 2/sqrt5 = (2sqrt5)/5#
#cot t = 1/(tan t) = 5/(2sqrt5)#.
#sec t = 1/(cos) = - 3/sqrt5 = - (3sqrt5)/5#.
#csc t = -3/2#

Science

Math

Humanities

... and beyond

How do you find the five remaining trigonometric function satisfying #csctheta=-3/2#, #costheta<0#?

1 Answer

Related questions

Impact of this question