How do you find the values of the six trigonometric functions given #sectheta=-2# and #sintheta>0#?

Question

How do you find the values of the six trigonometric functions given #sectheta=-2# and #sintheta>0#?

1 Answer

Impact of this question

6765 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-06T16:15:02

#sec t = 1/(cos t) = - 2# --> #cos t = - 1/2#
Find sin t.
#sin^2 t = 1 - cos^2 t = 1 - 1/4 = 3/4# --> #sin t = +- sqrt3/2#
Since sin t > 0, t is in Quadrant II, there for:
#sin t = sqrt3/2#.
#tan t = sin/(cos) = (sqrt3/2)(- 2/1) = - sqrt3#
#cot t = 1/(tan) = - 1/sqrt3 = - sqrt3/3#
#csc t = 1/(sin) = 2/sqrt3 = (2sqrt3)/3#

Science

Math

Humanities

... and beyond

How do you find the values of the six trigonometric functions given #sectheta=-2# and #sintheta>0#?

1 Answer

Related questions

Impact of this question