Question

How do you find the remaining trigonometric functions of `theta` given `tantheta=-1/2` and `sintheta>0`?

Answer 1

`tan t = - 1/2 --> cot t = - 2`
`sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 4) = 1/5`
`sin t = +- 1/sqrt5 = = +- sqrt5/5`
Take the positive value, because sin t > 0
`cos^2 t = 1 - sin^2 t = 1 - 1/5 = 4/5`
`cos t = +- 2/sqrt5`
Because tan t < 0, and sin t > o, then cos t < 0, take the negative value.
`sec t = 1/(cos) = - sqrt5/2`
`csc t = 1/(sin) = sqrt5`