Question

How do you find the exact values of `sin(u/2), cos(u/2), tan(u/2)` using the half angle formulas given `cscu=-5/3, pi<u<(3pi)/2`?

Answer 1

`sin (u/2) = 3sqrt10/10`
`cos (u/2) = - sqrt10/10`
`tan (u/2) = - 3`

Explanation:

`1/(sin u) = - 5/3` --> `sin u = - 3/5`
`cos u = 1 - sin^2 u = 1 - 9/25 = 16/25` --> `cos u = +- 4/5`
`cos u = - 4/5` because u is in Quadrant 3.
Use trig identity:
`2sin^2 a = 1 - sin 2a`.
In this case:
`2sin^2 (u/2) = 1 - cos u = 1 + 4/5 = 9/5`
`sin^2 (u/2) = 9/10`
`sin (u/2) = +- 3/sqrt10`
`sin (u/2) = 3/sqrt10` --> because `u/2` is in Quadrant 2
The same way -->
2cos^2 (u/2) = 1 + cos u = 1 - 4/5 = 1/5
`cos (u/2) = +- 1/sqrt10`
`cos (u/2) = - 1/sqrt10` --> because `u/2` is in Quadrant 2
`tan (u/2) = sin/(cos) = 3/-1 = - 3`