Question

If `costheta=-4/5` and `pi<theta<(3pi)/2`, how do you find `sin (theta/2)`?

Answer 1

`(3sqrt10)/10`

Explanation:

Use the trig identity:
`1 - cos 2a = 2sin^2 a`
Replacing a by `sin (t/2)`, we get:
`2sin^2 (t/2) = 1 - cos t = 1 + 4/5 = 9/5`
`sin^2 (t/2) = 9/10`
`sin (t/2) = +- 3/sqrt10 = +- (3sqrt10)/ 10`
Since t is Quadrant III `(pi < t < (3pi)/2)`, then `t/2` is in Quadrant II `(pi/2 < t/2 < pi)`. There for, `sin (t/2)` is positive.
`sin t = (3sqrt10)/10`