Question

If `sintheta=-(2sqrt11)/11` and `pi<theta<(3pi)/2`, how do you find `cos (theta/2)`?

Answer 1

`cos (t/2) = - 0.95`

Explanation:

`sin t = -(2sqrt11)/11 = - 2/sqrt11`
To find cos t, use trig identity: `cos^2 t = 1 - sin^2 t`
`cos^2 t = 1 - 4/11 = 7/11.`
`cos t = +- sqrt7/sqrt11 = 0.80`
To find `cos (t/2)`, use trig identity:
`2cos^2 a = 1 + cos 2a`
We get:
`2cos^2 (t/2) = 1 + cos a = 1 + 0.8 = 1.8`
`cos^2 (t/2) = (1.8)/2 = 0.9`
`cos (t/2) = +- 0.95`
Since t is in Quadrant III, `(t/2)` is in Quadrant II, and its cos is negative.
`cos (t/2) = - 0.95`