Question

How do you find the value of `tan(theta/2)` given `sintheta=(-3/5)` and `(3pi)/2<theta<2pi`?

Answer 1

`- 1/3`

Explanation:

First find cos t by using trig identity: `cos^2 t = 1 - sin^2 t`
`cos^2 t = 1 - 9/25 = 16/25` --> `cos t = +- 4/5`.
Since t is in Quadrant IV, then cos t is positive --> cos t = 4/5
Next find sin (t/2) and cos (t/2) by using trig identities:
`1 + cos 2t = 2cos^2 t`
`1 - cos 2t = 2sin^2 t`
Find `cos (t/2)`
`2cos^2 (t/2) = 1 + cos t = 1 + 4/5 = 9/5`
`cos^2 (t/2) = 9/10`
`cos (t/2) = 3/sqrt10 = 3sqrt10/10` (because cos (t/2) is positive).
Find sin (t/2)
`2sin^2 (t/2) = 1 - cos t = 1 - 4/5 = 1/5`
`sin^2 (t/2) = 1/10`
`sin (t/2) = +- 1/sqrt10 = +- sqrt10/10`
Since t is in Quadrant IV, sin t is negative
`tan (t/2) = sin/(cos) = (- sqrt10/10)(10/(3sqrt10)) = - 1/3`