Question

If Cos(t)=-7/8 with `pi<=t<=3pi/2`, how do you find the value of sin(t/2)?

Answer 1

`sqrt15/4`

Explanation:

Apply the trig identity: `cos 2t = 1 - 2sin^2 t` -->
`cos t = -7/8 = 1 - 2sin^2 (t/2)`
`2sin^2 (t/2) = 1 + 7/8 = 15/8`
`sin^2 (t/2) = 15/16`
`sin (t/2) = +- sqrt15/4`
t is in Quadrant III, then, t/2 is in Quadrant II. Its sin is positive.
Therefor, `sin (t/2) = sqrt15/4`
Check by calculator
`cos t = -7/8 = --0.875` --> `t = 151^@04`
`t/2 = 75^@52` --> `sin (t/2) = sin 75.52 = 0.968 = sqrt15/4`. OK