Question

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

Answer 1

`(7sqrt15)/32 = 0.85`

Explanation:

sin 2t = 2sin t.cos t
Knowing `cos t = -7/8`, find sin t.
Trig identity: `sin^2 a = 1 - cos^2 a` -->
`sin^2 t = 1 - cos^2 t = 1 - 49/64 = 15/64`
sin t = +- sqrt15/8.
Arc t is in Quadrant III, then, sin t is negative.
`sin 2t = 2(-7/8)(- sqrt15/8) = (7sqrt15)/32= 0.85`