Question

If `sin theta = 1/4`, theta in quadrant 2, how do you find the exact value of `sin(theta - pi/3)`?

Answer 1

`sin (t - pi/3) = (1 + 3sqrt5)/8`

Explanation:

`sin t = 1/4`. Find cos t
`cos^2 t = 1 - sin^2 t = 1 - 1/16 = 15/16`
`cos t = - sqrt15/4` (because t lies in Quadrant 2)
From trig identity, we get:
`sin (t - pi/3) = sin t.cos (pi/3) - sin (pi/3).cos t` (1)
Note.
`sin t = 1/4` --> `cos t = - sqrt15/4` -->
`cos (pi/3) = 1/2` --> `sin (pi/3) = sqrt3/2`
Equation (1) becomes:
`sin (t - pi/3) = 1/8 + (sqrt3.sqrt15)/8 = (1 + 3sqrt5)/8`
Check by calculator.
`sin t = 0.25` --> `t = 162^@52` --> `(t - 60) = 105^@22`
`sin (t - 60) = 0.96`
`(1 + 3sqrt5)/8 = 0.96`. Proved.