Question

How do you use the half angle formula to determine the exact values of the sine, cosine, and tangent `(19pi)/12`?

Answer 1

`sin ((19pi)/12) = - sqrt(2 + sqrt3)/2`
`cos ((19pi)/12) = sqrt(2 - sqrt3)/2`

Explanation:

First, call `(t/2) = (19pi)/12` -->
`t = (38pi)/12 = (2pi)/12 + 3pi = pi/6 + 3pi`
Next, use the half angle formulas:
`sin (t/2) = +- sqrt((1 - cos t)/2)`
`cos (t/2) = +- sqrt((1 + cos t)/2)`
Note that
`cos t = cos ((pi)/6 + 3pi) = - cos (pi/6) = - sqrt3/2`.
Therefor, since `(t/2)` is in Quadrant 4 -->
`sin (t/2) = - sqrt((1 + sqrt3/2)/2) = - sqrt(2 + sqrt3)/2`
By the same process, with `cos (t/2)` positive -->
`cos (t/2) = + sqrt((1 - sqrt3/2)/2) = sqrt(2 - sqrt3)/2`
From there -->
`tan (t/2) = sin/(cos) = - (2 + sqrt3)/(2 - sqrt3)`