Question

How do you evaluate `sin((23pi)/6) and cos((17pi)/6)`?

Answer 1

`(-1/2) and (-sqrt3/2)`

Explanation:

Trig unit circle and trig table -->
`sin ((23pi)/6) = sin (-pi/6 + (24pi)/6) = sin (-pi/6 + 4pi) =`
`= sin (-pi/6) = - sin (pi/6) = - 1/2`
Unit circle and property of supplemental arcs give -->
`cos ((17pi)/6) = cos (-pi/6 + (18pi)/6) = cos (-pi/6 + 3pi) =`
`= cos (-pi/6 + pi) = - cos (pi/6) = - sqrt3/2`