Question

How do you evaluate `sin((3pi)/4) - tan((5pi)/6)`?

Answer 1

`sqrt2/2 - sqrt3/3`

Explanation:

`S = sin ((3pi)/4) - tan ((5pi)/6)`
Trig table -->`sin ((3pi)/4) = sqrt2/2`
Unit circle and trig table -->
#tan ((5pi)/6) = tan ((6pi)/6 - (pi)/6) = tan (pi - pi/6) = tan (-pi/6) = - tan (pi/6) = -sqrt3/3.#
Therefor:
`S = sqrt2/2 - sqrt3/3`