Question

How do you find the exact value of `cos ( (2pi/3)-(pi/6))`?

Answer 1

Zero

Explanation:

Apply the trig identity: cos (a - b) = cos a.cos b + sin a.sin b
`cos ((2pi)/3 - pi/6) = cos ((2pi)/3).cos (pi/6) + sin ((2pi)/3).sin (pi/6)`
Trig table gives -->
`cos ((2pi)/3) = -1/2 ; cos (pi/6) = sqrt3/2`
`sin ((2pi)/3) = sqrt3/2 ; sin (pi/6) = 1/2.`
Therefor:
`cos ((2pi)/3 - pi/6) = (-1/2)(sqrt3/2) + (sqrt3/2)(1/2) = 0`