Question

How do you evaluate `tan ((7pi)/12)`?

Answer 1

`- sqrt(2 + sqrt3)/(sqrt2 - sqrt3)`

Explanation:

On the trig unit circle,
`tan ((7pi)/12) = tan (pi/12 + (6pi)/12) = tan (pi/12 + pi/2) = -cot (pi/12)`.
Find `cot (pi/12) = cos (pi/12)/sin (pi/12)`
Apply the trig identity: `cos 2a = 2cos^2 a - 1 = 1 - 2sin^2 a.`
`cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1.`
`cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/4`
`cos (pi/12) = sqrt(2 + sqrt3)/2` --> (since (pi/12) is in Quadrant I)
By the same way:
`sin (pi/12) = sqrt(2 - sqrt3)/2`
Finally:
`tan ((7pi)/12) = - cos/(sin) = - sqrt(2 + sqrt3)/(sqrt(2 - sqrt3))`