Question

How do you find the value of `tan((11pi)/12)`?

Answer 1

`-2 + sqrt3`

Explanation:

Trig unit circle -->
`tan ((11pi)/12 = - tan (pi.12)`
Find `tan (pi/12).`
Use the trig identity, and call tan (pi/12) = t
`tan 2a = (2tan a)/(1 - tan^2 a)`
`tan (pi/6) = 1/sqrt3 = (2t)/(1 - t^2)`
Cross multiply
`1 - t^2 = 2sqrt3t`
`t^2 + 2sqrt3t - 1 = 0`
Solve this quadratic equation for t by using improved quadratic formula (Socratic Search)
`D = d^2 = b^2 - 4ac = 12 + 4 = 15` --> `d = +- 4`
There are 2 real roots:
`t = -b/(2a) +- d/(2a) = - sqrt3 +- 2`
t1 = - sqrt3 + 2
t2 = - sqrt3 - 2 (rejected because tan (pi/12) is positive)
Finally:
tan ((11pi)/12) = - tan (pi/12) = - t1 = - 2 + sqrt3
Check by calculator.
tan ((11pi)/12) = - tan (pi/12) = - tan 15 = - 0.267. OK