Question

How do I find the value of tan(7pi/12)?

Answer 1

Find `tan ((7pi)/12)`

Ans: - (2 + sqrt3)

Explanation:

Call `tan ((7pi)/12) = t`
`tan 2t = tan ((14pi)/12) = tan ((2pi)/12 + pi) = tan (pi/6 + pi) = tan (pi/6) = 1/sqrt3`
Apply the trig identity: `tan 2t = (2tan t)/(1 - tan^2 t)`

`tan 2t = 1/sqrt3 = 2t/(1 - tan^2 t)`
`1 - t^2 = 2sqrt3t`-->
`t^2 + 2sqrt3t - 1 = 0`
`D = d^2 = b^2 - 4ac = 12 + 4 = 16 -`-> `d = +- 4`
`t = -2sqrt3/2 +- 4/2` -->
`t = tan ((7pi)/12) = - sqrt3 +- 2`

Since the arc (7pi)/12 is in Quadrant II, only the negative answer is accepted. `tan((7pi)/12) = - 2 - sqrt3 = -3.732`

Check by calculator
tan ((7pi)/12) = tan 105 = -3.732. OK