Question

How do you use the angle sum or difference identity to find the exact value of `tan((7pi)/12)`?

Answer 1

`- (2 + sqrt3)`

Explanation:

`tan ((7pi)/12) = tan (pi/12 + pi) = - tan (pi/12)`
Call `tan (pi/12) = tan t` -->`tan 2t = tan (pi/6) = 1/sqrt3`
Use trig identity:
`tan 2a = (2tan a)/(1 - tan^2 a)`
`1/sqrt3 = (2tan t)/(1 - tan^2 t)`
Cross multiply -->
`1 - tan^2 t = 2sqrt3tan t`
`- tan^2 t - 2sqrt3tan t + 1 = 0.`
Solve this quadratic equation for tan t by the improved quadratic equation (Socratic Search)
`D = d^2 = b^2 - 4ac = 12 + 4 = 16` --> `d = +- 4`
There are 2 real roots:
`tan t = -b/(2a) +- d/(2a) = (-2sqrt3)/-2 +- d/2 = sqrt3 +- 2`.
Since `pi/12` is in Quadrant I, its tan is positive
`tan t = tan (pi/12) = sqrt3 + 2`
There for:
`tan ((7pi)/12) = - tan t = - (sqrt3 + 2)`