Question

How do you find the exact value of `tan 150` using the half angle identity?

Answer 1

`tan 150 = - sqrt3/3`

Explanation:

Call tan 150 = tan t -->
`tan 2t = tan 300 = tan (-60 + 360) = tan (-60) = - sqrt3`
Use trig identity:
`tan 2t = (2tant)/(1 - tan^2 t)`
In this case:
`(2tan t)/(1 - tan^2 t) = - sqrt3`.
Cross multiply -->
`- sqrt3 + sqrt3tan^2 t = 2tan t`
`sqrt3tan^2 t - 2tan t - sqrt3 = 0`.
Solve this quadratic equation for tan t.
`D = d^2 = b^2 - 4ac = 4 + 12 = 16` --> `d = +- 4`
There are 2 real roots:
`tan t = -b/(2a) +- d/(2a) = 2/2sqrt3 +- 4/2sqrt3 = sqrt3/3 +- 2sqrt3/3`
a. `tan t = tan 150 = (3sqrt3)/3 = sqrt3` (rejected because tan 150 is negative)
b. `tan t = tan 150 = - sqrt3/3 = - 0.577`
Check by calculator.
`tan 150 = - 0.577 = - sqrt3/3`. Proved.