Question

How do you evaluate `tan (pi/12)` using the half angle formula?

Answer 1

`2 - sqrt3`

Explanation:

Use the trig identity:
`tan 2a = (2tan a)/(1 - tan^2 a)`
Trig table -->
`tan 2a = tan (pi/6) = 1/sqrt3`
Call tan (pi/12) = t, we get:
`1/sqrt3 = (2t)/(1 - t^2).` Cross multiply -->
`(1 - t^2) = 2sqrt3 t`
`t^2 + 2sqr3t - 1 = 0`
Solve this quadratic equation for t.
`D = d^2 - 4ac = 4(3) + 4 = 16` --> `d = +- 4`
There are 2 real roots:
`t = tan (pi/12) = -b/(2a) +- d/(2a) = -(2sqrt3)/2 +- 4/2 = - sqrt3 +- 2`.
Since `tan (pi/12)` is positive, therefor,
`tan (pi/12) = 2 - sqrt3`
Check by calculator.
`tan (pi/12) = tan 15^@ = 0.27`
`2 - sqrt3 = 2 - 1.73 = 0.27`. OK