Question

How do you use the half angle identity to find exact value of tan 15°?

Answer 1

Find `tan 15`

Ans: `2 - sqrt3`

Explanation:

Call tan 15 = tan t
`tan 2t = tan 30 = 1/sqrt3`
Apply the trig identity: `tan 2t = (2tan t)/(1 - tan^2 t)`
`tan 2t = 1/sqrt3 = (2tan t)/(1 - tan^2 t)`
`1 - tan^2 t = 2sqrt3.t`. Change side of the equation.
`tan^2 t + 2sqrt3.t - 1 = 0`
`D = d^2 = b^2 - 4ac = 12 + 4 = 16`--> `d = +- 4`
`tan t = - (2sqrt3)/2 +- 4/2 = -sqrt3 +- 2`
Since the arc 15 is located in Quadrant I, its tan is positive, therefor,
`tan 15 = 2 - sqrt3`

Check by calculator.
`2 - sqrt3` = 2 - 1.73 = 0.27. Calculator --> tan 15 = 0.27. OK