Question

How do you find the exact values of tan(5pi/12) & sin(5pi/12) using the half angle formula?

Answer 1

Find `tan ((5pi)/12)` and sin ((5pi)/12)
Answer: `+- (2 +- sqrt3) and +- sqrt(2 + sqrt3)/2`

Explanation:

Call tan ((5pi/12) = t.
Use trig identity: `tan 2a = (2tan a)/(1 - tan^2 a)`
`tan ((10pi)/12) = tan ((5pi)/6) = - 1/(sqrt3) = (2t)/(1 - t^2)`
`t^2 - 2sqrt3t - 1 = 0`

`D = d^2 = b^2 - 4ac = 12 + 4 = 16`--> `d = +- 4`

`t = tan ((5pi)/12) = (2sqrt3)/2 +- 4/2 = 2 +- sqrt3`

Call `sin((5pi)/12) = sin y`
Use trig identity: `cos 2a = 1 - 2sin^2 a`
`cos ((10pi)/12) = cos ((5pi)/6) = (-sqrt3)/2 = 1 - 2sin^2 y`
`sin^2 y = (2 + sqrt3)/4`
`sin y = sin ((5pi)/12) = +- sqrt(2 + sqrt3)/2`