Question

Algebraically determine the exact value of?

tan`(11pi)/12`

Answer 1

`tau= tan((11pi)/12) = -2 + sqrt(3)`

Explanation:

We seek:

`tau= tan((11pi)/12)`

Using the trigonometric identity:

`tan(A+-B) -= (tanA+-tanB)/(1 \ bar("+") \ tanAtanB)` ..... [A]

We can write

`tau = tan(pi-pi/12)`

`\ \ = (tan(pi)-tan(pi/12))/(1+tan(pi)tan(pi/12))`

`\ \ = (0-tan(pi/12))/(1+0)`

`\ \ = -tan(pi/12)`

And we can also use the same identity [A] to give:

`tan(2 (pi/12)) = (tan(pi/12)+tan(pi/12))/(1 - tan(pi/12)tan(pi/12))`

`:. tan(pi/6) = (2tan(pi/12))/(1 - tan^2(pi/12)`

And we know that `tan(pi/6) = sqrt(3)/3` and `tan(pi/12) = -tau` so:

`sqrt(3)/3 = (-2tau)/(1 - (-tau)^2)`

`:. 1 - tau^2= -6/sqrt(3)tau`

`:. 1 - tau^2= -2sqrt(3)tau`

`:. tau^2 - 2sqrt(3)tau - 1 = 0`

Which is quadratic in `tau`, so we can complete the square:

`:. (tau - sqrt(3))^2 - (sqrt(3))^2 - 1 = 0`

`:. (tau - sqrt(3))^2 - 3 - 1 = 0`

`:. (tau - sqrt(3))^2 = 4`

`:. tau - sqrt(3) = +-2`

`:. tau = +-2 + sqrt(3)`

Noting that `(11pi)/12` is in `QII`, then we know that `tau= tan((11pi)/12)` is negative, thus we deduce that

`tau = -2 + sqrt(3)`

Answer 2

`tan ((11pi)/12) = sqrt3 - 2`

Explanation:

`tan ((11pi)/12) = tan (-pi/12 + pi) = tan (-pi/12) = - tan (pi/12)`
First, find `tan (pi/12)` by using the trig identity:
`tan 2t = (2tan t)/(1 - tan^2 t)`
Call `tan (pi/12) = tan t`, --> `tan 2t = tan (pi/6) = 1/sqrt3`
we get:
`(2tan t)/(1 - tan^2 t) = 1/sqrt3`
Cross multiply
`2sqrt3tan t = 1 - tan^2 t`
`tan^2 t + 2sqrt3tan t - 1 = 0`.
Solve this quadratic equation for tan t:
`D = d^2 = b^2 - 4ac = 12 + 4 = 16` --> `d = +- 4`
The 2 real roots are:
`tan t = -b/(2a) +- d/(2a) = - sqrt3 +- 2`
a. `tan t = tan (pi/12) = -2 - sqrt3` (rejected since `tan (pi/12) > 0`)
b. `tan t = tan (pi/12) = 2 - sqrt3`
Finally,
`tan ((11pi)/12) = - tan (pi/12) = sqrt3 - 2`