Question

How do you find the value of `tan(-(5pi)/12)`?

Answer 1

`sqrt(2 - sqrt3)/(sqrt(2 + sqrt3)`

Explanation:

`sin ((-5pi)/12) = sin (((7pi)/12 - pi) = - sin ((7pi)/12) =`
`= - sin (pi/12 + pi/2) = - (-cos pi/12) = cos (pi/12)`
Find `cos (pi/12)` by using trig identity:
`2cos^2 a = 1 + cos 2a`.
In this case:
`2cos^2 (pi/12) = 1 + cos (pi/6) = 1 + sqrt3/2 = (2 + sqrt3)/2`
`cos^2 (pi/12) = (2 + sqrt3)/4`
`cos (pi/12) = +- sqrt(2 + sqrt3)/2`
cos pi/12 is positive, so, take the positive value.
sin^2 ([i/12) = 1 - cos^2 (pi/12) = 1 - (2 + sqrt3)/4 = (2 - sqrt3)/4
sin (pi/12) = +- sqrt(2 - sqrt3)/2
sin (pi/12) is positive, so, take the positive value.
`tan (-5pi)/12 = sin (pi/12)/(cos (pi/12) =`
#= (sqrt(2 - sqrt3)/2)/(sqrt( 2 + sqrt3)/2) = = sqrt(2 - sqrt3)/sqrt(2 + sqrt3)#