Algebraically determine the exact value of?
tan#(11pi)/12#
tan
2 Answers
# 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
# 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 - 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
# tau = -2 + sqrt(3)#
Explanation:
First, find
Call
we get:
Cross multiply
Solve this quadratic equation for tan t:
The 2 real roots are:
a.
b.
Finally,