How do you use the angle sum or difference identity to find the exact value of #tan((7pi)/12)#?

1 Answer
Nghi N.
Aug 18, 2016

#- (2 + sqrt3)#

Explanation:

#tan ((7pi)/12) = tan (pi/12 + pi) = - tan (pi/12)#
Call #tan (pi/12) = tan t# --># tan 2t = tan (pi/6) = 1/sqrt3#
Use trig identity:
#tan 2a = (2tan a)/(1 - tan^2 a)#
#1/sqrt3 = (2tan t)/(1 - tan^2 t)#
Cross multiply -->
#1 - tan^2 t = 2sqrt3tan t#
#- tan^2 t - 2sqrt3tan t + 1 = 0.#
Solve this quadratic equation for tan t by the improved quadratic equation (Socratic Search)
#D = d^2 = b^2 - 4ac = 12 + 4 = 16# --> #d = +- 4#
There are 2 real roots:
#tan t = -b/(2a) +- d/(2a) = (-2sqrt3)/-2 +- d/2 = sqrt3 +- 2#.
Since #pi/12# is in Quadrant I, its tan is positive
#tan t = tan (pi/12) = sqrt3 + 2#
There for:
#tan ((7pi)/12) = - tan t = - (sqrt3 + 2)#

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