How do you use a half-angle formula to find the exact value of #tan((9pi)/8)#?

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Answer 1 · 2015-05-30T21:30:45

Use trig identity: #tan 2a = (2tan a)/(1 - tan^2 a)#

#Call tan ((9pi)/8) = a -> tan ((18pi)/8)= 2a#

#tan 2a = tan ((18pi)/8) = tan ((9pi)/4) = tan (pi/4 + 2pi) = tan (pi/4) = 1#

#tan 2a = 1 = (2a)/(1 - a^2)# -> # a^2 + 2a - 1 = 0.#

#D = d^2 = 4 + 4 = 8 -> d = +-2sqrt2#

#a = -2/2 +- sqrt2/2 = -1 +- sqrt2#

#tan a = tan ((9pi)/8) = -1 - sqrt2# (Quadrant III).