How do you find the exact functional value sec (19Pi/12) using the cosine sum or difference identity?

1 Answer
Nghi N.
Aug 30, 2015

Find #sec ((19pi)/12)#

Ans: #- sqrt(2 + sqrt3)/2#

Explanation:

Call #(19pi/12) = t# --> #sec t = 1/sin t.#
#cos 2t = cos ((19pi)/6) = cos ((7pi)/6 + 2pi) = -cos pi/6 = -sqrt3/2#
Use the trig identity: #cos 2t = 1 - 2sin^2 t.#
cos 2t =# -sqrt3/2 = 1 - 2sin^2 t.#
2sin^2 t = 1 + sqrt3/2 = (2 + sqrt3)/2
sin^2 t = (2 + sqrt3)/4
sin t = #sin ((19pi)/12) = +- sqrt(2 + sqrt3)/2#
Since the arc #((19pi)/6)#= 285 deg is located in Quadrant IV, its sin is negative.
Then, #sin ((19pi)/12) = - sqrt(2 + sqrt3)/2#

Check by calculator: x = 285 --> sin 285 = -0.97.
#- sqrt(2 + sqrt3)/2 = -1.931/2 = -0.97# OK

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