How do you calculate #Cos^-1 [cos ((7pi)/6) ]#?

2 Answers
Rohit Deka
May 24, 2018

The answer will be #(5pi)/6#

Explanation:

You can write #(7pi)/6# as #(pi+pi/6)#
Thus we can clearly see the angle falls in the third quadrant. And the cosine value in third quadrant is always negative.

Hence, #cos(pi+pi/6) = -cos(pi/6)#

coming back to the question
#cos^-1[cos((7pi)/6)] = cos^-1[-cos(pi/6)]#
=#pi - cos^-1[cos(pi/6)]#
=#pi- pi/6#
=#(5pi)/6#

Dean R.
May 24, 2018

# arccos( cos ( {7 pi} / 6) ) = pm {7pi}/6 + 2pi k,# integer #k#

# text{Arc}text{cos}( cos ( {7 pi} / 6) ) = {5pi}/6#

Explanation:

I treat #arccos( cos ( {7 pi} / 6) ) # as a multivalued expression, all the angles whose cosine equals #cos ( {7 pi} / 6).#

In general #cos x = cos a # has solution #x=pm a + 2 pi k, # integer #k#

# x = arccos( cos ( {7 pi} / 6) ) #

# cos x = cos ( {7 pi} / 6) #

#x = pm {7pi}/6 + 2pi k#

# arccos( cos ( {7 pi} / 6) ) = pm {7pi}/6 + 2pi k,# integer #k#

The principal value is in the second quadrant, given here by the minus sign and #k=1.#

# text{Arc}text{cos}( cos ( {7 pi} / 6) ) = - {7pi}/6 + 2pi = {5pi}/6#

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