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

2 Answers
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

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