Question

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

Answer 1

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`

Answer 2

`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`