How do you evaluate #arccos (-1/3)#?

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Answer 1 · 2015-11-28T18:34:08

Think of the line #y=-1/3# crossing the cosine function as shown below. Where the two functions intersect, is the solution ...

The arccos function is only valid over the range #[0,pi]#.

Below is a graph of the two functions. As you can see, the two functions intersect at #x~~1.911# radians between #[0,pi]#.

Using a trigonometry calculator , you can also enter the inverse cosine at #-1/3# and get more accuracy:

#arccos(-1/3)=1.91063324# radians

Hope that helps

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Answer 2 · 2015-11-29T00:07:49

Evaluate arccos (-1/3)

Ans: #+- 109.47#

Use calculator.
#cos x = -1/3# --> arc # x = +- 109^@47#