How do you evaluate # arccos(-1/sqrt2)#?

1 Answer
Apr 15, 2016

As the range is not specified, the general value is given. The principal value of #arc cos(-1/sqrt2)# is #(3pi)/4#.
The general value is #2kpi+-(3pi)/4#, k = 0. 1. 2. 3.,...

Explanation:

Cosine is positive in the 1st quadrant [0, pi/2] and negative in the second #[pi/2, pi]#

The principal value for arc cos is defined to be in #[0, pi]#.

The general value is #2kpi+-#(principal value), k = 0. 1. 2. 3.,...

Here, the principal value of #arc cos (-1/sqrt2) = (3pi)/4#.