How do you find the exact value of #arccos(cos((5π)/4))#?

1 Answer
A. S. Adikesavan
Apr 14, 2016

#(5pi)/4#.. Even if a range is specified, use arc #cos( cos(theta)) = theta#, which is the result upon applying the operator 'arc cos' and its inverse 'cos', in succession.

Explanation:

Disambiguation:
If y = arc cos(x), cos(x) is single-valued.
Yet, y is many-valued in the form
#y=2kpi+-alpha. alpha# is the principal value of y in #[0, pi]#, k =, 0, 1, 2,...

If the range for y is specified, we have to pick value(s) from the general formula and give the answer.

This question does not arise at all, when we apply, in succession, an operator and its inverse, on an operand. The result is the operand.

Here, #arc cos (cos((5pi)/4))=(5pi)/4 and, similarly, arc cos (cos((3pi)/4))=(3pi)/4#.
In the other way about, #cos (arc cos (x))=x#..

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
2577 views around the world
You can reuse this answer
Creative Commons License