What is #arccos(cos(4))# ?
2 Answers
Feb 27, 2018
See other answer.
Feb 27, 2018
Explanation:
Note that
So the inverse of the function
The function
#cos(arccos(x)) = x#
for any
However, note that:
#arccos(cos theta) = theta#
only if
So what is the value of
It is some angle
Note that:
#cos (theta + pi) = - cos (theta)#
#cos (-theta) = cos (theta)#
Hence we find:
#cos (2pi - theta) = cos (theta - 2pi) = cos (theta)#
In particular:
#cos (2pi - 4) = cos (4)#
Now
#arccos(cos(4)) = 2pi-4#