What is #arccos(cos(4))# ?

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Answer 1 · 2018-02-27T20:58:31

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Answer 2 · 2018-02-27T21:32:57

#arccos(cos(4)) = 2pi - 4#

Note that #cos theta# is a periodic function with period #2pi#.

So the inverse of the function #cos theta# is not a function.

The function #arccos# satisfies:

#cos(arccos(x)) = x#

for any #x in [-1, 1]#

However, note that:

#arccos(cos theta) = theta#

only if #theta in [0, pi]#

So what is the value of #arccos(cos(4))# ?

It is some angle #theta in [0, pi]# such that #cos theta = cos 4#

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 #2pi-4 in [0, pi]#, so is in the range of #arccos# and we have:

#arccos(cos(4)) = 2pi-4#