Prove that for #-1 <= x < 1#, #"arc" cos(x) = k + arctan((sqrt(1-x^2))/(x))#, where #k# is a constant to be found?

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Answer 1 · 2018-08-12T04:09:33

#arccos values are restricted to #[ 0, pi ]# and

arctan values are restricted to # [ - pi/2, pi/2 ]#.

The common daomain is #[ 0, pi/2 ]#. Here, k = 0.

Elsewhere, the shift, from arctan to arc cos is by setting k = pi.

Example:
( i )# x = 1/2#.

#pi/3 = arccos ( 1/2 ) = arctan (sqrt3 )#. Here, k =0.

( ii ) #x = -1/2#.

#2/3pi = arccos ( -1/2 ) = pi +( - pi/3 ) = pi + arctan ( -sqrt3 )#.

If #arctan and arccos# are replaced by my piecewise wholesome

#( tan )^( - 1 ) and ( cos )^( - 1 )# operators, it is a matter for

pondering.