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.

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