How do you evaluate $cos^-1(cos((19pi)/10))$ ?

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Answer 1 · 2016-08-31T03:44:15

$19/10pi$

Use that, if f is a single valued function,

$f^(-1)f(x)=x$

Here, it is $cos^(-1)cos(19/10pi) = 19/10pi$

Disambiguation note on uniqueness, despite the convention

that the inverse trigonometric function values are defined as

principal values:

If f is a single-valued function operator, in $y =f(x)$ , the ordered and

coupled operation

$ff^(-1)(y)$ returns y

and the ordered couple operation

$f^(-1)f(x)$ returns x.

Despite that f might not be bijective, being single-valued, it is

bijective in an infinitesimal neighborhood

$(x-in, x+in)$ .

So there is unique answer for single-valued function under these

twin operations.

This note is important for applications wherein

$x= c$ (time t), $t in (-oo, oo)$ .

How do you evaluate cos^-1(cos((19pi)/10))cos^-1(cos((19pi)/10))?