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)# .