How do you evaluate #cos^-1(cos((17 pi)/5))#?

1 Answer
May 16, 2016

#(17pi)/5#

Explanation:

An inverse function might be single-valued or many-valued. But applying an inverse function and the function in succession, over an operand, returns the operand.

#O^(-1) O (c) =O^(-1) (O (c)) = c#.

#cos((17pi)/5)= cos (3pi + ((2pi)/5))=-cos((2pi)/5)= - -cos 72^o=- 0.3090#.

You can see the basic difference between evaluating directly

#cos^(-1)(-0.3090) = (3pi)/5 or - (3pi)/15 or (7pi)/5 or -(7pi)5, (17pi)/5 or -(17pi)/5#..... and

#cos^(-1)cos ((17pi)/5) = cos^(-1)(cos((17pi)/5)) = (1 7pi)/5#.

For that matter, if the expression is

#cos^(-1)(cos((3pi)/5))#, the value is #(3pi)/5#.

Note that #cos (+-(17/5)pi) = cos (+-(7/5)pi) = cos (+-(3/5)pi))= -0.3090#

Cosine is negative, in 2nd and 3rd quadrants..