How do you evaluate #cos^-1(cos((7pi)/10))#?

1 Answer
Aug 26, 2016

#(7pi)/10#

Explanation:

The whole purpose of the inverse cosine function (or of inverse functions in general) is to undo the action of the cosine function (or the parent function).

An easy example is the function #f(x)=x^2#. What "undoes" this function? That would be the inverse function #f^-1(x)=sqrtx#.

Inverse functions are useful when trying to figure out what values of #x# you need to input into #f# to get a certain output. For example, we could ask when #f(x)=x^2=16#. The answer to that would be #f^-1(16)=sqrt16=4#.

When you see #cos^-1(cos((7pi)/10))#, you should think of seeing something like #sqrt(4^2)#. The square root of the square of #4# is #4#, and the inverse cosine of the cosine of #(7pi)/10# is #(7pi)/10#.

As a general rule, #f(f^-1(x))=f^-1(f(x))=x#, for the same reasons discussed above.