How do you evaluate #cos(arccos(-0.1))#?

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Answer 1 · 2017-11-25T09:59:30

See explanation.

#cosx# and #arccosx# is a pair of function and its inverse. For such pair it is true that:

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

So here the value of the expression is the argument of inner function:

#cos(arccos(-0.1))=-0.1#