How do you simplify the function #f(x)=cos(arcsin(x))# and find the domain and range?

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Answer 1 · 2017-07-13T18:03:51

Given: #f(x)=cos(sin^-1(x))#

The domain for the inverse sine function is #-1<=x<=1# because this is the range for the sine function.

The range for the function is the same as the range for the cosine function, #-1<=f(x)<=1#

Use the identity #cos(x) = +-sqrt(1-sin^2(x))#

#f(x) = +-sqrt(1-sin^2(sin^-1(x)))#

The sine and the inverse sine function cancel; leaving us with #x^2#:

#f(x) = +-sqrt(1-x^2); -1<=x<=1#

We cannot know whether we use the + or the - sign without more information regarding the quadrant of the angle.