Question

How do you simplify the function $f\left( x \right) = \cos \left( {\arcsin \left( x \right)} \right)$ and find the domain and range?

Answer 1

Hint: Arcsin function is nothing but the representation of the arc created by the angle whose sine function we are using. In simple terms, it means the inverse function. Secondly, if sin inverse is used here then try to bring the whole equation in terms of sin so that the equation can be solved easily.

Formulas used:
$\
  \cos \theta = \sqrt {1 - {{\sin }^2}\theta } \\
  {\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta \\
\ $

Complete step by step solution:
In the above case, we are working with the arcsin function which is nothing but an inverse function. Thus, our first course of action should be to bring all of the equation into sin form only so that solving them will be easy. We have cos in the equation and we know that $\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $ so the equation becomes
$\
  f\left( x \right) = \cos \left( {\arcsin \left( x \right)} \right) \\
   = \cos \left( {{{\sin }^{ - 1}}x} \right) \\
   = \sqrt {1 - \left( {{{\sin }^2}\left( {{{\sin }^{ - 1}}x} \right)} \right)} \\
   = \sqrt {1 - {{\left( {\sin \left( {{{\sin }^{ - 1}}x} \right)} \right)}^2}} \\
   = \sqrt {1 - {x^2}} \\
\ $
Now, we have $f\left( x \right) = \sqrt {1 - {x^2}} $. Talking about its domain or the values it can have, we know that a root can never be negative or its result will be imaginary. As such for this function,
$\
  1 - {x^2} \geqslant 0 \\
  1 \geqslant {x^2} \\
   - 1 \leqslant x \leqslant 1 \\
\ $

So, the range will be $x \in \left[ { - 1,1} \right]$.
Now, talking about range, we need to find the maximum and minimum values which the function can have. It is evident that in $\sqrt {1 - {x^2}} $ when ${x^2}$increases, the value of root decreases and so when its value becomes 1, the value of the function will be zero when happens both while putting 1 and -1 as x in ${x^2}$. For the maximum value though, er need to look into the range as to where is the point before and after which ${x^2}$ starts increasing. This point occurs at 0 which gives us our maximum value which is 1. As such, range can be given by $f\left( x \right) \in \left[ {0,1} \right]$.

Note:
Here we have used the formula ${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta $ which is an extension of the relation that any function with a certain variable as input, when put as input in its own inverse will give the output as the same function. Also when finding a range of functions containing even powers of the variables such as ${x^2},{x^4},{x^6},{x^8}$, etc, domain ends generally don’t give a range so we need to observe the domain and find the points before and after which the function changes.