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Question: How do you simplify \(\cos \left( {{{\sin }^{ - 1}}x} \right)\)?...

How do you simplify cos⁡(sin⁡−1x)\cos \left( {{{\sin }^{ - 1}}x} \right)?

Explanation

Solution

Here we need to proceed by letting the value inside the bracket which is sin⁡−1x{\sin ^{ - 1}}x to be any variable say aa and then we will get sin⁡a=x\sin a = x and now we can easily find the value of cos⁡a\cos a which is required by using Pythagoras theorem.

Complete step by step solution:
Here we are given to simplify the term which is given as cos⁡(sin⁡−1x)\cos \left( {{{\sin }^{ - 1}}x} \right)
So let us consider the term inside the bracket which is sin⁡−1x{\sin ^{ - 1}}x to be any variable say aa
So we get sin⁡−1x=a{\sin ^{ - 1}}x = a
So we will get sin⁡a=x\sin a = x
We can also write it as sin⁡a=x1\sin a = \dfrac{x}{1}
Now we need to find the value of cos⁡(sin⁡−1x)\cos \left( {{{\sin }^{ - 1}}x} \right) which can be written as cos⁡a\cos aaccording to the variable which we have let sin⁡−1x=a{\sin ^{ - 1}}x = a
So let us consider the triangle ABCABCin which we can let the angle aa as ∠A\angle A and it is right angles atBB

Now we are given:
sin⁡a=x1\sin a = \dfrac{x}{1} −−−−(1) - - - - (1)
So we know that sin⁡a=perpendicularhypotenuse=BCAC\sin a = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} = \dfrac{{BC}}{{AC}} −−−−−(2) - - - - - (2)
So by comparing the equation (1) and (2) we will get:
BC=x AC=1  BC = x \\\ AC = 1 \\\
Now we know that by Pythagoras theorem we can say that in the right angles triangle:
AC2=AB2+BC2A{C^2} = A{B^2} + B{C^2}
Now we can put in it:
BC=x AC=1  BC = x \\\ AC = 1 \\\
We will get:
AC2=AB2+BC2 12=AB2+x2 AB=1−x2  A{C^2} = A{B^2} + B{C^2} \\\ {1^2} = A{B^2} + {x^2} \\\ AB = \sqrt {1 - {x^2}} \\\
Now we know that in the right angles triangle:
cos⁡a=basehypotenuse=ABAC\cos a = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} = \dfrac{{AB}}{{AC}}
Now we can substitute the values of AB,ACAB,AC in the above equation of cos⁡a\cos a
cos⁡a=basehypotenuse=ABAC\cos a = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} = \dfrac{{AB}}{{AC}}
cos⁡a=1−x21=1−x2\cos a = \dfrac{{\sqrt {1 - {x^2}} }}{1} = \sqrt {1 - {x^2}}
Now we can substitute the value of aa and get:
cos⁡(sin⁡−1x)\cos \left( {{{\sin }^{ - 1}}x} \right) =1−x2= \sqrt {1 - {x^2}}
Hence in this way by the use of Pythagoras theorem we can easily solve for such types of problems where we need to find the trigonometric function of the inverse function.

Note:
In such types of problems the student must keep in mind the basic trigonometric formula and the properties and also the use of Pythagoras theorem. We must know that cos⁡(sin⁡−1x)=sin⁡(cos⁡−1x)\cos \left( {{{\sin }^{ - 1}}x} \right) = \sin \left( {{{\cos }^{ - 1}}x} \right) because:
cos⁡(π2−cos⁡−1x)=sin⁡(cos⁡−1x)\cos \left( {\dfrac{\pi }{2} - {{\cos }^{ - 1}}x} \right) = \sin \left( {{{\cos }^{ - 1}}x} \right).