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Question: How do you find \[\sin \left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}y} \right)\] ?...

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

Explanation

Solution

Given are the trigonometric and inverse trigonometric functions in one expression. How we will solve this is by taking the help of substitution. We will put the inverse trigonometric functions as some variable. And we will apply the sum and difference formula of trigonometry for the sin function.

Complete step by step answer:
Given that,
sin⁡(sin⁡−1x+cos⁡−1y)\sin \left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}y} \right)
Now this expression above can be written as, sin⁡(A+B)=sin⁡A.cos⁡B−cos⁡A.sin⁡B\sin (A + B) = \sin A.\cos B - \cos A.\sin B
Now we will apply this formula on the expression above,
sin⁡(sin⁡−1x+cos⁡−1y)=sin⁡(sin⁡−1x).cos⁡(cos⁡−1y)−cos⁡(sin⁡−1x).sin⁡(cos⁡−1y)\sin \left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}y} \right) = \sin \left( {{{\sin }^{ - 1}}x} \right).\cos \left( {{{\cos }^{ - 1}}y} \right) - \cos \left( {{{\sin }^{ - 1}}x} \right).\sin \left( {{{\cos }^{ - 1}}y} \right)
We know that,
sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta and cos⁡(cos⁡−1θ)=θ\cos \left( {{{\cos }^{ - 1}}\theta } \right) = \theta
Thus the expression above with these type of inverse function becomes,
sin⁡(sin⁡−1x+cos⁡−1y)=x.y−cos⁡(sin⁡−1x).sin⁡(cos⁡−1y)\sin \left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}y} \right) = x.y - \cos \left( {{{\sin }^{ - 1}}x} \right).\sin \left( {{{\cos }^{ - 1}}y} \right)
Now for the terms after minus signs we will put the substitutions below.
A=sin⁡(cos⁡−1y)A = \sin \left( {{{\cos }^{ - 1}}y} \right)
Then, θ=cos⁡−1y\theta = {\cos ^{ - 1}}y
Thus, A=sin⁡θ&y=cos⁡θA = \sin \theta \& y = \cos \theta
On squaring and adding the above terms,
A2+y2=1{A^2} + {y^2} = 1
A=1−y2A = \sqrt {1 - {y^2}}
This will be done in similar way for the second term,
B=cos⁡(sin⁡−1x)B = \cos \left( {{{\sin }^{ - 1}}x} \right)
Then θ=sin⁡−1x\theta = {\sin ^{ - 1}}x
B=cos⁡θ&x=sin⁡θB = \cos \theta \& x = \sin \theta
On squaring and adding the above terms,
B2+x2=1{B^2} + {x^2} = 1
B=1−x2B = \sqrt {1 - {x^2}}
Substituting these values in the expression above,
sin⁡(sin⁡−1x+cos⁡−1y)=x.y−1−y2.1−x2\sin \left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}y} \right) = x.y - \sqrt {1 - {y^2}} .\sqrt {1 - {x^2}}
This is a simplified answer.

Note:
To solve this question we have to note two important things. One of them is that when we take inverse of the same function then only we get the angle mentioned so otherwise we have to use substitutions like we used in the above case. Also note that we are not asked to find any derivative or integral. So we have to find the x and y form of the equation.