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Question: If \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}\), then find the value of \(\left( {{x^2} +...

If sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}, then find the value of (x2+y2)\left( {{x^2} + {y^2}} \right).

Explanation

Solution

In the given problem, we are required to calculate the value of an expression given an inverse trigonometric equation. Such problems require basic knowledge of inverse trigonometric functions and formulae. Besides this, knowledge of algebraic identities and simplification rules is of significant use in the problem.

Complete step by step answer:
So, in the given problem, we are given the relation,
sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}
So, sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}
Shifting the sine inverse of y to right side of equation, we get,
⇒sin⁡−1x=π2−sin⁡−1y\Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}y
Now, we know the inverse trigonometric formula sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}. So, we get,
⇒sin⁡−1x=cos⁡−1y\Rightarrow {\sin ^{ - 1}}x = {\cos ^{ - 1}}y
Taking sine on both sides of the equation, we get,
⇒sin⁡(sin⁡−1x)=sin⁡(cos⁡−1y)\Rightarrow \sin \left( {{{\sin }^{ - 1}}x} \right) = \sin \left( {{{\cos }^{ - 1}}y} \right)
We know that sin⁡(sin⁡−1x)=x\sin \left( {{{\sin }^{ - 1}}x} \right) = x. So, we get,
⇒x=sin⁡(cos⁡−1y)−−−−(1)\Rightarrow x = \sin \left( {{{\cos }^{ - 1}}y} \right) - - - - \left( 1 \right)
Now, we have to find the sine of the angle whose cosine is given to us as y.
Let us assume θ\theta to be the concerned angle.
Then, θ=cos⁡−1y\theta = {\cos ^{ - 1}}y
Taking cosine on both sides of the equation, we get
⇒cos⁡θ=y\Rightarrow \cos \theta = y

To evaluate the value of the required expression, we must keep in mind the formulae of basic trigonometric ratios.
We know that, sin⁡θ=PerpendicularHypotenuse\sin \theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} and cos⁡θ=BaseHypotenuse\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}.
So, cos⁡θ=BaseHypotenuse=y1\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{y}{1}
Let the length of base be yy.
Then, length of hypotenuse =1 = 1.
Now, applying Pythagoras Theorem,
(Hypotenuse)2=(Base)2+(Perpendicular)2{\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}
⇒(1)2=(y)2+(Perpendicular)2\Rightarrow {\left( 1 \right)^2} = {\left( y \right)^2} + {\left( {Perpendicular} \right)^2}
⇒1−y2=(Perpendicular)2\Rightarrow 1 - {y^2} = {\left( {Perpendicular} \right)^2}
⇒(Perpendicular)=1−y2\Rightarrow \left( {Perpendicular} \right) = \sqrt {1 - {y^2}}
So, we get the length of perpendicular as 1−y2\sqrt {1 - {y^2}} .

Now, sin⁡θ=PerpendicularHypotenuse\sin \theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}
⇒sin⁡θ=1−y21=1−y2\Rightarrow \sin \theta = \dfrac{{\sqrt {1 - {y^2}} }}{1} = \sqrt {1 - {y^2}}
Hence, sin⁡(cos⁡−1y)=1−y2\sin \left( {{{\cos }^{ - 1}}y} \right) = \sqrt {1 - {y^2}}
So, putting in the equation (1)\left( 1 \right), we get,
⇒x=1−y2\Rightarrow x = \sqrt {1 - {y^2}}
Squaring both sides of the equation, we get,
⇒x2=(1−y2)2\Rightarrow {x^2} = {\left( {\sqrt {1 - {y^2}} } \right)^2}
⇒x2=1−y2\Rightarrow {x^2} = 1 - {y^2}
Shifting the term consisting y to left side of the equation, we get,
∴x2+y2=1\therefore {x^2} + {y^2} = 1

Hence, the value of x2+y2{x^2} + {y^2} is 11.

Note: The given problem can also be solved by use of some basic inverse trigonometric identities such as sin⁡−1(x)+cos⁡−1(x)=π2{\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}. For finding a trigonometric ratio for an angle given in terms of an inverse trigonometric ratio, we have to first assume that angle to be some unknown, let's say θ\theta . Then proceeding further, we have to find the value of a trigonometric function of that unknown angle θ\theta . Then we find the required trigonometric ratio with help of basic trigonometric formulae and definitions of trigonometric ratios. Such questions require clarity of basic concepts of trigonometric functions as well as their inverse.