Answer:
\(sin(sin^{-1}(x) + cos^{-1}(y)) = x y + \sqrt{1 - y^2} \sqrt{1- x^2}\)
Step-by-step explanation:
Given
\(sin(sin^{-1}(x) + cos^{-1}(y))\)
Let's define
\(\alpha = sin^{-1}(x)\)
\(\beta = cos^{-1}(y)\)
Replacing
\(sin(\alpha + \beta)\)
\(sin(\alpha) cos(\beta) + sin(\beta) cos(\alpha)\)
But
\(sin(\alpha) = sin(sin^{-1}(x))=x\)
\(cos(\beta) = cos(cos^{-1}(y))=y\)
From trigonometric identity
\(sin^2(\beta) + cos^2(\beta) = 1\)
\(sin(\beta) = \sqrt{1 - cos^2(\beta)} = \sqrt{1 - y^2}\)
\(sin^2(\alpha) + cos^2(\alpha) = 1\)
\(cos(\alpha) = \sqrt{1 - sin^2(\alpha)} = \sqrt{1- x^2}\)
Replacing
\(sin(sin^{-1}(x) + cos^-1 (y)) = x y + \sqrt{1 - y^2} \sqrt{1- x^2}\)
