Question

\(\sin^{-1}x+\sin^{-1}y=\) (principal values)

• A. \(\sin^{-1}\!\left(x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}}\right)\)
• B. \(\sin^{-1}\!\left(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\right)\)
• C. \(\sin^{-1}\!\left(x\sqrt{1+y^{2}}+y\sqrt{1+x^{2}}\right)\)
• D. \(\sin^{-1}\!\left(x\sqrt{1+y^{2}}-y\sqrt{1+x^{2}}\right)\)

Hint:

Convert inverse sines to angles, use $\sin(\alpha+\beta)$, then return via $\sin^{-1}$.

Answer 1

The Correct Option is B

Let \(\alpha=\sin^{-1}x,\ \beta=\sin^{-1}y\Rightarrow \sin\alpha=x,\ \sin\beta=y\) with \(\alpha,\beta\in[-\pi/2,\pi/2]\). Then \[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta =x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}. \] Since \(\alpha+\beta\) lies in \([-\pi,\pi]\) and under usual exam assumptions \(\alpha+\beta\in[-\pi/2,\pi/2]\), we write \(\alpha+\beta=\sin^{-1}\!\big(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\big)\).