Question

If cos^-1x + cos^-1 y = $\frac{\mathrm{\pi }}{3}$, then sin^-1 x + sin^-1 y =____________________.

Answer 1

We know

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}$, for all x ∈ R .....(1)

Also, ${\sin }^{-1}y+{\cos }^{-1}y=\frac{\mathrm{\pi }}{2}$, for all y ∈ R .....(2)

Adding (1) and (2), we get

${\sin }^{-1}x+{\cos }^{-1}x+{\sin }^{-1}y+{\cos }^{-1}y=\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}$

$\Rightarrow {\sin }^{-1}x+{\sin }^{-1}y+{\cos }^{-1}x+{\cos }^{-1}y=\mathrm{\pi }$

$\Rightarrow {\sin }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{y}+\frac{\mathrm{\pi }}{3}=\mathrm{\pi }$ (Given)

$\Rightarrow {\sin }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{y}=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3}$

If cos⁻¹ x + cos⁻¹ y = $\frac{\mathrm{\pi }}{3}$, then sin⁻¹ x + sin⁻¹ y = $\underline{\frac{2\mathrm{\pi }}{3}}$.