- #1

brotherbobby

- 700

- 163

- Homework Statement
- If ##\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi##, show that ##\boxed{x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz}##

- Relevant Equations
- 1. ##\sin^{-1}x+\sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})##

2. ##\sin^{-1}0=\pi##

(I must confess that, in spite of working through the chapter on

##\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) = \sin^{-1}(0\sqrt{1-z^2}-z\sqrt{1-0^2})=\sin^{-1}(-z)\Rightarrow \small{x\sqrt{1-y^2}+y\sqrt{1-x^2} = -z}## ##\\[10 pt]##

Upon squaring both sides, we obtain ##x^2(1-y^2)+y^2(1-x^2)+2xy\sqrt{(1-x^2)(1-y^2)}=z^2##.

This is clearly looking hopeless far as the required solution (in box above) is concerned.

**inverse circular functions**, I could barely proceed with this problem. Note what it asks to prove : ##x\sqrt{1-x^2}+\ldots## and how much is that at odds with the formula (1 above) of adding two ##sin^{-1}##'s, where you have ##x\sqrt{1-y^2}+\ldots##)**Attempt :**From the given equation, ##\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi\Rightarrow \sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\sin^{-1}0-\sin^{-1}z##, which simplifies to##\\[10pt]####\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) = \sin^{-1}(0\sqrt{1-z^2}-z\sqrt{1-0^2})=\sin^{-1}(-z)\Rightarrow \small{x\sqrt{1-y^2}+y\sqrt{1-x^2} = -z}## ##\\[10 pt]##

Upon squaring both sides, we obtain ##x^2(1-y^2)+y^2(1-x^2)+2xy\sqrt{(1-x^2)(1-y^2)}=z^2##.

This is clearly looking hopeless far as the required solution (in box above) is concerned.

**Any help or hint will be welcome.**