If $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z = \pi$, then $x\sqrt{1-x^2} + y\sqrt{1-y^2} +z\sqrt{1-z^2} = 2xyz$.

Question

If $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z = \pi$, prove that $$x\sqrt{1-x^2} + y\sqrt{1-y^2} +z\sqrt{1-z^2} = 2xyz$$

I tried rewriting $z$ in terms of $x$ and $y$. I got $$z=x\sqrt{1-y^2} + y\sqrt{1-x^2}$$ I further tried simplifying this term, but I just couldn't get the desired result.

Answer 1

Let $$\sin^{-1}x=A,\sin^{-1}y=B,\sin^{-1}z=C$$and $A+B+C=\pi$

It is enough to prove that $$\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$$ Let us take $LHS$$$\implies2\sin(A+B)\cos(A-B)+2\sin C\cos C$$$$=2\sin C\cos(A-B)+2\sin C\cos C$$$$=2\sin C[\cos(A-B)+\cos C]$$$$=2\sin C[\cos(A-B)-\cos(A+B)]$$$$=2\sin C\ 2\sin B \ \sin A$$$$=4\sin A\sin B\sin C$$$$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2\sin A\sin B\sin C=2xyz$$

Answer 2

Proof.$\blacktriangleleft$ Let $A = \arcsin(x), B = \arcsin(y), C = \arcsin(z)$, then $A,B,C \in [-\pi/2, \pi/2]$, thus $\cos(A), \cos(B), \cos(C) \geqslant 0$. Therefore we are requested to prove $$ \sin(A)\cos(A) + \sin(B)\cos(B) + \sin(C) \cos(C) = 2\sin(A) \sin(B) \sin(C)\quad [A + B + C = \pi]. $$ Now calculate: \begin{align*} \mathrm {LHS} &= \sin(A)\cos(A) + \sin(B)\cos(B) - \sin(A+B) \cos(A+B) \\ &= \frac 12 (\sin(2A) + \sin(2B)) - \sin(A+B) \cos(A+B)\\ &= \sin(A+B) \cos(A-B) - \sin(A+B) \cos(A+B)\\ &= \sin(A+B) (\cos(A-B) - \cos(A+B))\\ &= 2\sin(C)\sin(A) \sin(B) = \mathrm {RHS}. \end{align*} Hence the equation. $\blacktriangleright$

Answer 3

Good start. Indeed, on the one hand: $$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z = \pi \iff \\ \sin^{-1}x=\pi-(\sin^{-1}y+\sin^{-1}z) \iff \\ \sin (\sin^{-1}x)=\sin(\sin^{-1}y+\sin^{-1}z) \iff \\ x=y\cos(\sin^{-1}z)+z\cos (\sin^{-1}y) =y\sqrt{1-z^2}+z\sqrt{1-y^2}.$$ Similarly: $$y=x\sqrt{1-z^2}+z\sqrt{1-x^2};\\ z=x\sqrt{1-y^2}+y\sqrt{1-x^2}.$$ Multiply the three equations by $\sqrt{1-x^2},\sqrt{1-y^2},\sqrt{1-z^2}$, respectively, and add up: $$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\\ 2\left(\color{blue}{x\sqrt{(1-y^2)(1-z^2)}+y\sqrt{(1-x^2)(1-z^2)}+z\sqrt{(1-x^2)(1-y^2)}}\right)$$ On the other hand: $$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z = \pi \iff \\ \sin{(\sin^{-1}x+\sin^{-1}y+\sin^{-1}z)}=0 \iff \\ \sin{(\sin^{-1}x)}\cdot \cos{(\sin^{-1}y+\sin^{-1}z)}\color{red}+\\ \cos{(\sin^{-1}x)}\cdot \sin{(\sin^{-1}y+\sin^{-1}z)}=0 \iff \\ x\cdot (\cos{(\sin^{-1}y)}\cos{(\sin^{-1}z)}-\sin{(\sin^{-1}y)}\sin{(\sin^{-1}z)})\color{red}+\\ \sqrt{1-x^2}\cdot (y\cos{(\sin^{-1}z)}+z\cos{(\sin^{-1}y)})=0 \iff \\ x\cdot (\sqrt{(1-y^2)(1-z^2)}-yz)\color{red}+\sqrt{1-x^2}\cdot (y\sqrt{1-z^2}+z\sqrt{1-y^2})=0 \iff \\ xyz=\color{blue}{x\sqrt{(1-y^2)(1-z^2)}+y\sqrt{(1-x^2)(1-z^2)}+z\sqrt{(1-x^2)(1-y^2)}}.$$