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

Question

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

My Attempt: $$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$ $$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1-x^2})+\sin^{-1} (z)=\dfrac {\pi}{2}$$ $$\sin^{-1} (x\sqrt {1-y^2} + y\sqrt {1-x^2})=\dfrac {\pi}{2} - \sin^{-1} (z)$$

Answer 1

Let: $\sin^{-1}x=\alpha$, $\sin^{-1}y=\beta$, $\sin^{-1}z=\gamma$. We know that $\alpha+\beta+\gamma={\pi\over2}$, that is: $\gamma={\pi\over2}-\alpha-\beta$ and $$\sin\gamma=\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$ We have then: $$ \begin{align} &x^2+y^2+z^2+2xyz=\\ &\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=\\ &\sin^2\alpha+\sin^2\beta+(\cos\alpha\cos\beta-\sin\alpha\sin\beta)^2 +2\sin\alpha\sin\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=\\ &\sin^2\alpha+\sin^2\beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=\\ &\sin^2\alpha(1-\sin^2\beta)+\sin^2\beta+\cos^2\alpha\cos^2\beta=\\ &\sin^2\alpha\cos^2\beta+\sin^2\beta+\cos^2\alpha\cos^2\beta=\\ &(\sin^2\alpha+\cos^2\alpha)\cos^2\beta+\sin^2\beta=\\ &\cos^2\beta+\sin^2\beta=1 \end{align} $$

Answer 2

Put differently, if $\alpha+\beta+\gamma=\frac \pi2$, then $$ \tag1\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=1.$$ Let's investigate how the left hand side aries if we keep $\gamma $ fixed, that is, we consider $$\tag2 \sin^2(\alpha+t)+\sin^2(\beta-t)+\sin^2\gamma+2\sin(\alpha+t)\sin(\beta-t)\sin\gamma$$ and compute its derivative $\frac d{dt}$ at $t=0$: $$ \begin{align}&2\sin \alpha\cos \alpha-2\sin \beta\cos\beta+2(\cos\alpha\sin\beta-\sin\alpha\cos\beta)\sin\gamma\\ ={}&\sin2\alpha-\sin 2\beta-2\sin(\alpha-\beta)\sin\gamma\\ {}={}&2\cos(\alpha+\beta)\sin(\alpha-\beta)-2\sin(\alpha-\beta)\sin\gamma\\ ={}&2\sin\gamma\sin(\alpha-\beta)-2\sin(\alpha-\beta)\sin\gamma\\={}&0\end{align}$$ We conclude that $(2)$ is constant as a function of $t$. Therefore, it suffices to show $(1)$ for the case $\beta=0$ (i.e., $t=\beta$), i.e., we are reduced to showing

If $\alpha+\gamma=\frac\pi2$ then $\sin^2\alpha+\sin^2\gamma=1$.

But that is of course true.

Answer 3

I want to show a slight variation of Aretino's nice proof that does not presuppose knowledge of the identity we want to prove, but rather derives it from the given equation. We need the identity $\cos(\sin^{-1}(a)) = \sqrt{1-a^2}$ as well as

$$ \sin(a+b+c) = \sum\limits_{cyc} \Big(\sin(a)\cos(b)\cos(c)\Big)-\sin(a)\sin(b)\sin(c) $$

Thus by applying $\sin$ on both sides of $\frac \pi2 = \sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)$ we deduce

$$ 1 = \sum_{cyc} \Big( x\sqrt{1-y^2}\sqrt{1-z^2}\Big)- xyz $$

On the other hand if we solve the original equation for any of the variables, say, $x$ we obtain:

$$ x = \sqrt{1-y^2}\sqrt{1-z^2} - yz $$

Which we can plug into the former equation to eliminate the roots:

$$ 1 = \sum_{cyc} \Big( x(x+yz)\Big)- xyz = x^2+y^2+z^2 +2xyz$$