# If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi-\arcsin z$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
Similarly,$y\sqrt{1-z^2}+z\sqrt{1-y^2}=x$
Similarly,$x\sqrt{1-z^2}+z\sqrt{1-x^2}=y$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\frac{x+y+z}{2}$

Let $\sin^{-1}(x) = A\Rightarrow x=\sin A$ and $\sin^{-1}(y)=B\Rightarrow y=\sin (B)$

and $\sin^{-1}(z)=C\Rightarrow z=\sin C$

So above we have given $A+B+C = \pi$

Now we have to prove $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$

So for $\bf{L.H.S}$

$$\sin 2A+\sin 2B+\sin 2C = 2\sin (A+B)\cdot \sin (A-B)+2\sin C\cos C$$

$$=2\sin C\cdot \sin (A-B)+2\sin C\cos C$$

$$=2\sin C\left[\cos(A-B)+\cos(\pi-(A+B))\right]$$

$$=4\sin C\cdot \left[\cos(A-B)-\cos(A+B)\right]$$

$$=4\sin A\sin B\sin C$$

• This proof is fatally flawed. To begin, $$\sin (2A)+\sin (2B)=2\sin(A+B)\cos(A-B)\ne 2 \sin (A+B)\sin (A-B)$$ Dec 22, 2015 at 3:40
• @Dr.MV, The correct Proof : math.stackexchange.com/questions/154505/…. For the sake of completeness, we should mention here $$0\le\arcsin x\le\dfrac\pi2\implies\cos(\arcsin x)=+\sqrt{1-x^2}$$ Dec 23, 2015 at 4:57
• @labbhattacharjee Lab, I posted the correct proof yesterday herein. I'd like to hear your thoughts. Dec 23, 2015 at 5:52

Hint:

Write

$$x = \sin X \qquad y = \sin Y \qquad z = \sin Z$$

Where $X + Y + Z = \pi$ (meaning that those are the angles of a triangle).

All we need to facilitate analysis is the double angle formula

$$\sin(2a)=2\sin(a)\,\cos(a) \tag 1$$

$$\cos (a-b)-\cos(a+b)=2\sin (a)\,\sin (b) \tag 2$$

along with the Reverse Prosthaphareis Identity

$$\sin (a)+\sin (b)=2\sin\left(\frac{a+b}{2}\right)\,\cos\left(\frac{a-b}{2}\right) \tag 3$$

Then, letting $x=\sin \alpha$, $y=\sin \beta$, and $z=\sin \gamma$ gives

$$\alpha +\beta +\gamma = \pi \tag 4$$

We wish now to transform the function $F(\alpha,\beta,\gamma)$ as given by

\begin{align} F(\alpha,\beta,\gamma)&=\frac12 \sin (2\alpha)+\frac12 \sin (2\beta)+\frac12 \sin (2\gamma) \tag 5\\\\ &=x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2} \end{align}

First, use $(1)$ with $a=2\gamma$, $(3)$ with $a=2\alpha$ and $b=2\beta$, and $(4)$ to write $(5)$ as

\begin{align} F(\alpha,\beta,\gamma)&=\sin(\alpha+\beta)\,\cos(\alpha-\beta)+\sin(\gamma)\,\cos (\gamma) \\\\ &=\sin(\pi-\gamma)\,\cos(\alpha-\beta)+\sin(\gamma)\,\cos (\pi-\alpha-\beta) \end{align}

Next, substituting $(4)$ into $(6)$ and recalling that $\sin (\pi -a)=\sin(a)$ and $\cos (\pi-a)=-\cos(a)$ reveals

$$F(\alpha,\beta,\gamma)=\sin(\gamma)\left(\cos(\alpha-\beta)-\cos (\alpha+\beta)\right) \tag 7$$

Finally, using $(2)$ in $(7)$ yields

\begin{align} F(\alpha,\beta,\gamma)&=\sin(\gamma)\left(2\sin(\alpha)\,\sin(\beta)\right)\\\ &=2\sin(\alpha)\,\sin(\beta)\,\sin)\gamma)\\\\ &=2xyz \end{align}

Therefore, we obtain the identity

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

as was to be shown!