Prove an equation involving inverse circular functions

[brotherbobby]

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 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.

 

[pbuk]

Any help or hint will be welcome.

Whenever you see a function of $1 - x^2$ this should alert you that a trigonometric substitution may be useful and from the first equation $x = \sin \alpha$ etc. looks promising...

 

[WWGD]

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 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.

Do you have additional context, i.e., are x,y,z any 3 Real numbers, etc.?

 

[pbuk]

Do you have additional context, i.e., are x,y,z any 3 Real numbers, etc.?

I think we can start by assuming that $x, y, z \in [0, 1]$ and then see if any loosening of this makes sense.

Spoiler: Hint

Does $\alpha + \beta + \gamma = \pi$ remind you of a geometrical figure?

 

[brotherbobby]

Do you have additional context, i.e., are x,y,z any 3 Real numbers, etc.?

Hi, I am the creator of this thread or OP as you call them on here in PP.

I do not understand. We know that $-1\le \sin\theta \le 1$, so if $\sin\theta=x\Rightarrow \theta=\sin^{-1}x$, of course we must have $-1\le x \le 1$.

Please correct me if I am making a silly mistake somewhere.

 

[pbuk]

Please correct me if I am making a silly mistake somewhere.

No you are right, but this is not really an important point. Have you tried the $x = \sin \alpha$ substitution yet?

 

[brotherbobby]

No you are right, but this is not really an important point. Have you tried the $x = \sin \alpha$ substitution yet?

Yes. I am the creator of this thread. Let me mention the problem statement and solution.

Problem statement : If we have $\boldsymbol{\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi}$, show that $\boxed{x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz}$.

Revised attempt : As mentioned by @pbuk above, let me take the L.H.S. of what is asked to be proven in blue colour above and substitute in it the variables $x=\sin\alpha, y = \sin\beta\; \text{and}\; z = \sin\gamma$. That would make the given problem statement in bold italicised characters above as :
\begin{equation}
\alpha+\beta+\gamma=\pi
\end{equation}
The L.H.S. of the required statement becomes= $\sin\alpha \cos\alpha+\sin\beta \cos\beta+\sin\gamma \cos\gamma = \dfrac{1}{2}(\sin 2\alpha+\sin 2\beta+\sin 2\gamma)$ $\\[10 pt]$
$=\dfrac{1}{2}\left[2\sin(\alpha+\beta)\cos(\alpha-\beta)+2\sin\gamma \cos\gamma \right]\;\;\;\;$ (using $\sin(C+D)$ rule and $\sin 2A$ rule) $\\[5pt]$
$=\sin\gamma \cos(\alpha-\beta)+\sin\gamma \cos\gamma\;\;\;\;$(since $\alpha+\beta=\pi-\gamma$)
$=\sin\gamma[\cos(\alpha-\beta)+\cos\gamma]=z . 2 \cos\dfrac{\alpha-\beta+\gamma}{2} \cos\dfrac{\alpha-\beta-\gamma}{2}= 2z.\cos\dfrac{\pi-2\beta}{2}\cos\dfrac{\alpha-(\pi-\alpha)}{2}$ $\\[5 pt]$
$=2z\sin\beta\cos(\alpha-\pi/2)=\boxed{2xyz}$ = R.H.S.