Prove an equation involving inverse circular functions

In summary, the given problem statement states that 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##. By substituting ##x = \sin \alpha, y = \sin \beta, z = \sin \gamma## and simplifying, the L.H.S. of the required statement is shown to be equal to the R.H.S., proving the statement.
  • #1
brotherbobby
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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.
 
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  • #2
brotherbobby said:
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...
 
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  • #3
brotherbobby said:
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.?
 
  • #4
WWGD said:
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.
Does ## \alpha + \beta + \gamma = \pi ## remind you of a geometrical figure?
 
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  • #5
WWGD said:
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.
 
  • #6
brotherbobby said:
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?
 
  • #7
pbuk said:
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.
 
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FAQ: Prove an equation involving inverse circular functions

1. What are inverse circular functions?

Inverse circular functions are functions that can be used to find the angle in a right triangle when the lengths of the sides are known. They are also known as arc functions and are the inverse of the trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant).

2. How do you prove an equation involving inverse circular functions?

To prove an equation involving inverse circular functions, you can use the properties of inverse functions, trigonometric identities, and algebraic manipulation. You can also use a graphing calculator or software to graph both sides of the equation and see if they intersect at the same points.

3. Can all equations involving inverse circular functions be proven?

No, not all equations involving inverse circular functions can be proven. Some equations may not have a solution, while others may have multiple solutions. It is important to check the domain and range of the inverse circular functions to determine if an equation can be proven.

4. What is the importance of proving equations involving inverse circular functions?

Proving equations involving inverse circular functions is important because it helps to verify the accuracy of mathematical statements and equations. It also allows for a deeper understanding of the relationships between different trigonometric functions and their inverses.

5. Are there any tips for proving equations involving inverse circular functions?

Yes, some tips for proving equations involving inverse circular functions include simplifying both sides of the equation, using trigonometric identities, and being aware of the domain and range of the inverse circular functions. It can also be helpful to draw a diagram or use a graphing calculator to visualize the problem.

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