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

[utkarshakash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I assume the inverse functions to be θ, α, β respectively. Rearranging and taking tan of both sides

$tan(\theta + \alpha) = tan(\pi - \beta) \\ tan(\theta + \alpha) = -tan(\beta)$

After simplifying I get something like this
$x\sqrt{(1-y^2)(1-z^2)}+y\sqrt{(1-x^2)(1-z^2)}+z\sqrt{(1-x^2)(1-y^2)} = xyz$

I know it's close but it is not yet the final result.

 

[Mandelbroth]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I assume the inverse functions to be θ, α, β respectively. Rearranging and taking tan of both sides

$tan(\theta + \alpha) = tan(\pi - \beta) \\ tan(\theta + \alpha) = -tan(\beta)$

After simplifying I get something like this
$x\sqrt{(1-y^2)(1-z^2)}+y\sqrt{(1-x^2)(1-z^2)}+z\sqrt{(1-x^2)(1-y^2)} = xyz$

I know it's close but it is not yet the final result.

Hello again, utkarshakash! Starting from your second step...

$\frac{tan\theta + tan\alpha}{1-tan\alpha tan\theta} = -tan\beta \\ tan\theta + tan\alpha = -tan\beta + tan\alpha \ tan\theta \ tan\beta \\ tan\theta + tan\alpha + tan\beta = tan\alpha \ tan\theta \ tan\beta \\ tan(arcsinx) + tan(arcsiny) + tan(arcsinz) = tan(arcsinx) \ tan(arcsiny) \ tan(arcsinz) \\ \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} + \frac{z}{\sqrt{1-z^2}} = \frac{xyz}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}}$

Here, I think, is where you went wrong. You want each term on the left hand side to be of the form "n√(1-n^2)". Thus, we multiply each term by 1 .

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

Got it from here?

 

[haruspex]

There's probably a more elegant way, but if you simply eliminate z from each side (= x√(1-y²) + y√(1-x²)) then it should become reasonably evident.

 

[utkarshakash]

Hello again, utkarshakash! Starting from your second step...

$\frac{tan\theta + tan\alpha}{1-tan\alpha tan\theta} = -tan\beta \\ tan\theta + tan\alpha = -tan\beta + tan\alpha \ tan\theta \ tan\beta \\ tan\theta + tan\alpha + tan\beta = tan\alpha \ tan\theta \ tan\beta \\ tan(arcsinx) + tan(arcsiny) + tan(arcsinz) = tan(arcsinx) \ tan(arcsiny) \ tan(arcsinz) \\ \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} + \frac{z}{\sqrt{1-z^2}} = \frac{xyz}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}}$

Here, I think, is where you went wrong. You want each term on the left hand side to be of the form "n√(1-n^2)". Thus, we multiply each term by 1 .

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

Got it from here?

I am still not getting it. What I have to do after the last step?

 

[physicsernaw]

I'm sorry for somewhat hijacking this thread, but how do you guys (the homework helpers and the like) help people with problems such as this on a whim? I just got done with Calc1 and I wouldn't even know where to begin with this proof really. For example, I completely forgot that tan(a+b) = tan(a)+tan(b) / (1 - tan(a)tan(b)). How do you guys keep these identities fresh in your mind? Are you teachers or mathematics degree students?

I don't mention the OP because it's different learning something and applying it directly in a problem that you know involves applying what you have recently learned; I'm talking about learning something and being able to retain it long after you have learned it.