Prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$.

Question

We have

$$x\cos \theta+y\cos \phi = -z\cos \psi \tag 1$$

$$x\sin \theta+y\sin \phi = -z\sin \psi \tag 2$$

$$x\sec \theta+y\sec \phi = -z\sec \psi \tag 3$$

and we have to prove that $$(x^2 + y^2 - z^2)^2 = 4x^2y^2$$

squaring (1) & (2), and adding them we have

$$x^2 + y^2 - z^2 = -2xy \cos(\theta - \phi)$$

multiplying (1) & (3), we have

$$x^2 + y^2 - z^2 = -xy (\cos\theta \sec\phi + \cos\phi \sec\theta)$$

from here I can't move forward

help me

Are $\phi$, $\theta$ and $\psi$ related to $x$, $y$ and $z$ in any way? In general that claim won't hold. — Henrik supports the community, Aug 14 '17 at 18:18
Taking the square root of both sides on $(x^2 + y^2 - z^2)^2 = 4x^2y^2$ immediately gives $z^2=(x-y)^2$ — James Arathoon, Aug 14 '17 at 18:29
In general, the only solution is zero but if all angles are equal then $x+y+z=0$. It may help to know $(x^2+y^2-z^2)^2-4x^2y^2=(x - y - z) (x + y - z) (x - y + z) (x + y + z)$. — Somos, Aug 14 '17 at 19:36

score 1 · Answer 1 · answered Aug 14 '17 at 19:54

We denote $a,b,c$ the angles $\theta, \phi, \psi$. The condition is $$\begin{bmatrix} \cos a &\cos b &\cos c\\\sin a &\sin b &\sin c\\\sec a &\sec b &\sec c\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ The determinant of the matrix is equal to $$\frac{\sin(b-a)}{\cos c}+\frac{\sin(a-c)}{\cos b}+\frac{\sin(c-b)}{ \cos a}$$ This determinant is in general non-zero as it is easy to verify.

Therefore the solution of the linear system above has a unique solution which is the obvious one $x=y=z=0$ in which case the equality to prove is trivially true.

The conclusion is that it remains to add some condition on the angles.

James Arathoon · Answer 2 · 2017-08-14T19:57:18.093

multiplying (2) & (3), we have

$$x^2\tan\theta+y^2\tan\phi+xy\left(\sin\theta \sec\phi+\sin\phi\sec\theta \right)=z^2\tan\psi$$

Therefore in simplest terms $\tan\theta=1$, $\tan\phi=1$ and $\tan\psi=1$ whilst at the same time $\cos(\theta - \phi) =-1$ from squaring (1) and (2).

One solution that meets these constraints is $\theta=\frac{\pi}{4}$, $\phi=-\frac{3\pi}{4}$ and $\psi=-\frac{3\pi}{4}$ which just indicates a fixed phase relationship between $\theta$, $\phi$ and $\psi$ I think.

If there is a shared angle of rotation, $\alpha$, the actual solution will be in the form $\theta=\frac{\pi}{4}+\alpha$, $\phi=-\frac{3\pi}{4}+\alpha$ and $\psi=-\frac{3\pi}{4}+\alpha$.

score 0 · Answer 3 · answered Aug 09 '23 at 10:20

Multiply your last two eequations and you have:

$$(x^2 + y^2 - z^2)^2 = 2x^2y^2\cos(\theta-\phi)(\cos\theta \sec\phi + \cos\phi \sec\theta)$$

So to get the result you want, you need:

$$\cos(\theta-\phi)(\cos\theta \sec\phi + \cos\phi \sec\theta)=2$$

This is an implicit relationship between $\theta$ and $\phi$. It includes the line $\theta=\phi$ but also more. Without this relation, your desired result does not hold.

Prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$.

3 Answers3