I can't follow this simplification: arccos(z sqrt[x^2+y^2+z^2]/sqrt[(x^2+z^2)(y^2+z^2

[Jarama]

Could someone explain to me how the author of the book I'm reading managed to perform this simplification?

\(\displaystyle \cos^{-1}\, \left(\, \dfrac{z\, \cdot\, \sqrt{\strut x^2\, +\, y^2\, +\, z^2\,}}{\sqrt{\strut (x^2\,+\, z^2)\, \cdot\, (y^2\,+\, z^2)}}\, \right)\, =\, \sin^{-1}\, \left(\, \dfrac{x\, \cdot\, y}{\sqrt{\strut (x^2\, +\, z^2)\, \cdot\, (y^2\, +\, z^2)}}\, \right)\)

It is the last step in a large series of simplifications that I all managed to understand, but here he lost me. Unfortunately I need to understand it completely, because I need to perform it myself in a slightly different context.

Thanks in advance!

 

[Deleted member 4993]

Could someone explain to me how the author of the book I'm reading managed to perform this simplification?

It is the last step in a large series of simplifications that I all managed to understand, but here he lost me. Unfortunately I need to understand it completely, because I need to perform it myself in a slightly different context.

Thanks in advance!

Please define x, y and z.

 

[Jarama]

Please define x, y and z.

They are distances. I use the formula to calculate the solid angle of a rectangle with a width x and a height y at a distance z perpendicular to one of the vertices. See this sketch (but here x is called l and y is called w​):

 

[Steven G]

Could someone explain to me how the author of the book I'm reading managed to perform this simplification?

\(\displaystyle \cos^{-1}\, \left(\, \dfrac{z\, \cdot\, \sqrt{\strut x^2\, +\, y^2\, +\, z^2\,}}{\sqrt{\strut (x^2\,+\, z^2)\, \cdot\, (y^2\,+\, z^2)}}\, \right)\, =\, \sin^{-1}\, \left(\, \dfrac{x\, \cdot\, y}{\sqrt{\strut (x^2\, +\, z^2)\, \cdot\, (y^2\, +\, z^2)}}\, \right)\)

It is the last step in a large series of simplifications that I all managed to understand, but here he lost me. Unfortunately I need to understand it completely, because I need to perform it myself in a slightly different context.

Thanks in advance!

You know that the arc cos( . ) = theta, ie cos (theta) = ( . ). Well if you know what the cos (theta) equals you should be able to compute the sin (theta) and .....

 

[Bob Brown MSEE]

Right Triangle

Could someone explain to me how the author of the book I'm reading managed to perform this simplification?

\(\displaystyle \cos^{-1}\, \left(\, \dfrac{z\, \cdot\, \sqrt{\strut x^2\, +\, y^2\, +\, z^2\,}}{\sqrt{\strut (x^2\,+\, z^2)\, \cdot\, (y^2\,+\, z^2)}}\, \right)\, =\, \sin^{-1}\, \left(\, \dfrac{x\, \cdot\, y}{\sqrt{\strut (x^2\, +\, z^2)\, \cdot\, (y^2\, +\, z^2)}}\, \right)\)

It is the last step in a large series of simplifications that I all managed to understand, but here he lost me. Unfortunately I need to understand it completely, because I need to perform it myself in a slightly different context.

Thanks in advance!

Consider a triangle with sides...
\(\displaystyle A=z \sqrt{x^2+y^2+z^2}\)
\(\displaystyle B=x y\)
\(\displaystyle C=\sqrt{\left(x^2+z^2\right) \left(y^2+z^2\right)}\)

Prove that it is a right triangle by showing that...
\(\displaystyle C^2=A^2+B^2\)

 

[Jarama]

Consider a triangle with sides...
\(\displaystyle A=z \sqrt{x^2+y^2+z^2}\)
\(\displaystyle B=x y\)
\(\displaystyle C=\sqrt{\left(x^2+z^2\right) \left(y^2+z^2\right)}\)

Prove that it is a right triangle by showing that...
\(\displaystyle C^2=A^2+B^2\)

Thank you very much! I can't believe that it was so simple... I was lost in using all kinds of crazy goniometric rules.

 

[pka]

Could someone explain to me how the author of the book I'm reading managed to perform this simplification?

\(\displaystyle \cos^{-1}\, \left(\, \dfrac{z\, \cdot\, \sqrt{\strut x^2\, +\, y^2\, +\, z^2\,}}{\sqrt{\strut (x^2\,+\, z^2)\, \cdot\, (y^2\,+\, z^2)}}\, \right)\, =\, \sin^{-1}\, \left(\, \dfrac{x\, \cdot\, y}{\sqrt{\strut (x^2\, +\, z^2)\, \cdot\, (y^2\, +\, z^2)}}\, \right)\)

It is the last step in a large series of simplifications that I all managed to understand, but here he lost me. Unfortunately I need to understand it completely, because I need to perform it myself in a slightly different context.

It seems that you have asked us to explain something with only being put down into the middle of an explanation. We should know the whole context. You should read this webpage. Note that the \(\displaystyle \arccos~ \&~\arcsin\) have different ranges that only intersect on \(\displaystyle [0,\frac{\pi}{2}]\).