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I am attempting to find the inverse of the function: $f:S^2 \to S^2$, defined as:

$f(x,y,z)=(x\cos(z)+y\sin(z), x\sin(z)-y\cos(z),z)$

My approach so far has been to use linearity and express the function as a sum of scaled basis vectors:

$f(x,y,z)=x(\cos(z),\ sin(z),0)+y(\sin(z), -\cos(z),0)+z(0,0,1)$

However, I'm unsure how to proceed from this point. Any guidance or assistance would be greatly appreciated! Thanks.

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2 Answers 2

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Your transformation can be written under a matrix form :

$$\pmatrix{x'\\y'\\z'}=\pmatrix{\cos(z) & \sin(z) & 0\\ \sin(z) & -\cos(z) & 0\\0&0&1}\pmatrix{x\\y\\z}$$

Its inverse transformation is therefore given by the inverse matrix (the inverse being obtained by inverting the diagonal blocks) :

$$\pmatrix{x\\y\\z}=\pmatrix{\cos(z) & \sin(z) & 0\\ \sin(z) & -\cos(z) & 0\\0&0&1}\pmatrix{x'\\y'\\z'}$$

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    $\begingroup$ ... which are in fact the same formulas ! The transformation is its own inverse. $\endgroup$
    – Jean Marie
    Dec 9, 2023 at 13:22
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    $\begingroup$ Geometric-geographical explanation : this transformation sends a point onto its opposite point on the same meridian line. $\endgroup$
    – Jean Marie
    Dec 9, 2023 at 15:20
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If we write $x = \cos(\theta)$, $y = \sin(\theta)$, then

$f(x,y,z)=(\cos(z+\theta), \sin(z-\theta),z)$

which is an involution (is equal to its own inverse).

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