Prove that $g(r, \alpha, \beta) = (r\cos(\alpha)\cos(\beta), r\sin(\alpha)\cos(\beta), r\sin(\beta))$ is an injection.

Question

Prove that

$g: (0,∞)×(0,2π)×(−π/2,π/2)→R^3,$

$g(r, \alpha, \beta) = (r\cos(\alpha)\cos(\beta), r\sin(\alpha)\cos(\beta), r\sin(\beta))$

is an injection.

Exactly what it says on the tin. I haven't gone beyond assuming $g(a,b,c) = g(c,d,e)$. After constructing the relevant system of equations I am stuck trying to equate the variables from there. I might have forgotten too much from trigonometry. This is homework, so I would appreciate if someone could walk me through the process, but not the solution.

Answer 1

hint

Put $$g(a,b,c)=(A,B,C)$$ $$g(d,e,f)=(D,E,F)$$

then

$$g(a,b,c)=g(d,e,f) \implies$$

$$(A,B,C)=(D,E,F)\implies$$

$$A^2+B^2+C^2=D^2+E^2+F^2\implies$$ $$a^2=d^2\implies a=d$$

Answer 2

First consider $f(x,y,z)=x^2+y^2+z^2$ and observe that $$ f(g(r,\alpha,\beta))=r^2\cos^2\alpha\cos^2\beta+r^2\sin^2\alpha\cos^2\beta+r^2\sin^2\beta=r^2 $$ Therefore if $g(r,\alpha,\beta)=g(s,\gamma,\delta)$, then applying $f$ to both sides yields $r^2=s^2$, hence $r=s$.

Now the equations are simpler: $$ \begin{cases} \cos\alpha\cos\beta=\cos\gamma\cos\delta \\[1ex] \sin\alpha\cos\beta=\sin\gamma\cos\delta \\[1ex] \sin\beta=\sin\delta \end{cases} $$ Square the first two equations and sum them, getting $\cos^2\beta=\cos^2\delta$. Since $\beta,\delta\in(-\pi/2,\pi/2)$, we obtain $\cos\beta=\cos\delta$ and now with the third equation we can conclude that $\beta=\delta$.

Finally, from $\cos\beta=\cos\delta\ne0$, we obtain $\cos\alpha=\cos\gamma$ and $\sin\alpha=\sin\gamma$, so $\alpha=\gamma$.