Understanding why $\sqrt{x^2+y^2} \cos z=x$ and $\sqrt{x^2+y^2} \sin z=y$

Question

Consider the map $T:(0, \infty) \times \mathbb{R} \rightarrow \mathbb{R}^3$, given by $$\tag{1} T(a, b)=(a \cos b, a \sin b, b) $$ and the map $V$ defined from the image of $T$ into $(0,\infty)$ by $$\tag{2} V(x, y, z)=\left(\sqrt{x^2+y^2}, z\right) $$ My teacher uses in a proof that for any $p\in (0,\infty)$ and any $q$ in the image of $T$ the following holds: $$\tag{3} V(T(p))=p,\qquad T(V(q))=q $$ The left equality is simple to prove using $\sin^2+\cos^2=1$. Consider then an arbitrary $q=(x,y,z)$ in the image of $T$, and the right equality becomes: $$\tag{4} T(V(q))=T\left(\sqrt{x^2+y^2}, z\right)=\left(\sqrt{x^2+y^2} \cos z, \sqrt{x^2+y^2} \sin z, z\right) $$ Eq. $(3)$ then implies that we must have $$\tag{5} \sqrt{x^2+y^2} \cos z=x,\quad \sqrt{x^2+y^2} \sin z=y $$ Can someone help me understand why the relations in eq. $(5)$ hold?

Answer 1

This is because (3) on the right tells you that $T(V(q)) = q$ by setting $q=(x,y,z)$ you get $T(V(x,y,z)) = (x,y,z)$ then using the equality in (4) you get $(x,y,z) = T(V(x,y,z)) = (\sqrt{x^2+y^2}\cos{z},\sqrt{x^2+y^2}\sin{z},z)$

Looking at the components this gives you $x = \sqrt{x^2+y^2}\cos{z}$ and $y = \sqrt{x^2+y^2}\sin{z}$

This of course relies on eq. (3)

If you want to prove (3) you just have to take q in the Image of T then $q = (x,y,z) = (a \cos{b} , a \sin{b}, b)$, this is because q is not any value in $\mathbb{R}^3$ but in the image of T.

This gives you by (4) that $T(V(q)) = (\sqrt{(a \cos{b})^2+(a \sin{b})^2}\cos{b},\sqrt{(a \cos{b})^2+(a \sin{b})^2}\sin{b},b)$ Now $sin^2 + cos ^2 = 1$ gives you by the distributivity ($(a \cos{b})^2+(a \sin{b})^2 = a^2*((\cos{b})^2+(\sin{b})^2$) $T(V(q)) = (\sqrt{a^2}\cos{b},\sqrt{a^2}\sin{b},b) = (a\cos{b},a\sin{b},b)$