Coordinate systems/Derivation of formulas/Spherical from Cartesian unit vectors

From Wolfram Mathworld, we have the following relations for the unit vectors in a Spherical coordinate system:

$\begin{array}{cc}\widehat{𝐫}& =\cos \varphi \sin \theta \widehat{𝐱}+\sin \varphi \sin \theta \widehat{𝐲}+\cos \theta \widehat{𝐳}\\ \hat{\mathit{\theta }}& =\cos \varphi \cos \theta \widehat{𝐱}+\sin \varphi \cos \theta \widehat{𝐲}-\sin \theta \widehat{𝐳}\\ \hat{\mathit{\varphi }}& =-\sin \varphi \widehat{𝐱}+\cos \varphi \widehat{𝐲}\end{array}$

Next, we need to change all the $\varphi$ and $\theta$ to $x,y,z$. We use the following relations, also available on the same page:

$\begin{array}{cc}r& =\sqrt{x^2+y^2+z^2}\\ \theta & =\arccos (\frac{z}{r})\\ \varphi & =\arctan (\frac{y}{x})\end{array}$

Using these, we can find the following quantities:

$\begin{array}{cc}\cos \theta & =\frac{z}{r}=\frac{z}{\sqrt{x^2+y^2+z^2}}\\ \sin \theta & =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{z^2}{x^2+y^2+z^2}}=\sqrt{\frac{x^2+y^2}{x^2+y^2+z^2}}\\ \cos \varphi & =\frac{x}{\sqrt{x^2+y^2}}\\ \sin \varphi & =\frac{y}{\sqrt{x^2+y^2}}\end{array}$

Lastly, we simply need to substitute these relations into our first set of equations to obtain our desired result:

$\begin{array}{cc}\widehat{𝐫}& =\frac{x}{\sqrt{x^2+y^2}}\sqrt{\frac{x^2+y^2}{x^2+y^2+z^2}}\widehat{𝐱}+\frac{y}{\sqrt{x^2+y^2}}\sqrt{\frac{x^2+y^2}{x^2+y^2+z^2}}\widehat{𝐲}+\frac{z}{\sqrt{x^2+y^2+z^2}}\widehat{𝐳}=\frac{x\widehat{𝐱}+y\widehat{𝐲}+z\widehat{𝐳}}{\sqrt{x^2+y^2+z^2}}\\ \hat{\mathit{\theta }}& =\frac{x}{\sqrt{x^2+y^2}}\frac{z}{\sqrt{x^2+y^2+z^2}}\widehat{𝐱}+\frac{y}{\sqrt{x^2+y^2}}\frac{z}{\sqrt{x^2+y^2+z^2}}\widehat{𝐲}-\sqrt{\frac{x^2+y^2}{x^2+y^2+z^2}}\widehat{𝐳}=\frac{xz\widehat{𝐱}+yz\widehat{𝐲}-(x^2+y^2)\widehat{𝐳}}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\\ \hat{\mathit{\varphi }}& =-\frac{y}{\sqrt{x^2+y^2}}\widehat{𝐱}+\frac{x}{\sqrt{x^2+y^2}}\widehat{𝐲}=\frac{-y\widehat{𝐱}+x\widehat{𝐲}}{\sqrt{x^2+y^2}}\end{array}$