B To calculate the polar unit vectors using rotated coordinates

[brotherbobby]

TL;DR

To derive the polar unit vectors $$\boxed{\begin{align*}\hat{\rho}&=\cos\theta \;\hat{i}+\sin\theta \;{\hat j}\\

\hat{\phi}&=-\sin\theta \; \hat i+\cos\theta \; \hat j

\end{align*}}\quad\quad\large{\mathbf{(b)}}$$

using coordinates of a point under rotation : $$\begin{align*}x'&=\cos\theta \;x+\sin\theta \;y\\

y'&=-\sin\theta \; x+\cos\theta \; y

\end{align*}\quad\quad\large{\mathbf{(a)}}$$

We know that if cartesian coordinates $(x,y)$ (see figure alongside) are rotated to $(x',y')$ about the origin by an angle $\theta$ counter-clockwise as shown, the rotated coordinates are given by $$\begin{align*}x'&=\cos\theta \;x+\sin\theta \;y\\
y'&=-\sin\theta \; x+\cos\theta \; y
\end{align*}\quad\quad\large{\mathbf{(a)}}$$

$\small{\texttt{Can these (the above) be used to derive the (familiar) unit vectors using polar coordinates}}$ :

$$\boxed{\begin{align*}\hat{\rho}&=\cos\theta \;\hat{i}+\sin\theta \;{\hat j}\\
\hat{\phi}&=-\sin\theta \; \hat i+\cos\theta \; \hat j
\end{align*}}\quad\quad\large{\mathbf{(b)}}$$ I ask because the equations look similar. However, as I show in my workings below, it is far from straightforward. Their (actual) derivation is quite different and doesn't use the rotated co-ordinates shown in $\mathbf{(a)}$ above.



Discussion : The way to go from $\text{Fig. (2)}\rightarrow\text{Fig. (1)}$ is to let $\text{A}'\rightarrow\text{P}$. That would make $y'=0\Rightarrow x\sin\theta=y\cos\theta$ from $\mathbf{(a)}$ above. But this doesn't lead me anywhere towards deriving $\mathbf{(b)}$.?

Can I use the equations of $\mathbf{(a)}$ to write $x'\hat{i}'=\cos\theta x\hat i+\sin\theta y\hat j?$ If so, then dividing throughout by $x'$, we get : $\hat{i}'=\frac{x}{x'}\cos\theta\hat i+\frac{y}{x'}\sin\theta \hat j$. This would imply $\frac{x}{x'}=1=\frac{y}{x'}$ which is clearly not true.

Request : A hint as to how to derive $\mathbf{(b)}$ from $\mathbf{(a)}$.

 

[anuttarasammyak]

In Fig.1 choose $\theta$ so that P is on x' axis then x',y' are $\hat{\rho}$, $\hat{\phi}$, with renaming $\theta$ with $\phi$.