B To calculate the polar unit vectors using rotated coordinates

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    Polar Unit Vectors
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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)}}$$
1726582830959.png
We know that if cartesian coordinates ##(x,y)## (see figure alongside) are rotated to ##(x',y')## about the origin by an angle ##\theta## counter-clockwise as shown, the rotated coordinates are given by $$\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}}## :


1726582890040.png
$$\boxed{\begin{align*}\hat{\rho}&=\cos\theta \;\hat{i}+\sin\theta \;{\hat j}\\
\hat{\phi}&=-\sin\theta \; \hat i+\cos\theta \; \hat j
\end{align*}}\quad\quad\large{\mathbf{(b)}}$$ 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)}##.
 
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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##.
 
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