Visualizing why a right-angle rotation formula works in polar coordinates

I am trying to get a solid and intuitive handle on polar and spherical coordinates, and I'm getting stuck with what I think should be simple geometry:

To find the unit vector in Cartesian coordinates $\hat{\rho}$. It's very intuitive, I draw a y-axis and x-axis, and I see that $\hat{\rho} = \cos(\phi)\hat{x} + \sin(\phi)\hat{y}$

But I'm sitting here trying to reach the known representation of unit vector $$\hat{\phi} = -\sin(\phi)\hat{x} + \cos(\phi)\hat{y}$$ and I'm not getting it with simple geometry.

I often reach $-\cos(\phi)\hat{x} + \sin(\phi)\hat{y}$ which is just the negative of the $x$ unit vector.

I know how to verify this using the known relationship between $\hat{\phi}$ unit vector and $\hat{\rho}$ unit vector being perpendicular, and therefore their scalar product must equal $0$. And yet why is it not as simple and intuitive finding the $\hat{\phi}$ unit vector as it was $\hat{\rho}$? Could be I've just reached a mental block here.

I don't want to reach this by "guessing" coefficients for $\hat{x}$ and $\hat{y}$ and then verifying with scalar product equaling zero, I want to see this geometrically.

Thanks!

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I just give a geometric explanation:

the figure clearly shows that $p$ is a rotation of $\hat\rho$ by $90^\circ$, and the relation between their component are marked.

I would also like to share the code the the picture: \begin{tikzpicture}[scale=2.5] \draw[->,thick] (-3,0)--(3,0) node[right]{$\hat x$}; \draw[->,thick] (0,0)--(0,3)node[right]{$\hat y$}; \draw[->,thick] (0,0)--(30:2cm) node[anchor=south west]{$\hat\rho=\cos\phi \hat x+\sin\phi\hat y$}; \draw[dotted,thick] (1.73,0)node[anchor=north]{$\cos\phi$}--(30:2cm)--(0,1)node[left]{$\sin\phi$}; \draw[->] (1cm,0) arc (0:30:1cm); \draw (15:1cm)node[right]{$\phi$}; \draw[->,thick, blue] (0,0)--(120:2cm)node[anchor=south east]{$p=-\sin\phi\hat x+\cos\phi\hat y$}; \draw[->] (.7cm,0) arc (0:120:.7cm); \draw (60:.5cm)node{$\theta$}; \draw[dotted,thick] (0,1.73)node[right]{$\cos\phi$}-- (120:2cm)--(-1,0)node[anchor=north]{$-\sin\phi$}; \draw (-2,-1)node[right]{$\theta=90^\circ+\phi$}; \draw (-2,-1.2)node[right] {$p=\cos\theta\hat x+\sin\theta\hat y =-\sin\phi\hat x+\cos\phi\hat y$}; \end{tikzpicture}

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The usual way to define the unit vector $\hat\phi$ is as the normalized derivative of the position vector with respect to $\phi$. The position vector is $\cos\phi\hat x+\sin\phi\hat y$, and differentiating with respect to $\phi$ yields $-\sin\phi\hat x+\cos\phi\hat y$ as expected.

If you want to understand this more geometrically, this is the direction along which the position moves if you change $\phi$. This direction must be counterclockwise and at right angles to the radial direction. You don't have to guess to find such a direction; the unit vectors perpendicular to a unit vector $(a,b)$ are $(b,-a)$ and $(-b,a)$, and the sign must be as it is for the direction to be counterclockwise.

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How about I draw a picture?

$\hskip 1.2in$

The green line segment on the right represents $\hat{\rho}$, and the one on the left $\hat{\phi}$ (which is $\rho$ rotated $90^\circ$ counterclockwise). The blue lines are added haphazardly to show the two green lines are part of an imaginary square, hence the angle between them is indeed right. In order to go from $\hat{\rho}$ to $\hat{\phi}$, we first reflect it across the gray line (which is the graph of $y=x$), and then reflect it across the $y$-axis.

The first action switches the $x$ and $y$ coordinates, while the second action negates the $x$-coordinate.

$$(\cos\phi,\sin\phi) \mapsto (\sin\phi,\cos\phi) \mapsto (-\sin\phi,\cos\phi).$$

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