# All the Steps in Transforming the Laplace Equation ($\Delta f=0$)from Rectangular to Cylindrical and to Spherical Coordinates

Way back in January of 1972, it would have been foolish for me to begin another semester at college, since I had to report to Great Lakes Naval Training Center for Boot Camp; nor did I want to try finding some meaningless short (two months at best) job — if I even could. So, apart from spending more time with friends, I was often in the library of Harper College reading whatever interested me; which at that time was mostly about the orbitals of atoms (I had been a Chemistry major) and trying to gain a better understanding of how the Schrödinger Equation was used to arrive at these oribtals.

While doing some research on only a part of the Schrödinger Equation (called the Laplace equation), I came across this rather crazy footnote (for a textbook on mathematics) concerning the direct conversion of the Laplace Equation from rectangular to spherical coordinates:

Doing this can make you forget your troubles the next time you have a toothache at an airport and are informed that your plane is 3 hours late.

Well, with all the time I had, I took that as a challenge and proceeded to carry out (and record) each step of the conversion! All of those steps will be shown at the end of this presentation.  Much of what follows is the same as you will find in that or some other textbook on differential equations, but many intermediary steps have been included here which authors often leave out, because they were considered too trivial or basic for such a textbook.

### The Del Operator ($\nabla$), the Laplace (or Laplacian) Operator ($\nabla^2$ or $\Delta$) and the Laplace Equation

The Del Operator (represented by the nabla symbol, $\nabla$) is simply a convenient way of referring to the sum of the partial derivatives of some function with respect to a system of coordinates. For the coordinates $x, y, z,$ Del would refer to: $\newcommand{\pdif}{\frac{\partial #1}{\partial #2}}$ $\pdif{}{x} +\pdif{}{y} +\pdif{}{z}$. The Laplace Operator (represented by either $\nabla^2$ or $\Delta$) is the sum of the second order partial derviatives of some function, so for our three dimensions of $x, y, z,$ it would refer to: $\pdif{^2}{x^2} +\pdif{^2}{y^2} +\pdif{^2}{z^2}$.  And, thus, the Laplace Equation for a function $f$ (often written as $\nabla^2 f = 0$ or only $\Delta f = 0$) in rectangular (or cartesian) coordinates ($x, y, z$) when expanded is:

$$\Delta f =\pdif{^2 f}{x^2} \ +\ \pdif{^2 f}{y^2} \ +\ \pdif{^2 f}{z^2} = 0\ .$$

And here is the Laplace equation after transforming it into cylindrical ($\rho, \phi, z$) and spherical ($r, \theta, \phi$) coordinates; respectively:
\begin{align*} \nabla^{2} f & =\pdif{^2 f}{\rho^2}\ +\ \frac{1}{\rho }\ \pdif{f}{\rho }\ +\ \frac{1}{\rho^2}\ \pdif{^2 f}{\phi^2} +\pdif{^2 f}{z^2} = 0 \quad\textbf{(Equation C)}\\ & \\ \nabla^2 f & =\pdif{^2 f}{\ r^2}\ +\ \frac{2}{r}\ \pdif{f}{\ r}\ +\frac{1}{r^2}\ \pdif{^2 f}{\theta^2}\ +\frac{cos\thinspace \theta }{r^2\thinspace sin\thinspace \theta }\ \pdif{f}{\theta}\ +\ \frac{1}{r^2\thinspace sin^2\thinspace \theta}\ \pdif{^2 f}{\phi^2} = 0 \quad\ \textbf{(Equation S)} \end{align*}

## Transformations of the Laplace Operator ($\nabla^2$)

In the following pages, two different approaches will be used in transforming the Laplace operator from cartesian to cylindrical coordinates; and from there, on to spherical coordinates. Although the approaches are quite similar, almost identical in parts, they begin in a way that is opposite to each other for solving the problem. As a whole, neither approach seems to have any great advantage over the other.

## Transforming from Cartesian to Cylindrical - Approach 1

Assuming that $f$ is a function of three rectangular coordinates ($x, y, z$), which is continuous and has continuous partial derivatives of the first and second orders, we can define $\nabla^2$ by: $\nabla^2 f = \pdif{^2 f}{x^2} \ +\ \pdif{^2 f}{y^2} \ +\ \pdif{^2 f}{z^2}$, and it will first be transformed into cylindrical coordintates ($\rho, \phi, z$), which are related to the cartesian as follows:

$x = \rho \cos\phi ,\quad y = \rho \sin\phi ,\quad z=z\quad$ and $\quad\phi =\arctan\frac{y}{x} ,\quad\rho =\sqrt{x^2 + y^2}\ .$

Since a transformation is actually a special case of parametric equations; in this case: $x=x(\rho ,\phi)$; $y=y(\rho ,\phi)$; $z=z$, the chain rule can be applied, producing the following equations:

$$\pdif{f}{x} = \pdif{f}{\rho} \pdif{\rho}{x} +\pdif{f}{\phi} \pdif{\phi}{x}\ \ \textbf{(1.a)}\quad \text{and}\quad \pdif{f}{y} = \pdif{f}{\rho} \pdif{\rho}{y} +\pdif{f}{\phi} \pdif{\phi}{y}\ \ \textbf{(1.b)}$$

By differentiating $\rho$ and $\phi$ with respect to $x$ and $y$ we also have the following:

$$\pdif{\rho}{x} =\cos \phi\ ,\quad \pdif{\phi}{x} =\frac{-\sin\phi}{\rho}\ \ \textbf{(2.a1, 2.a2)}\quad \text{and}\quad \pdif{\rho}{y} =\sin \phi\ ,\quad\pdif{\phi}{y} =\frac{\cos\phi}{\rho}\ \ \textbf{(2.b1, 2.b2)}$$

Upon substitution of these derivatives above in equations (1.a) and (1.b), when $\rho >0$, we obatin:

$$\pdif{f}{x} =\cos\ \phi\ \pdif{f}{\rho} -\frac{\sin \phi}{\rho}\ \pdif{f}{\phi}\ \ \textbf{(3.a)}\quad \text{and}\quad \pdif{f}{y} =\sin\ \phi\ \pdif{f}{\rho} +\frac{\cos \phi}{\rho}\ \pdif{f}{\phi}\ \ \textbf{(3.b)}$$

To find the second order derivatives of $f$ with respect to $x$ and $y$, we use the principles of higher-order partial differentiation and the Chain Rule. By observing these principles, we can derive the necessary equations from a second differentiation of (1.a) and (1.b) as follows:

\begin{align*} \pdif{^2 f}{x^2} & =\pdif{}{x} \left(\pdif{ f}{ x}\right)_{\rho} +\pdif{}{x} \ \left(\pdif{ f}{ x}\right)_{\phi}\\& \\ & =\pdif{}{x} \left(\pdif{ f}{ \rho} \ \pdif{\rho}{x}\right) +\pdif{}{x} \left(\pdif{ f}{\phi} \ \pdif{\phi}{x}\right)\\& \\ & =\left(\pdif{^2 f}{\rho\ \partial x}\right)\ \pdif{\rho}{x} +\left(\pdif{^2 f}{\phi\ \partial x}\right)\ \pdif{\phi}{x}\\& \\ & =\left(\dfrac{\partial}{\partial\rho}\ \pdif{f}{x}\right)\ \pdif{\rho}{x} +\left(\dfrac{\partial}{\partial\phi}\ \pdif{f}{x}\right)\ \pdif{\phi}{x} \quad\textbf{(4.a)} \end{align*}

In a similar manner, the second derivative of (1.b) becomes:

$$\pdif{^2 f}{y^2} =\left(\dfrac{\partial}{\partial\rho}\ \pdif{f}{y}\right)\ \pdif{\rho}{y} +\left(\dfrac{\partial}{\partial\phi}\ \pdif{f}{y}\right)\ \pdif{\phi}{y} \quad\textbf{(4.b)}$$

By substituting the right sides of (2.a1, 2.a2) and (2.b1, 2.b2) for their left side terms in equations (4.a) and (4.b), we obtain:

\begin{align*} \pdif{^2 f}{x^2} & =\cos \phi \left(\dfrac{\partial}{\partial\rho}\ \pdif{f}{x}\right) -\ \frac{\sin \phi}{\rho} \left(\dfrac{\partial}{\partial\phi}\ \pdif{f}{x}\right) \quad\textbf{(5.a)} \quad\text{and}\\& \\ \pdif{^2 f}{y^2} & =\sin \phi \left(\dfrac{\partial}{\partial\rho}\ \pdif{f}{y}\right) +\ \frac{\cos \phi}{\rho} \left(\dfrac{\partial}{\partial\phi}\ \pdif{f}{y}\right) \quad\textbf{(5.b)} \end{align*}

And upon substitution of all the right side terms of (3.a) and (3.b) into equations (5.a) and (5.b); for which all the intermediary steps are shown further below, when $\rho > 0$, we obtain (6.a) and (6.b):

\begin{align*} \pdif{^2 f}{x^2} & =\cos^2 \phi\ \ \pdif{^2 f}{\rho^2} \colorbox{pink}{$-\frac{2\ \sin\phi\ \cos\phi}{\rho}\ \ \pdif{^2 f}{\rho\ \partial\phi}$}\ \colorbox{pink}{$+\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi}$} +\frac{\sin^2 \phi}{\rho}\ \pdif{f}{\rho} +\frac{\sin^2 \phi}{\rho^2}\ \pdif{^2 f}{\phi^2} \colorbox{pink}{$+\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi}$}\quad\textbf{(6.a)}\\& \\ \pdif{^2 f}{y^2} & =\sin^2 \phi\ \ \pdif{^2 f}{\rho^2} \colorbox{pink}{$+\frac{2\ \sin\phi\ \cos\phi}{\rho}\ \ \pdif{^2 f}{\rho\ \partial\phi}$}\ \colorbox{pink}{$-\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi}$} +\frac{\cos^2 \phi}{\rho}\ \pdif{f}{\rho} +\frac{\cos^2 \phi}{\rho^2}\ \pdif{^2 f}{\phi^2} \colorbox{pink}{$-\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi}$}\quad\textbf{(6.b)} \end{align*}

Notice that many of the terms on the right sides of the two equations above cancel each other out upon adding them together (more details may be found at the end of this section), which results in what we will call the Fundamental Identity (or Equation 7):

$$\pdif{^2 f}{x^2} +\pdif{^2 f}{y^2} \colorbox{lightgreen}{=} \pdif{^2 f}{\rho^2} +\frac{1}{\rho}\ \pdif{f}{\rho} +\frac{1}{\rho^2}\ \pdif{^2 f}{\phi^2}\quad \textbf{(7)}$$

After adding the $z$ coordinate term to Equation 7, we have the Laplace Equation in terms of cylindrical coordinates ($\rho, \phi, z$):

$$\large{\colorbox{lightblue}{\nabla^2 f =\pdif{^2 f}{\rho^2} +\frac{1}{\rho}\ \pdif{f}{\rho} +\frac{1}{\rho^2}\ \pdif{^2 f}{\phi^2} +\pdif{^2 f}{z^2} = 0}} \quad\textbf{(Equation C)}$$

## All the Intermediate and Detailed Steps for the Equations Above:

As mentioned above and in the footnotes, here are the detailed steps for the derivatives shown at (2.a1, 2.b1):

 \begin{align*} \colorbox{yellow}{$\pdif{\rho}{x}$} & =\pdif{\sqrt{x^2 + y^2}}{\ x} =\left(\dfrac{1}{2}\right) \ (2x)\ \ \left( x^{2} \ +\ y^{2}\right)^{-1/2}& \\ & =\left(\dfrac{x}{\sqrt{x^2 + y^2}}\right) =\frac{\rho \cos\phi}{\rho} = \colorbox{yellow}{$\cos\phi$}\ , \quad\text{(2.a1)} \end{align*} \begin{align*} \colorbox{yellow}{$\pdif{\rho}{y}$} & =\pdif{\sqrt{x^2 + y^2}}{\ y} =\left(\dfrac{1}{2}\right) \ (2y)\ \ \left( x^{2} \ +\ y^{2}\right)^{-1/2}& \\ & =\left(\dfrac{y}{\sqrt{x^2 + y^2}}\right) =\frac{\rho \sin\phi}{\rho} = \colorbox{yellow}{$\sin\phi$}\ . \quad\text{(2.b1)} \end{align*}

Where we have, of course, used: $x =\rho \cos \phi ,\quad y =\rho \sin \phi$  and  $\rho =\sqrt{x^2 + y^2}$  and the more complicated details for (2.a2, 2.b2); using $\phi =\arctan (y/x)$ and some trigonometric identities, along with these differentiation rules:

$$\frac{d}{dx} (\arctan\ u) =\frac{1}{1 + u^2}\ \frac{du}{dx}, \quad\frac{d}{dx} \left(\frac{u}{w}\right) =\frac{w\frac{du}{dx} - u\frac{dw}{dx}}{w^2}, \quad\colorbox{pink}{\frac{dx}{dy}(y) =\frac{dy}{dx}(x) = 0}\quad\text{allows us to arrive at:}$$
 \begin{align*} \colorbox{yellow}{$\pdif{\phi}{x}$} & =\pdif{\ [\arctan\ (y/x)]}{\ x} =\colorbox{lightgreen}{$\frac{1}{1 + (y/x)^2}$}\ \pdif{\ (y/x)}{\ x} =\colorbox{lightgreen}{$\frac{1}{1 + \tan^2 \phi}$}\ \pdif{\ (y/x)}{\ x}\quad\\ & =\colorbox{lightgreen}{$\frac{1}{\sec^2 \phi}$}\ \pdif{\ (y/x)}{\ x} =\frac{\colorbox{lightgreen}{$\cos^2 \phi$}\ (\colorbox{pink}{$x\ \pdif{y}{x}$} - y)}{x^2} =\frac{\cos^2\phi\ (-\rho\sin\phi)}{x^2}\\ & =\frac{-\rho\ \sin\phi\ \cos^2 \phi}{x^2} =\frac{(- \sin \phi)\ \rho\ \cos^2 \phi}{\rho^2\ \cos^2 \phi} = \colorbox{yellow}{$-\frac{\sin\ \phi}{\rho}$}, \quad\textbf{(2.a2)} \end{align*} \begin{align*} \colorbox{yellow}{$\pdif{\phi}{y}$} & =\frac{(\cos^2\ \phi)\ (x - \colorbox{pink}{$y \pdif{\ x}{\ y}$})}{x^2}\quad\\ & =\frac{(\cos^2 \phi)\ \rho\cos\phi}{\rho^2\ \cos^2\phi}\\ & = \colorbox{yellow}{$\frac{\cos\ \phi}{\rho}$} .\quad\textbf{(2.b2)}\end{align*}

The following rule and derivative substitutions were applied in obtaining Equation (6.a):

$$\colorbox{yellow}{\frac{d}{dx}\ (uw) = u\ \frac{dw}{dx} + w\ \frac{du}{dx}}\ , \quad\pdif{\ (-\sin \phi/\rho)}{\ \rho} =\frac{(-1) (-\sin\phi)}{\rho^2} =\frac{\sin \phi}{\rho^2} ,$$ $$\colorbox{pink}{\pdif{\ (\cos \phi)}{\ \rho} = 0}\ , \quad\pdif{\ (\cos \phi)}{\ \phi} = -\sin\phi\ ,\quad\pdif{\ (-\sin\phi/\rho)}{\ \phi} = -\frac{\cos \phi}{\rho}\ .$$
 \begin{align*} \pdif{^2 f}{x^2} & = \cos \phi\ \left[\pdif{}{\rho} \left(\cos \phi\ \pdif{f}{\rho} - \frac{\sin \phi}{\rho}\ \pdif{f}{\phi}\right) \right] - \frac{\sin \phi}{\rho} \left[\pdif{}{\phi} \left(\cos\phi\ \pdif{f}{\rho} - \frac{\sin\phi}{\rho} \pdif{f}{\phi}\right) \right]\\ & =\cos\phi\ \left(\cos\phi\ \pdif{^2 f}{\rho^2} - \frac{\sin\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi} + \frac{\sin\phi}{\rho^2}\ \pdif{f}{\phi} \right) - \frac{\sin\phi}{\rho} \left(\cos\phi\ \pdif{^2 f}{\rho\ \partial\phi} - \sin\phi\ \pdif{f}{\rho} - \frac{\sin\phi}{\rho}\ \pdif{^2 f}{\phi^2} - \frac{\cos\phi}{\rho}\ \pdif{f}{\phi}\right)\\ & =\cos^2 \phi\ \pdif{^2 f}{\rho^2} \colorbox{lightgreen}{$-\frac{\sin\phi\ \cos\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi}$} + \frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi} + \frac{\sin^2\phi}{\rho}\ \pdif{f}{\rho} \colorbox{lightgreen}{$-\frac{\sin\phi\ \cos\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi}$} + \frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi} + \frac{\sin^2\phi}{\rho^2}\ \pdif{^2 f}{\phi^2}\\ & =\cos^2 \phi\ \pdif{^2 f}{\rho^2} \colorbox{lightgreen}{$-\frac{2\ \sin\phi\ \cos\phi}{\rho}\ \ \pdif{^2 f}{\rho\ \partial\phi}$} +\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi} +\frac{\sin^2 \phi}{\rho}\ \pdif{f}{\rho} +\frac{\sin^2 \phi}{\rho^2}\ \pdif{^2 f}{\phi^2} +\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi} \quad\textbf{(6.a)} \end{align*}

Likewise, Equation (6.b) was obtained by applying the same differentiation rule above and these substitutions in the equations below:

$$\pdif{\ (\cos\phi / \rho)}{\rho} = \frac{(-1)(\cos\phi)}{\rho^2} = -\frac{\cos\phi}{\rho^2}\ , \quad\colorbox{pink}{\pdif{\ (\sin\phi)}{\ \rho} = 0}\ ,$$ $$\pdif{\ (\sin\phi)}{\ \phi} = \cos\phi\ , \quad\pdif{\ (\cos\phi / \rho)}{\ \phi} = - \frac{\sin \phi}{\rho}\ .$$
 \begin{align*} \pdif{^2 f}{y^2} & =\sin\phi \left[\pdif{}{\rho} \left(\sin\phi\ \pdif{f}{\rho} + \frac{\cos\phi}{\rho}\ \pdif{f}{\phi}\right)\right] + \frac{\cos\phi}{\rho} \left[\pdif{}{\phi} \left(\sin\phi\ \pdif{f}{\rho} + \frac{\cos\phi}{\rho}\ \pdif{f}{\phi}\right)\right]\\ & =\sin\phi \left(\sin\phi\ \pdif{^2 f}{\rho^2} + \frac{\cos\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi} - \frac{\cos\phi}{\rho^2}\ \pdif{f}{\phi}\right) + \frac{\cos\phi}{\rho} \left(\sin\phi\ \pdif{^2 f}{\rho\ \partial\phi} + \cos\phi\ \pdif{f}{\rho} + \frac{\cos\phi}{\rho}\ \pdif{^2 f}{\phi^2} - \frac{\sin\phi}{\rho}\ \pdif{f}{\phi}\right)\\ & =\sin^2\phi\ \pdif{^2 f}{\rho^2} \colorbox{lightgreen}{$+ \frac{\sin\phi\ \cos\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi}$} - \frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi} \colorbox{lightgreen}{$+ \frac{\sin\phi\ \cos\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi}$} + \frac{\cos^2\phi}{\rho}\ \pdif{f}{\rho} + \frac{\cos^2\phi}{\rho^2}\ \pdif{^2 f}{\phi^2} - \frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi}\\ & =\sin^2 \phi\ \ \pdif{^2 f}{\rho^2} \colorbox{lightgreen}{$+\frac{2\ \sin\phi\ \cos\phi}{\rho}\ \pdif{^2 f}{\rho\ \partial\phi}$} -\frac{\sin\phi\ \cos\phi}{\rho^2}\ \ \pdif{f}{\phi} +\frac{\cos^2 \phi}{\rho}\ \pdif{f}{\rho} +\frac{\cos^2 \phi}{\rho^2}\ \pdif{^2 f}{\phi^2} -\frac{\sin\phi\ \cos\phi}{\rho^2}\ \pdif{f}{\phi}\ \quad\textbf{(6.b)} \end{align*}

Note: Apart from the terms which cancel out (already shown in pink in the body of the text above), the Fundamental Identity (Equation 7) is arrived at using the trigonometric identity $\colorbox{lightgreen}{$\sin^2\phi +\cos^2\phi = 1$}$ on three more sets of those terms:

 \begin{align*} \pdif{^2 f}{x^2} +\pdif{^2 f}{y^2} & \colorbox{yellow}{$=$}\left(\cos^2\phi +\sin^2\phi\right)\ \pdif{^2 f}{\rho^2} +\frac{\colorbox{lightgreen}{$\sin^2\phi +\cos^2\phi$}}{\rho}\ \pdif{f}{\rho} +\frac{\colorbox{lightgreen}{$\sin^2\phi +\cos^2\phi$}}{\rho^2}\ \pdif{^2 f}{\phi^2}\\ \\ \pdif{^2 f}{x^2} +\pdif{^2 f}{y^2} & \colorbox{yellow}{$=$}\pdif{^2 f}{\rho^2} +\frac{1}{\rho}\ \pdif{f}{\rho} +\frac{1}{\rho^2}\ \pdif{^2 f}{\phi^2} \quad\textbf{Equation (7)} \end{align*}

# Continue to Approach 2

 For those interested in how the math symbols (such as: $\nabla^2 f = \Delta f = 0$) are being displayed, I recently started learning how to use LaTeX commands with MathJaX. None of the equations or math symbols above are made from pictures or text, but if you right-click on them and choose "Show Math As" and select "TeX Commands," you’ll see the code which displays them! Note: The \pdif{}{} code on these pages is not a LaTeX command, but rather a macro I defined with the \newcommand to remove almost all occurrences of \partial from the code.

Footnotes

1[Return to Text] For those interested in the Schrödinger Equation, there are two webpages I’d recommend for further study:

Schrödinger’s equation — what is it? (from a Maths perspective at Cambridge University), and:
The Schrödinger equation, particle in a box, and atomic wavefunctions (from an interactive Chemistry Textbook at Cal State, Fullerton).

One of the books I was reading at that time was Inorganic Chemistry by R.T. Sanderson; no longer in that library, and in both that work and my textbook from 1970, Chemistry Principles, on page 147, a form of the Schrödinger Equation for the time independent version of the wave function ($\psi$) in three dimensions ($x, y, z$) for the electron in a Hydrogen atom is: $$\pdif{^2\psi}{x^2}\ +\pdif{^2\psi}{y^2}\ +\pdif{^2\psi}{z^2}\ +\frac{8\pi^2 m}{h^2}\ (E\ +\frac{e^2}{r})\ \psi\ = 0$$ where $m$ is the mass of the electron, $h$ is Planck’s constant, $E$ the total energy, $e$ the charge of the electron and $r$ its distance from the nucleus. One can clearly see upon comparison that the first 3 terms are exactly the same as the Laplace Equation; thus, the reason Schrödinger’s equation can also be converted to spherical coordinates.

2[Return to Text] Ralph Palmer Agnew, Differential Equations, 2nd Ed. (New York: McGraw-Hill Book Co., Inc., 1960), p. 148.

3[Return to Text] The differentiation steps in Approach 1 are of a more complex nature than Approach 2; where the first order derivatives ($\pdif{f}{x}, \pdif{f}{y}$) are derived from sets of equations. However, the steps for obtaining the second order derivatives ($\pdif{^2 f}{x^2}, \pdif{^2 f}{y^2}$) are the same for both approaches.

5[Return to Text] The specific trigonometric identities we made use of were:  $(\sin\phi / \cos\phi) =\tan\phi$,  $1 +\tan^2\phi =\sec^2\phi$  and  $\sec\phi = (1 / \cos\phi)$, so:  $\sec^2\phi = (1 / \cos^2\phi)$  or:  $(1 /\sec^2\phi) =\cos^2\phi$.

First Posted on: 30 October 2023 (2023.10.30).
Updated on:
4 NOV 2023 (2023.11.04); minor corrections, typos and began adding an Index, 5 NOV 2023 (2023.11.05); revised page layout for Index section, 7 NOV 2023 (2023.11.07); clarified a few statements and added color highlighting, 8 NOV 2023 (2023.11.08); further clarifications, page finished, 10 NOV 2023 (2023.11.10); added specific trig identities used and more clarifications through color highlighting, 17 NOV 2023 (2023.11.17); reduced page size by using macro to replace "{\partial}" code, 18 NOV 2023 (2023.11.18); minor clarifications.

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