RajRaizada

Hello. I'm a math and computer science educator based in NYC. For several years, I was a faculty member in cognitive neuroscience at the University of Rochester, but I gradually got fed up with that and moved into school teaching, and I'm very glad that I did.

I recently (Spring 2022) started making some additions and edits to math pages on Wikipedia, with the aim of making them clearer and easier to understand. Quite a few of the math pages could benefit from some improvement on that front, as they tend to be a bit too dry and technical, and they often don't have enough illustrative pictures.

Although Wikipedia isn't meant to be a tutorial textbook, it is supposed to be accessible to a broad general audience. So, I'm trying to help in a small way with that. Also, it's a nice feeling to be improving (or at least trying to improve!) the quality of webpages that get tens of thousands of page views a month, and which often show up at the top of a Google search on the particular topic in question.

Here is a collection of the edits and additions that I've made:

Pythagorean theorem

Visual proof of the Pythagorean theorem by area-preserving shearing.

Proof by area-preserving shearing

As shown in the accompanying animation, area-preserving shear mappings and translations can transform the squares on the sides adjacent to the right-angle onto the square on the hypotenuse, together covering it exactly. Each shear leaves the base and height unchanged, thus leaving the area unchanged too. The translations also leave the area unchanged, as they do not alter the shapes at all. Each square is first sheared into a parallelogram, and then into a rectangle which can be translated onto one section of the square on the hypotenuse.

(Also on this page, there was an animated SVG of the rearrangement proof that wasn't actually showing up as animated until you clicked on it. It was a nice animation, so this struck me as unfortunate. Using some web-based tools, I turned the animated SVG into an animated gif, so now the webpage shows that instead.)

Angle bisector theorem

Animated illustration of the angle bisector theorem.

As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle $\triangle ABC$ gets reflected across a line that is perpendicular to the angle bisector $AD$, resulting in the triangle $\triangle AB_{2}C_{2}$ with bisector $AD_{2}$. The fact that the bisection-produced angles $\angle BAD$ and $\angle CAD$ are equal means that $BAC_{2}$ and $CAB_{2}$ are straight lines. This allows the construction of triangle $\triangle C_{2}BC$ that is similar to $\triangle ABD$. Because the ratios between corresponding sides of similar triangles are all equal, it follows that $|AB|/|AC_{2}|=|BD|/|CD|$. However, $AC_{2}$ was constructed as a reflection of the line $AC$, and so those two lines are of equal length. Therefore, $|AB|/|AC|=|BD|/|CD|$, yielding the result stated by the theorem.

Ptolemy's theorem

Visual proof

Animated visual proof of Ptolemy's theorem, based on Derrick & Herstein (2012).

The animation here shows a visual demonstration of Ptolemy's theorem, based on Derrick & Herstein (2012).

List of trigonometric identities

(I first looked at this page while I was teaching Precalculus, which includes trig identities as part of its curriculum. I noticed that the page had far too few pictures on it. Any topic involving geometry really needs pictures. I enjoy making figures in Desmos, so I added some made with that, along with explanatory captions. If you click through on the diagrams to their homes on Wikimedia Commons, the descriptions there should link to the Desmos graphs that were used to make them.)

Ptolemy's theorem

Diagram illustrating the relation between Ptolemy's theorem and the angle sum trig identity for sine.

Main article: Ptolemy's theorem

Ptolemy's theorem is important in the history of trigonometric identities, as it is how results equivalent to the sum and difference formulas for sine and cosine were first proved (see the section on classical antiquity in the page History of trigonometry). It states that in a cyclic quadrilateral $ABCD$, as shown in the accompanying figure, the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. In the special cases of one of the diagonals or sides being a diameter of the circle, this theorem gives rise directly to the angle sum and difference trigonometric identities. The relationship follows most easily when the circle is constructed to have a diameter of length one, as shown here.

By Thales's theorem, $\angle DAB$ and $\angle DCB$ are both right angles. The right-angled triangles $DAB$ and $DCB$ both share the hypotenuse ${\overline {BD}}$ of length 1. Thus, the side ${\overline {AB}}=\sin \alpha$, ${\overline {AD}}=\cos \alpha$, ${\overline {BC}}=\sin \beta$ and ${\overline {CD}}=\cos \beta$.

By the inscribed angle theorem, the central angle subtended by the chord ${\overline {AC}}$ at the circle's center is twice the angle $\angle ADC$, i.e. $2(\alpha +\beta )$. Therefore, the symmetrical pair of red triangles each has the angle $\alpha +\beta$ at the center. Each of these triangles has a hypotenuse of length ${\frac {1}{2}}$, so the length of ${\overline {AC}}$ is $2\times {\frac {1}{2}}\sin(\alpha +\beta )$, i.e. simply $\sin(\alpha +\beta )$. The quadrilateral's other diagonal is the diameter of length 1, so the product of the diagonals' lengths is also $\sin(\alpha +\beta )$.

When these values are substituted into the statement of Ptolemy's theorem that $|{\overline {AC}}|\cdot |{\overline {BD}}|=|{\overline {AB}}|\cdot |{\overline {CD}}|+|{\overline {AD}}|\cdot |{\overline {BC}}|$, this yields the angle sum trigonometric identity for sine: $\sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$. The angle difference formula for $\sin(\alpha -\beta )$ can be similarly derived by letting the side ${\overline {CD}}$ serve as a diameter instead of ${\overline {BD}}$.

Inscribed angle theorem

Animated gif of proof of the inscribed angle theorem. The large triangle that is inscribed in the circle gets subdivided into three smaller triangles, all of which are isosceles because their upper two sides are radii of the circle. Inside each isosceles triangle the pair of base angles are equal to each other, and are half of 180° minus the apex angle at the circle's center. Adding up these isosceles base angles yields the theorem, namely that the inscribed angle, $\psi$, is half the central angle, $\theta$.

Other trig identities

(The figure showing sin, cos, tan, cot, csc and sec on the unit circle is on two pages, the Pythagorean identities section of List_of_trigonometric_identities, and also the page on the Pythagorean trigonometric identity.)

Trigonometric functions and their reciprocals on the unit circle. All of the right-angled triangles are similar, i.e. the ratios between their corresponding sides are the same. For sin, cos and tan the unit-length radius forms the hypotenuse of the triangle that defines them. The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse. The triangle shaded blue illustrates the identity $1+\cot ^{2}\theta =\csc ^{2}\theta$, and the red triangle shows that $\tan ^{2}\theta +1=\sec ^{2}\theta$.	Visual demonstration of the double-angle formula for sine. The area, 1/2 × base × height, of an isosceles triangle is calculated, first when upright, and then on its side. When upright, the area = $\sin \theta \cos \theta$. When on its side, the area = 1/2 $\sin 2\theta$. Rotating the triangle does not change its area, so these two expressions are equal. Therefore, $\sin 2\theta =2\sin \theta \cos \theta$.
Cosine power-reduction formula: an illustrative diagram. The red, orange and blue triangles are all similar, and the red and orange triangles are congruent. The hypotenuse ${\overline {AD}}$ of the blue triangle has length $2\cos \theta$. The angle $\angle DAE$ is $\theta$, so the base ${\overline {AE}}$ of that triangle has length $2\cos ^{2}\theta$. That length is also equal to the summed lengths of ${\overline {BD}}$ and ${\overline {AF}}$, i.e. $1+\cos(2\theta )$. Therefore, $2\cos ^{2}\theta =1+\cos(2\theta )$, yielding the power-reduction formula when both sides are divided by $2$. The half-angle formula for cosine can be obtained by replacing $\theta$ with $\theta /2$ and taking the square-root of both sides.	Sine power-reduction formula: an illustrative diagram. The shaded blue and green triangles, and the red-outlined triangle $EBD$ are all right-angled and similar, and all contain the angle $\theta$. The hypotenuse ${\overline {BD}}$ of the red-outlined triangle has length $2\sin \theta$, so its side ${\overline {DE}}$ has length $2\sin ^{2}\theta$. The line segment ${\overline {AE}}$ has length $\cos 2\theta$ and sum of the lengths of ${\overline {AE}}$ and ${\overline {DE}}$ equals the length of ${\overline {AD}}$, which is 1. Therefore, $\cos 2\theta +2\sin ^{2}\theta =1$. Subtracting $\cos 2\theta$ from both sides and dividing by 2 by two yields the power-reduction formula for sine. The half-angle formula for sine can be obtained by replacing $\theta$ with $\theta /2$ and taking the square-root of both sides. Note that this figure also illustrates, in the vertical line segment ${\overline {EB}}$, that $\sin 2\theta =2\sin \theta \cos \theta$.
Diagram illustrating sum-to-product identities for sine and cosine. The blue right-angled triangle has angle $\theta$ and the red right-angled triangle has angle $\varphi$. Both have a hypotenuse of length 1. Auxiliary angles, here called $p$ and $q$, are constructed such that $p=(\theta +\varphi )/2$ and $q=(\theta -\varphi )/2$. Therefore, $\theta =p+q$ and $\varphi =p-q$. This allows the two congruent purple-outline triangles $AFG$ and $FCE$ to be constructed, each with hypotenuse $\cos q$ and angle $p$ at their base. The sum of the heights of the red and blue triangles is $\sin \theta +\sin \varphi$, and this is equal to twice the height of one purple triangle, i.e. $2\sin p\cos q$. Writing $p$ and $q$ in that equation in terms of $\theta$ and $\varphi$ yields the sum-to-product identity for sine. Similarly, the sum of the widths of the red and blue triangles yields the corresponding identity for cosine.	Diagram showing the angle difference identities for $\sin(\alpha -\beta )$ and $\cos(\alpha -\beta )$.

Exponentiation

(I added some text to the intro of this page, to emphasise that the exponentiation rules follow as necessary consequences from the basic fact that $b^{n}$ means $n$ occurrences of $b$ all multiplied by each other. The previous text stated these rules as definitions, potentially giving the false impression that they might have been arbitrary choices.)

Starting from the basic fact stated above that, for any positive integer $n$, $b^{n}$ is $n$ occurrences of $b$ all multiplied by each other, several other properties of exponentiation directly follow. In particular:

${\begin{aligned}b^{n+m}&=\underbrace {b\times \dots \times b} _{n+m{\text{ times}}}\\[1ex]&=\underbrace {b\times \dots \times b} _{n{\text{ times}}}\times \underbrace {b\times \dots \times b} _{m{\text{ times}}}\\[1ex]&=b^{n}\times b^{m}\end{aligned}}$

In other words, when multiplying a base raised to one exponent by the same base raised to another exponent, the exponents add. From this basic rule that exponents add, we can derive that $b^{0}$ must be equal to 1, as follows. For any $n$, $b^{0}\cdot b^{n}=b^{0+n}=b^{n}$. Dividing both sides by $b^{n}$ gives $b^{0}=b^{n}/b^{n}=1$.

The fact that $b^{1}=b$ can similarly be derived from the same rule. For example, $(b^{1})^{3}=b^{1}\cdot b^{1}\cdot b^{1}=b^{1+1+1}=b^{3}$. Taking the cube root of both sides gives $b^{1}=b$.

The rule that multiplying makes exponents add can also be used to derive the properties of negative integer exponents. Consider the question of what $b^{-1}$ should mean. In order to respect the "exponents add" rule, it must be the case that $b^{-1}\cdot b^{1}=b^{-1+1}=b^{0}=1$. Dividing both sides by $b^{1}$ gives $b^{-1}=1/b^{1}$, which can be more simply written as $b^{-1}=1/b$, using the result from above that $b^{1}=b$. By a similar argument, $b^{-n}=1/b^{n}$.

The properties of fractional exponents also follow from the same rule. For example, suppose we consider ${\sqrt {b}}$ and ask if there is some suitable exponent, which we may call $r$, such that $b^{r}={\sqrt {b}}$. From the definition of the square root, we have that ${\sqrt {b}}\cdot {\sqrt {b}}=b$. Therefore, the exponent $r$ must be such that $b^{r}\cdot b^{r}=b$. Using the fact that multiplying makes exponents add gives $b^{r+r}=b$. The $b$ on the right-hand side can also be written as $b^{1}$, giving $b^{r+r}=b^{1}$. Equating the exponents on both sides, we have $r+r=1$. Therefore, $r={\frac {1}{2}}$, so ${\sqrt {b}}=b^{\frac {1}{2}}$.

Penrose tiling

Added a pair of animations here: Penrose tiling#Deflation for P2 and P3 tilings

One step of the substitution rules for kite and dart Penrose tiles.

Four iterations of the kite and dart substitutions.