Category: O-Level Math

The Parabolic Reflector

The diagram below shows a parabola with equation $y =x^2$ . Recall that its directrix has equation $y = -1/4$ .

A light ray traveling downward along $x = a$ reflects off the tangent $T$ to the curve at $P(a, a^2)$ at an angle of $\alpha$ , and intersects the $y$ -axis at $F_0$ .

Problem 1. Calculate the $y$ -intercept $H$ of $T$ in terms of $a$ .

(Click for Solution)

Solution. Recall that $T$ has gradient $2a$ and hence equation

$y = 2a(x-a) + a^2.$

At the $y$ -intercept, $x = 0$ , so that $y = -a^2$ . Hence, $T$ has a $y$ -intercept of $(0, -a^2)$ .

Problem 2. Show that $F_0 P H D$ is a rhombus.

(Click for Solution)

Solution. Denote $H$ below as per Problem 1.

By vertically opposite angles, $\angle HPD = \alpha$ .

Since $PD \parallel F_0H$ , the alternate angles equal:

$\angle F_0 H P = \angle HPD = \alpha = \angle F_0 PH.$

Since the base angles of $\Delta F_0 HP$ are equal, $\Delta F_0 H P$ is isosceles, so that

$F_0 H = F_0 P.$

By direct computation,

$PD^2 = (a^2 + 1/4)^2 = a^4 + a^2/2 + 1/16$

and

$\begin{aligned} HD^2 &= a^2 + (a^2 - 1/4)^2 \\ &= a^2 + (a^4 - a^2/2 + 1/16) \\ &= a^4 + a^2/2 + 1/16 = PD^2, \end{aligned}$

so that $HD = PD$ . Hence, $\Delta DPH$ is isosceles and

$\angle DHP = \angle DPH = \alpha = \angle F_0 PH = \angle F_0HP.$

With the common side $PH$ , $\angle DPH = \angle F_0PH$ , and $\angle DHP = \angle F_0 HP$ , the ASA Criterion yields

$\Delta DPH \equiv \Delta F_0 PH.$

Therefore,

$DP = PF_0 = F_0 H = HD,$

implying that $F_0 PHD$ is a rhombus.

Problem 3. Deduce that $F_0$ coincides with the focus of the parabola.

(Click for Solution)

Solution. Since $F_0 H = PD = a^2 + 1/4$ , the $y$ -coordinate of $F_0$ is

$-a^2 + (a^2+1/4) = 1/4.$

Hence, $F_0(0, 1/4)$ , coinciding with the focus of $y = x^2$ .

Remark 1. Our arguments generalise to other parabolas, requiring extra book-keeping.

Remark 2. A similar calculation can establish the converse: given that light starts at $F_0$ and reflects off $T$ at $P$ , the resulting light ray travels upward, and parallel to the $y$ -axis.

—Joel Kindiak, 10 Jan 26, 2026H

April 3, 2026
The Chain Rule

Previously, we have defined differentiation, roughly speaking, as gradient-calculation. That is, the function $y = f(x)$ has a derivative $f'(t)$ at $x = t$ if the tangent line to the curve $y = f(x)$ at $(t, f(t))$ has equation

$y = f'(t) (x-t) + f(t).$

In this case, we write

$\displaystyle \frac{\mathrm d }{ \mathrm dx }(f(x)) = f'(x).$

The simplest kind of function would be the powers of $x$ (i.e. the power rule): for any rational number $n$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = nx^{n-1}.$

Indeed, this result is meaningful by adapting the calculations in this exercise.

Differentiation is “splittable” over addition:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(x) + g(x)) = \frac{\mathrm d}{\mathrm dx}(f(x)) + \frac{\mathrm d}{\mathrm dx}(g(x)).$

It even works for functions scaled by a constant:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(c \cdot f(x) ) = c \cdot \frac{\mathrm d}{\mathrm dx}(f(x)).$

That is, differentiation satisfies linearity.

But as discussed previously, we do not get splitting over products

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(x) \cdot g(x)) \neq \frac{\mathrm d}{\mathrm dx}(f(x)) \cdot \frac{\mathrm d}{\mathrm dx}(g(x)).$

Nor do we get splitting over function-in-function combinations (i.e. compositions),

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f( g(x) )) \neq f'( g(x) ).$

However, it is still possible to evaluate their derivatives.

Example 1. Define $f(x) = x^2$ and $g(x) = x^3 + 1$ . Show that

$\displaystyle f( g(x) ) = (x^3 + 1)^2,$

and hence, check that

$\displaystyle \frac{\mathrm d}{\mathrm dx} ( f( g(x) ) ) = f'( g(x) ) \cdot g'(x).$

Proof. By definition of the individual functions,

$\begin{aligned} f( g(x) ) &= g(x)^2 \\ &= ( x^3 + 1 )^2 \\ &= (x^3)^2 + 2 \cdot x^3 \cdot 1 + 1^2 \\ &= x^6 + 2x^3 + 1. \end{aligned}$

By the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( f( g(x) ) ) &= \frac{\mathrm d}{\mathrm dx} (x^6 + 2x^3 + 1) \\ &= \frac{\mathrm d}{\mathrm dx}( x^6 ) + 2 \cdot \frac{\mathrm d}{\mathrm dx} ( x^3 ) + \frac{\mathrm d}{\mathrm dx} ( 1 ) \\ &= 6x^5 + 2 \cdot 3x^2 + 0 \\ &= 6x^5 + 6x^2.\end{aligned}$

On the other hand,

$f'(x) = \displaystyle \frac{\mathrm d}{\mathrm dx}(f(x)) = \frac{\mathrm d}{\mathrm dx}(x^2) = 2x$

and using linearity,

$\begin{aligned} g'(x) = \frac{\mathrm d}{\mathrm dx}(g(x)) &= \frac{\mathrm d}{\mathrm dx}(x^3+1) \\ &= \frac{\mathrm d}{\mathrm dx}(x^3) + \frac{\mathrm d}{\mathrm dx}(1) \\ &= 3x^2 + 0 = 3x^2. \end{aligned}$

Therefore,

$\begin{aligned} f'( g(x) ) \cdot g'(x) &= 2 \cdot g(x) \cdot g'(x) \\ &= 2 \cdot (x^3 + 1) \cdot 3x^2 \\ &= 6x^2 \cdot (x^3 + 1) \\ &= 6x^2 \cdot x^3 + 6x^2 \cdot 1 \\ &= 6x^5 + 6x^2. \end{aligned}$

Hence,

$\displaystyle \frac{\mathrm d}{\mathrm dx}( f( g(x) ) ) = 6x^5 + 6x^2 = f'(g(x)) \cdot g(x).$

This result is true in general, and known as the chain rule.

Theorem 1 (Chain Rule). For functions $f(x), g(x)$ with derivatives $f'(x), g'(x)$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx} ( f( g(x) ) ) = f'( g(x) ) \cdot g'(x).$

Writing $u = g(x)$ so that $\displaystyle \frac{ \mathrm du }{ \mathrm dx } = g'(x)$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx} ( f( u ) ) = f'( u ) \cdot \frac{ \mathrm du }{ \mathrm dx }.$

Proof. See this post.

Example 2. Show that $\displaystyle \frac{\mathrm d}{\mathrm dx}(h(x)^n) = n \cdot h(x)^{n-1} \cdot h'(x)$ .

Solution. Setting $f(x) = x^n$ and $g(x) = h(x)$ , using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(h(x)^n)= \frac{\mathrm d}{\mathrm dx}( f(g(x)) ) &= f'(g(x)) \cdot g'(x) \\ &= f'(h(x)) \cdot h'(x). \end{aligned}$

Using the power rule, $f'(x) = nx^{n-1}$ implies that $f'(h(x)) = n \cdot h(x)^{n-1}$ :

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(h(x)^2) &= f'(h(x)) \cdot h(x) \\ &= n \cdot h(x)^{n-1} \cdot h'(x). \end{aligned}$

Remark 1. In particular, setting $n = 2$ and $n = -1$ respectively,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(h(x)^2) &= 2 \cdot h(x) \cdot h'(x), \\ \frac{\mathrm d}{\mathrm dx} \left( \frac 1{ h(x) }\right) &= - \frac{ h'(x) }{ h(x)^2 }. \end{aligned}$

Example 3. Define $h(x) = f(x) + g(x)$ . Show that

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( h(x)^2 ) &= \frac{\mathrm d}{\mathrm dx}(f(x)^2 ) + \frac{\mathrm d}{\mathrm dx}(g(x)^2 ) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ).\end{aligned}$

Solution. Using Remark 1,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( h(x)^2 ) &= 2 \cdot h(x) \cdot h'(x). \end{aligned}$

Using linearity,

$\begin{aligned} h'(x) = \frac{\mathrm d}{\mathrm dx}(h(x)) &= \frac{\mathrm d}{\mathrm dx}( f(x) + g(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}(f(x)) + \frac{\mathrm d}{\mathrm dx}(g(x)) \\ &= f'(x) + g'(x). \end{aligned}$

Together with $h(x) = f(x) + g(x)$ ,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( h(x)^2 ) &= 2 \cdot (f(x) + g(x)) \cdot (f'(x) + g'(x)) \\ &= 2 \cdot f(x) \cdot f'(x) + 2 \cdot g(x) \cdot g'(x) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x)). \end{aligned}$

By Remark 1 again,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(x)^2) = 2 \cdot f(x) \cdot f'(x), \quad \frac{\mathrm d}{\mathrm dx}(g(x)^2) = 2 \cdot g(x) \cdot g'(x).$

Hence,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( h(x)^2 ) &= \frac{\mathrm d}{\mathrm dx}(f(x)^2 ) + \frac{\mathrm d}{\mathrm dx}(g(x)^2 ) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ).\end{aligned}$

Remark 2. Example 3 helps us prove the product rule, which, in turn, together with the second result in Remark 1, helps us prove the quotient rule. We will visit both results next time.

The chain rule empowers us to differentiate all sorts of functions.

Example 4. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}((x^{67} + 89)^{100})$ .

Solution. While terrifying and tragically anti-funny, the chain rule renders this problem trivial. By Example 2 and linearity,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}((x^{67} + 89)^{100}) &= 100 (x^{67} + 89)^{99} \cdot \frac{\mathrm d}{\mathrm dx}( (x^{67} + 89) ) \\ &= 100 (x^{67} + 89)^{99} \cdot \left( \frac{\mathrm d}{\mathrm dx}( x^{67} ) + \frac{\mathrm d}{\mathrm dx} ( 89) \right) \\ &= 100 (x^{67} + 89)^{99} \cdot ( 67x^{66} + 0 ) \\ &= 100 (x^{67} + 89)^{99} \cdot 67x^{66} \\ &= 6700 x^{66} (x^{67} + 89)^{100}. \end{aligned}$

Example 5. For any positive constant $r > 0$ , evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}(\sqrt{r^2 - x^2})$ .

Solution. Recall that using the power rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\sqrt{x}) &= \frac{\mathrm d}{\mathrm dx}(x^{1/2}) = \frac 12 x^{-1/2} = \frac 1{2 \sqrt x}. \end{aligned}$

Hence, using the chain rule (or Example 2) and linearity,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(\sqrt{r^2 - x^2}) &= \frac 1{ 2\sqrt{ r^2 - x^2} } \cdot \frac{\mathrm d}{\mathrm dx}(r^2-x^2) \\ &= \frac 1{ 2\sqrt{ r^2- x^2} } \cdot \left( \frac{\mathrm d}{\mathrm dx}(r^2)-\frac{\mathrm d}{\mathrm dx}(x^2) \right) \\ &= \frac 1{ 2\sqrt{ r^2 - x^2} } \cdot \left( 0 - 2x \right) \\ &= -\frac{2x}{ 2\sqrt{ r^2 - x^2} } \\ &= -\frac{x}{\sqrt{ r^2-x^2}}.\end{aligned}$

Remark 3. Example 5 gives us yet another proof that the radius of a circle must be perpendicular to its tangent.

The chain rule is, arguably, the most powerful theorem pertaining differentiation. We can use it to prove the product rule and the quotient rule, and these latter rules help us compute expressions such as

$\displaystyle \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) \quad \text{and} \quad \frac{\mathrm d}{\mathrm dx} \left( \frac{ f(x) }{ g(x) } \right)$

correctly. These we visit next time.

—Joel Kindiak, 8 Jan 26, 1925H

April 2, 2026
The Flavours of Average

Consider the following dot diagram that displays the heights, measured in cm, of 9 Gen Z humans.

Question 1. What is the average height of these 9 individuals?

You might think that the answer is automatically obtained by adding all heights then dividing by 9. We will touch base with this kind of average later on. The problem is that the word “average” can take on multiple meanings.

One meaning means, what is the height that most people would have? Clearly, the height with the largest number of dots is 167 cm.

This means that the height that the majority of this set of 7 persons has is 167 cm. We call this the mode of the data set.

An element in a data set is called a data point.

Definition 1. The mode of a data set is the value taken on by majority of the data points.

We could formalise this notion using more technical symbols, but I don’t think that is helpful for us, since we are not terribly interested in any rigorous analysis of the data set.

Example 1. Consider the pie chart below illustrating the favourite music artists in 2025.

Then the most popular artist, namely Drake, is the mode of the data set. I am surprised that Taylor Swift isn’t on this list. Don’t flame me.

Example 2. Consider the dot diagram we started with.

According to the diagram, the mode of the data set is 167 cm. If, however, we add another data set at 162 cm, we get the dot diagram below.

Then both 162 cm and 167 cm are modes of this data set. In this case, we call the data set bimodal.

This ambiguity could be a problem. We would like our answer to the “average” question to produce a unique answer.

To do that, we could interpret our data set as balancing on a beam with a pivot.

If the pivot is positioned at the 160 cm data point, then the entire beam would fall down. Likewise with the 170 cm data point.

Question 2. Where would we position the pivot to balance the beam?

Intuitively, the further apart the data set is positioned from the pivot, the greater the rotating effect (i.e. the moment). Assume each data point has equal “mass”. Then we would like to compute some “balance point” $\bar x$ such that the sum of $(x - \bar x)$ equals 0.

Denoting the data set by $\{x_1,\dots, x_n\}$ , we require

$\displaystyle (x_1 - \bar x) + (x_2 - \bar x) + \cdots + (x_n - \bar x) = 0.$

Collecting the data points together,

$\displaystyle (x_1 + x_2 + \cdots + x_n) - n\bar x = 0.$

By writing $\bar x$ in terms of the data points,

$\displaystyle \bar x = \frac{x_1 + x_2 + \cdots + x_n}{n} \equiv \frac{ \Sigma x }{n},$

where the Greek letter $\Sigma$ (read ‘Sigma’) denotes the sum

$\Sigma x \equiv x_1 + x_2 + \cdots + x_n.$

Definition 2. The mean of a data set $\{x_1, \dots, x_n\}$ is defined by $\bar x := (\Sigma x)/n$ .

Example 3. Consider the dot diagram we started with again.

By evaluating their sum, $\Sigma x = 1315$ . You can compute this result using either manual addition or by using the spreadsheet function SUM(...). Hence, $\bar x = 1485/9 = 165\ \text{cm}$ .

Example 4. Consider the same dot diagram, but now, a Gen Z human with height 140 cm is included.

By evaluating their sum, $\Sigma x = 1670$ . Hence, $\bar x = 1630/10 = 163\ \text{cm}$ .

Clearly, however, the 140 cm human is an exceptional case (i.e. an outlier) among this group of humans. However, since the mean incorporates all possible heights, $\bar x$ changed from 165 cm to 163 cm. The point of Example 4 is this: the mean is incredibly sensitive to outliers (though thankfully, there are various strategies to mitigate this effect).

Question 3. Can we obtain an average that is less sensitive to outliers?

Return to the original data set.

If we arranged the data points in non-decreasing order, we obtain the following non-decreasing sequence:

160 ≤ 162 ≤ 162 ≤ 165 ≤ 165 ≤ 167 ≤ 167 ≤ 167 ≤ 170

In this ordered sense, the average height is 165 cm.

If we did include the outlier data point 140 cm, we get two middle values:

140 ≤ 160 ≤ 162 ≤ 162 ≤ 165 ≤ 165 ≤ 167 ≤ 167 ≤ 167 ≤ 170

In this latter case, the middle-of-the-middle is the simple average:

$\displaystyle \frac {165 + 165}{2} = 165,$

and the average height unchanged at 165 cm. We call this value the median height.

Definition 3. The median $Q_2 \equiv Q_2(x_1,\dots, x_n)$ of a sorted data set

$x_1 \leq x_2 \leq \cdots \leq x_n$

is defined by $Q_2 := x_{(n+1)/2}$ if $n$ is odd, and $Q_2 := \frac 12 (x_{n/2} + x_{( n/2 ) + 1})$ if $n$ is even. We will explain the $Q_2$ notation in the next post.

Using our data set, the median height remains unchanged when given the extra data point 140 cm. However, the median height can change; if instead we had an additional data point 190 cm, then we get two new middle values:

160 ≤ 162 ≤ 162 ≤ 165 ≤ 165 ≤ 167 ≤ 167 ≤ 167 ≤ 170 ≤ 190.

In this case, our new median is $Q_2 = \frac 12 (165 + 167) = 166\, \text{cm}$ .

Intuitively, however, we would prefer the median to the mean for its relative resilience against outliers.

The mean and median have modifications that allow us to discuss the relative spread of data. We will explore this idea next time.

—Joel Kindiak, 9 Feb 26, 1346H

April 1, 2026
Scientific Notation
Let u denote any agreed-upon unit for a quantity. For example, when measuring distance, u refers to 1 metre, denoted 1 m. When measuring information, u refers to 1 byte, denoted 1 B.

Definition 1. Given a unit u of quantity,
- one kilo-unit, denoted ku, is defined to be 1000 units of that quantity,
- one mega-unit, denoted Mu, is defined to be 1000 kilo-units of that quantity,
- one giga-unit, denoted Gu, is defined to be 1000 mega-units of that quantity,
- one tera-unit, denoted Tu, is defined to be 1000 giga-units of that quantity.
Example 1. 1 kB = 1000 B.

Problem 1. Explain why 1 kB = (1 × 10³) B. Deduce the integer α such that

1 TB = (1 × 10^α) B.

(Click for Solution)

Solution. We remark that

$1000 = 10 \times 10 \times 10 = 10^3.$

Therefore,

$\begin{aligned}1\, \mathrm{TB} &= 1000\, \mathrm{GB} = (1 \times 10^3)\, \mathrm{GB}.\end{aligned}$

Repeating the pattern,

$\begin{aligned}1\, \mathrm{TB} &= (1 \times 10^3)\, \mathrm{GB} \\ &= (1 \times 10^3 \cdot 10^3)\, \mathrm{MB} \\ &= (1 \times 10^3 \cdot 10^3 \cdot 10^3)\, \mathrm{kB} \\ &= (1 \times 10^3 \cdot 10^3 \cdot 10^3 \cdot 10^3)\, \mathrm{B} \\ &= (1 \times 10^{12})\, \mathrm B. \end{aligned}$

Therefore, $\alpha = 12$ .

Example 2. Letting the dollar, $1, denote the unit of measurement for money, one million dollars is equal to one mega-dollar. One billion dollars is equal to one giga-dollar. One trillion dollars is equal to one tera-dollar.

Definition 2. Given a unit u of quantity,
- one deci-unit, denoted du, is defined to be 0.1 units of that quantity,
- one centi-unit, denoted cu, is defined to be 0.01 units of that quantity,
- one milli-unit, denoted mu, is defined to be 0.001 units of that quantity,
- one micro-unit, denoted μu, is defined to be 0.001 milli-units of that quantity,
- one nano-unit, denoted nu, is defined to be 0.001 micro-units of that quantity,
- one pico-unit, denoted pu, is defined to be 0.001 nano-units of that quantity.
Example 3. 1 m = 100 cm.

Problem 2. Explain why 1 mm = (1 × 10^–3) m. Deduce the integer α such that

1 pm = (1 × 10^α) m.

(Click for Solution)

Solution. We remark that

$1\, \mathrm{mm} = 0.001\, \mathrm{m}.$

Therefore,

$1\, \mathrm{m} = 1000\, \mathrm{mm} = 10^3\, \mathrm{mm}.$

Dividing both sides by $10^3$ ,

$1\, \mathrm{mm} = (1/10^3)\, \mathrm m = (1 \times 10^{-3})\, \mathrm m.$

Similar as to Problem 1,

$\begin{aligned}1\, \mathrm{pm} &= (1 \times 10^{-3})\, \mathrm{nm} \\ &= (1 \times 10^{-3} \cdot 10^{-3})\, \mathrm{\text{\textmu} m} \\ &= (1 \times 10^{-3} \cdot 10^{-3} \cdot 10^{-3})\, \mathrm{mm} \\ &= (1 \times 10^{-3} \cdot 10^{-3} \cdot 10^{-3} \cdot 10^{-3})\, \mathrm{m} \\ &= (1 \times 10^{-12})\, \mathrm m. \end{aligned}$

Therefore, $\alpha = -12$ .

Theorem 1. For any positive real number N, there exists a unique integer α_N and a unique real number 1 ≤ N₀ < 10 such that

N = N₀ × 10^–α_N.

The right-hand side is called the scientific form of N. We call α_N the order of magnitude of N.

Proof. Using real analysis, there must exist some smallest (and thus, unique) integer α_N such that

1 ≤ N₀ / 10^α_N < 10

Define N₀ := N₀ / 10^α_N, so that N = N₀ × 10^–α_N as required.

Problem 3. Write down the order of magnitude for 1 TB and 1 pm respectively. What do you notice?

(Click for Solution)

Solution. The orders of magnitude are 12 and –12 respectively. The order of magnitude is positive if the quantity is greater than or equal to 1, and negative if the quantity is smaller than 1.

Theorem 2. Suppose α_N ≤ α_M. We have the following properties for orders of magnitude:
- α_M + α_N ≤ α_MN ≤ α_M + α_N + 1,
- α_M – α_N – 1 ≤ α_M/N ≤ α_M – α_N,
- α_M ≤ α_{M + N} ≤ α_M + 1,
- α_M – 1 ≤ α_{M – N} ≤ α_M.
Proof. Left as an exercise for the motivated student.

Problem 4. In chemistry, the accepted Avogadro constant N_A, measured in units of mol^–1, is given approximately by the number

N_A ≈ 602 214 076 000 000 000 000 000.

That is, 1 mol is defined to be the number N_A.

The accepted Boltzmann’s constant k_B, measured in units of J K^–1, given approximately by the number

k_B ≈ 0.000 000 000 000 000 000 000 013 806 490.

Express N_A and k_B in scientific form. Hence, evaluate the ideal gas constant R, measured in units of J K^–1 mol^–1, defined by R := N_A · k_B.

(Click for Solution)

Solution. We observe that

$\begin{aligned} N_{\mathrm A} &\approx 602\, 214\, 076\times 10^{15}\\ &= 6.022\, 140\, 760 \times 10^8 \times 10^{15} \\ &= 6.022\, 140\, 760 \times 10^{23} \\ &\approx 6.02 \times 10^{23}. \end{aligned}$

Similarly,

$\begin{aligned} k_{\mathrm B} &= 0.013\, 806\, 490 \times 10^{-21} \\ &= 1.380\, 649 \times 10^{-2} \times 10^{-21} \\ &= 1.380\, 649 \times 10^{-23} \\ &\approx 1.38 \times 10^{-23}. \end{aligned}$

Therefore,

$\begin{aligned}R = N_{\mathrm A} \cdot k_{\mathrm B} &= (6.022\, 140\, 760 \times 10^{23} ) \cdot (1.380\, 649 \times 10^{-23}) \\ &= 6.022\, 140\, 760 \cdot 1.380\, 649 \\ &= 8.314\, 462\, 618\, 200 \\ & \approx 8.31. \end{aligned}$

Problem 5. The ideal gas law states the following, given:
- the number N of molecules of a gas,
- the temperature T of the gas (measured in Kelvin, denoted K),
- the pressure p of the gas (measured in Pascals, denoted Pa),
- and the volume V of the gas (measured in m³),
the equation p · V = N · k_B · T holds.

Determine the volume of 0.400 mol of a gas with a pressure of 101 325 Pa and a temperature of 284 K, giving your answer in scientific notation.

(Click for Solution)

Solution. Using Problem 4, $N = 0.400 \cdot N_{\mathrm A}$ .

Substituting the values,

$101\, 325 \cdot V = 0.400 \cdot \underbrace{ N_{\mathrm A} \cdot k_{\mathrm B}}_R \cdot 284.$

Making $V$ the subject,

$\begin{aligned} V &= \frac{0.400 \cdot R \cdot 284}{101\, 325} \\ &= \frac{0.400 \cdot 8.314\, 462\, 618\, 200 \cdot 284}{101\, 325} \\ &= 0.009\, 321\, 716\, 787 \\ &= 9.321\, 716\, 787 \times 10^{-3} \\ &\approx 9.32 \times 10^{-3}. \end{aligned}$

Problem 6. In physics and astronomy, the accepted universal gravitational constant G, measured in units of m³ kg^–1 s^–2, is given approximately by the quantity

G ≈ 0.000 000 000 066 743.

Newton’s law of universal gravitation states that the magnitude F of the gravitational force between two masses M, m (measured in kg) that are separated by a distance of r metres is given by the equation

$\displaystyle F = \frac {GMm}{r^2}.$

Express G in scientific form. Furthermore, given that the Earth has a mass of

5 972 000 000 000 000 000 000 000 kg

and a radius of 6371 km, determine the gravitational acceleration g of an object with 1 kg near the surface of the Earth, measured in units of m s^–2, defined by g := F /m.

(Click for Solution)

Solution. Expressing $G$ in scientific form,

$\begin{aligned} G&\approx 0.000\, 000\, 000\, 066\, 743 \\ &= 0.066\, 743\times 10^{-9} \\ &= 6.674\, 300 \times 10^{-2} \times 10^{-9} \\ &= 6.674\, 300 \times 10^{-11} \\ &\approx 6.67 \times 10^{-11}. \end{aligned}$

Similarly, expressing $M$ in scientific form,

$\begin{aligned} M&= 5\, 972\, 000\, 000\, 000\, 000\, 000\, 000\, 000 \\ &= 5972 \times 10^{21} \\ &= 5.972 \times 10^3 \times 10^{21} \\ &= 5.972 \times 10^{24}. \end{aligned}$

Since the object is near the surface of the Earth,

$\begin{aligned} r &\approx 6371 \times 10^3 \\ &= 6.371 \times 10^3 \times 10^3 \\ &= 6.371 \times 10^6. \end{aligned}$

Substituting the quantities,

$\begin{aligned} g = \frac{F}{m} = \frac{GM}{r^2} &= \frac{(6.674\, 300 \times 10^{-11}) \cdot (5.972 \times 10^{24})}{(6.371 \times 10^6)^2} \\ &= \frac{6.674\, 300 \cdot 5.972}{6.371^2} \times \frac{10^{-11} \cdot {10}^{24}}{(10^6)^2} \\ &= 0.981\, 997\, 342\, 600 \times 10 \\ &= 9.819\, 973\, 426 \\ &\approx 9.82. \end{aligned}$

Problem 7. The Sun has a volume of 1.412 × 10¹⁸ m³, while the Earth has a volume of 1.083 × 10¹² m³. Disregarding issues related to unused empty space (i.e. sphere-packing), how many Earths could fit inside the Sun?

(Click for Solution)

Solution. The total number of Earths that could fit inside the Sun is simply the ratio below:

$\begin{aligned} \frac{1.412 \times 10^{18}}{1.083 \times 10^{12}} &= 1.303\, 785\, 780 \times 10^6 \\ &= 1\, 303\, 785. 780 \\ &\approx 1\, 303\, 785. \end{aligned}$

Therefore, approximately 1 303 785 Earths can fit in the Sun. Surprisingly, after accounting for sphere-packing, at least 982 334 Earths can still fit into the Sun. The Sun is not small.

Problem 8. The Schwarzchild radius is the radius r_S of the largest black hole that could be formed by an object with mass M. It is calculated using the formula

$\displaystyle r_{\mathrm S} = \frac {2GM}{c^2},$

where c ≈ 299 792 458, measured in m s^–1, denotes the speed of light in a vacuum. Determine the Schwarzchild radius for a black hole whose mass is equal to the mass of the Earth.

(Click for Solution)

Solution. Expressing $c$ in scientific notation,

$\begin{aligned}c &\approx 299\, 792\, 458 \\ &= 299.792\, 458 \times 10^6 \\ &= 2.997\, 924\, 580 \times 10^2 \times 10^6 \\ &= 2.997\, 924\, 580 \times 10^8.\end{aligned}$

Therefore, using the data in Problem 6,

$\begin{aligned} r_{\mathrm S} &= \frac{ 2 \cdot 6.674\, 300 \times 10^{-11} \cdot 5.972 \times 10^{24} }{(2.997\, 924\, 580 \times 10^8)^2} \\ &= \frac{ 2 \cdot 6.674\, 300 \cdot 5.972 }{2.997\, 924\, 580^2} \times \frac{10^{-11} \cdot {10}^{24}}{(10^8)^2} \\ &= 8.869\, 805\, 825\, 400 \times 10^{-3} \\ &\approx 8.87 \times 10^{-3} \\ &= 8.87\, \text{mm}. \end{aligned}$

Thus, the black hole would have a radius of 8.87 mm.

Problem 9. The gross domestic product (GDP) of a country is the total market value of all final goods and services in that country. The GDP per capita of a country is the GDP per person in that country. For simplicity, we will measure GDP in US dollars (USD).

Given that Singapore has a GDP of 572.47 billion USD and a population of 6.04 million as of June 2024, calculate Singapore’s GDP per capita as of June 2022.

(Click for Solution)

Solution. Using scientific notation, the required GDP per capita is given by

$\begin{aligned} \frac{572.47 \times 10^9}{6.04 \times 10^6} &= \frac{5.7247}{6.04} \times \frac{10^2 \cdot 10^9}{10^6} \\ &= 0.9477980132 \times 10^5 \\ &= 94\, 779.801\, 320 \\ &\approx 94\, 800\, \text{USD}. \end{aligned}$

In contrast, the GDP per capita of Australia in 2024 is approximately 65 000 USD.

—Joel Kindiak, 30 Mar 26, 0007H
March 30, 2026
The Triangle Inequality
Problem 1. Consider the triangle below with $a < b$ .

Show that $c \leq a + b$ . This result is known as the triangle inequality.

(Click for Solution)

Solution. Draw the altitude $h$ of the triangle with base $c$ and write $c = r + s$ .

Using Pythagoras’ theorem,

$\begin{aligned} c = r + s & \leq \sqrt{a^2 - h^2} + \sqrt{b^2 - h^2} \\ &\leq \sqrt{a^2} + \sqrt{b^2} \\ &= a + b. \end{aligned}$

Problem 2. Show that $a-b \leq c$ . This result is known as the reverse triangle inequality.

(Click for Solution)

Solution. Apply Problem 1 to obtain the inequality

$a \leq b + c.$

Subtracting $b$ on both sides, $a-b \leq c$ .

Problem 3. For any real number $x$ , define the absolute value of $x$ by

$|x| := \begin{cases} x, & x \geq 0, \\ -x, & x < 0. \end{cases}$

Show that $|x| = \sqrt{x^2}$ . Deduce that $|x| = |{-x}|$ and $-|x| \leq x \leq |x|$ .

(Click for Solution)

Solution. For $x \geq 0$ ,

$\sqrt{x^2} = x = |x|.$

For $x < 0$ , $|x| = -x > 0$ . Since

$(-x)^2 = (-x) \cdot (-x) = x\cdot x = x^2,$

we have

$\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|.$

Therefore, $|x| = \sqrt{x^2}$ . In particular,

$|{-x}| = \sqrt{(-x)^2} = \sqrt{x^2} = |x|.$

For the last inequality, the case $x \geq 0$ yields

$-|x| \leq -0 = 0 \leq x = |x| \leq |x|.$

The case $x < 0$ yields $-x > 0$ , so applying the previous result gives

$-|{-x}| \leq -x \leq |{-x}|.$

By negating all sides of the inequality,

$-|x| = -|{-x}| \leq x \leq |{-x}| = |x|.$

Problem 4. Using the definition in Problem 3, show that for any real number $a, b$ ,

$|a \pm b| \leq |a| + |b|.$
(Click for Solution)

Solution. Using Problem 3, we have the inequalities

$-|a| \leq a \leq |a|,\quad -|b| \leq b \leq |b|.$

Adding them together,

$-(|a| + |b|) \leq a + b \leq |a| + |b|.$

Negating all sides,

$-(|a| + |b|) \leq -(a+b) \leq -(-(|a| + |b|)) = |a| + |b|.$

There are two cases to consider:
- If $a+b \geq 0$ , then $|a+b| = a+b \leq |a| + |b|$ .
- If $a + b < 0$ , then $|a+b| = -(a+b) \leq |a|+|b|$ .
Therefore, we always have $|a+b| \leq |a| + |b|$ . Finally,

$|a-b| = |a+(-b)| \leq |a| + |{-b}| = |a| + |b|.$
—Joel Kindiak, 17 Jan 26, 1303H
March 27, 2026
Number Patterns

Problem 1. Consider the following equations.

$\begin{aligned}T_1 &= 1 = \textstyle \frac 12 \times 1 \times 2, \\ T_2 &= 1 + 2 = \textstyle \frac 12 \times 2 \times 3, \\ T_3 &= 1+2+3 = \textstyle \frac 12 \times 3 \times 4.\end{aligned}$

Write down equations for $T_4, T_5$ . Hence, write down an equation for $T_n$ in terms of $n$ and prove that this equation is correct.

(Click for Solution)

Solution. We check that

$\begin{aligned} T_4 &= 1 + 2 + 3 + 4 = \textstyle \frac 12 \times 4 \times 5 = 10, \\ T_5 &= 1 + 2 + 3 + 4 + 5= \textstyle \frac 12 \times 5 \times 6 = 15.\end{aligned}$

Hence, we conjecture $T_n = \frac 12 \times n \times (n+1)$ , and prove it as follows:

$\begin{aligned} T_n &= 1 + 2 + 3 + \cdots + (n-2) + (n-1) + n \\ T_n &= n + (n-1) + (n-2) + \cdots + 3 + 2 + 1 \\ T_n +T_n &= \underbrace{ (n+1) + (n+1) + \cdots + (n+1) + (n+1) }_n \\ 2T_n &= n \times (n+1) \\ T_n &= \textstyle \frac 12 \times n \times (n+1).\end{aligned}$

Problem 2. Let $a, d$ be real numbers. For each $k$ , define

$u_k = a + (k-1) d.$

Given a positive integer $n > 1$ , use Problem 1 to evaluate

$S_n := u_1 + u_2 + \cdots + u_n$

in terms of $n$ .

(Click for Solution)

Solution. By expanding the terms,

$\begin{aligned} S_n &= a + (a+d) + \cdots + (a+(n-1)d) \\ &= (\underbrace{a + a + \cdots + a}_n) + ( 1 + 2 + \cdots + (n-1) )d \\ &= na + T_{n-1} d \\ &= na + \textstyle \frac 12 (n-1)nd \\ &= \textstyle \frac 12n(2a + (n-1)d) \\ &= \textstyle \frac 12 n(u_1 + u_n). \end{aligned}$

Remark 1. The sequence $\{ u_k \}$ in Problem 2 is called an arithmetic progression with first term $a$ and common difference $d$ . The sum of the first $n$ terms of such a progression is given by $S_n$ .

Problem 3. Fix a real number $r \neq 1$ . Consider the following equations.

$\begin{aligned}T_1 &= 1 + r = \frac{1 - r^2}{1 - r}, \\ T_2 &= 1+r+r^2 = \frac{1 - r^3}{1 - r} ,\\ T_3 &= 1+r+r^2+r^3 = \frac{1 - r^4}{1 - r}.\end{aligned}$

Write down an equation for $T_n$ in terms of $n$ and prove that this equation is correct.

(Click for Solution)

Solution. By following the pattern, we conjecture that

$\displaystyle T_n = 1 + r + r^2 + \cdots + r^n = \frac{ 1 - r^{n+1} }{1-r}.$

We prove this result as follows:

$\begin{aligned} T_n &= 1 + r + r^2 + \cdots + r^n \\ rT_n &= \phantom{1+.\!} r + r^2 + \cdots + r^n + r^{n+1} \\ T_n - rT_n &= 1 - r^{n+1} \\ (1-r)T_{n} &= 1 - r^{n+1} \\ T_{n} &= \frac{ 1 - r^{n+1} }{1 - r}. \end{aligned}$

Problem 4. Let $a \neq 0$ and $r \neq 1$ be real numbers. For each $k$ , define

$v_k = a r^{k-1}$

Given a positive integer $n > 1$ , use Problem 2 to evaluate

$W_n := v_1 + v_2 + \cdots + v_n$

in terms of $n$ .

(Click for Solution)

Solution. By expanding the terms,

$\begin{aligned} W_n &= u_1 + u_2 + \cdots + u_n \\ &= a + ar + \cdots + ar^{n-1} \\ &= a(1 + r + \cdots + r^{n-1}) \\ &= a \cdot T_{n-1} \\ &= \frac{a(1 - r^n)}{1-r}. \end{aligned}$

Remark 1. The sequence $\{ u_k \}$ in Problem 4 is called a geometric progression with first term $a$ and common ratio $r$ . The sum of the first $n$ terms of such a progression is given by $T_n$ .

Problem 5. Show that for any $n > 1$ , $u_n - u_{n-1}$ is constant and $v_n/v_{n-1}$ is constant.

(Click for Solution)

Solution. By Problems 2 and 4,

$\begin{aligned} u_n - u_{n-1} &= a + (n-1) d - (a + (n-2)d) \\ &= ((n-1) - (n-2))d \\ &= d, \\ \frac{ v_n }{ v_{n-1}} &= \frac{ar^{n-1}}{ar^{n-2}} = r. \end{aligned}$

Problem 6. Let $\{x_n\}$ be a sequence of nonzero numbers such that for any $n > 1$ , $x_n - x_{n-1}$ is constant and $x_n / x_{n-1}$ is constant. Show that $x_i = x_j$ for any $i, j$ .

(Click for Solution)

Solution. Suppose there exists real numbers $d, r$ such that for any $n > 1$ ,

$\displaystyle x_n - x_{n-1} = d,\quad \frac{x_n}{x_{n-1}} = r.$

Since $x_n = r \cdot x_{n-1}$ , we have

$\begin{aligned} (r-1) \cdot x_{n-1} &= r \cdot x_{n-1} - x_{n-1} \\ &= d. \end{aligned}$

Using the same logic but replacing $n$ with $n+1$ ,

$\displaystyle (r-1) \cdot x_{n} = d.$

Subtracting both equations,

$\begin{aligned} (r-1) \cdot d &= (r-1) \cdot (x_n - x_{n-1}) \\ &= d-d \\ &= 0. \end{aligned}$

Therefore, either $r = 1$ or $d = 0$ . In both cases, $x_n = x_{n-1}$ . Since this reasoning works for any $n > 1$ , we have $x_n = x_1$ for any $n > 1$ . In particular, $x_i = x_j$ for any $i, j$ .

—Joel Kindiak, 16 Jan 26, 1739H

March 25, 2026
Stationary Points
Differentiation finds one of its greatest powers in optimisation; that is maximising or minimising some constrained quantity.

Consider the graph of $y = f(x)$ below with minimum point $P(c, f(c))$ .

Since the tangent to the curve at $P$ is horizontal, it has a gradient of zero. That is, $f'(c) = 0$ .

Theorem 1 (Zero Derivative Condition). If $(c, f(c))$ is a local minimum or local maximum, then $f'(c) = 0$ . Denoting $y = f(x)$ ,

$\displaystyle \frac{\mathrm d y}{\mathrm dx} \Big|_{x = c} \equiv f'(c) = 0.$

Proof. We illustrate the proof for the local minimum case.

Denote $c^- = c-\delta$ and $c^+ = c+\delta$ for sufficiently small $\delta$ . The rough idea is that using the diagram,

$f'(c^-) < 0 < f'(c^+)$

so that by taking $\delta \approx 0$ ,

$f'(c) \leq 0 \leq f'(c) \quad \Rightarrow \quad f'(c) = 0.$

For more rigorous details, see this post.

Example 1. Calculate the turning points of the graph of $y = \frac 13x^3 - 4x + 2$ .

Solution. To determine the turning points, we use the zero derivative condition:

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = 0.$

We first evaluate the left-hand side:

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} \left( \frac 13x^3 - 4x + 2 \right) \\ &= \frac 13 \cdot \frac{\mathrm d}{\mathrm dx}(x^3) - 4 \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(2) \\ &= \frac 13 \cdot 3x^2 - 4 \cdot 1 + 0 \\ &= x^2 - 4.\end{aligned}$

Hence, we solve the equation

$\begin{aligned} x^2 - 4 &= 0 \\ x^2 &= 4 \\ x &= \pm 2. \end{aligned}$

And we resolve two cases:
- At $x = -2$ , $y = \frac 13 \cdot (-2)^3 - 4 \cdot (-2) + 2 = 22/3$ .
- At $x = 2$ , $y = \frac 13 \cdot 2^3 - 4 \cdot 2 + 2 = -10/3$ .
Therefore, the two turning points have coordinates $(-2, 22/3)$ , $(2,-10/3)$ .

Remark 1. Using software, the graph of $y = \frac 13x^3 - 4x + 2$ is given as follows.

Hence, the power of calculus arises in calculating the turning points even without technology or visual intuition.

Example 2. Given constants $a,b,c,d$ with $a \neq 0$ , show that the graph

$y = ax^3 + bx^2 + cx + d$

has two distinct stationary points if and only if $b^2 - 3ac > 0$ .

Solution. We take the first derivative:

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = 3ax^2 + 2bx + c.$

By the zero derivative condition, the graph has two distinct stationary points if and only if the equation $\displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0$ has two real and distinct roots. Now the equation

$3ax^2 + 2bx + c = 0$

has two real and distinct roots if and only if its discriminant is positive:

$\begin{aligned} (2b)^2 - 4 \cdot 3a \cdot c &> 0 \\ 4b^2 - 4 \cdot 3ac & > 0 \\ 4 \cdot (b^2 - 3ac) & > 0 \\ b^2 - 3ac & > 0. \end{aligned}$

Each step holds bi-directionally so the proof holds.

Suppose now we know that $f'(c) = 0$ . How do we know what kind of turning point $(c, f(c))$ is? Sadly, there is a third situation in which $f'(c) = 0$ .

By considering the graph above, there are three instances in which $f'(c) = 0$ occurs:
- $c=c_1$ : at a local maximum,
- $c=c_2$ : at a stationary point of inflection,
- $c=c_3$ : at a local minimum.
How do we distinguish between these three types? Graphically, but expressed in equations.

Theorem 2 (First Derivative Test). Suppose $f'(c) = 0$ . For small $\delta > 0$ , define $c^- := c - \delta$ and $c^+ := c + \delta$ . Suppose $f'(c^-) \neq 0$ and $f'(c^+) \neq 0$ . Then $P(c, f(c))$ is a:
- local minimum if $f'(c^-) < 0$ and $f'(c^+) > 0$ ,
- local maximum if $f'(c^-) > 0$ and $f'(c^+) < 0$ ,
- stationary point of inflection if $f'(c^-) f'(c^+) > 0$ .
Proof. The diagram above illustrates all three scenarios. For details, see this post.

Example 3. Determine the nature of the turning points calculated in Example 1.

It turns out that we can take a short-cut to Theorem 1 by considering the second derivative, defined by the derivative of the first derivative:

$f''(x) := (f')'(x).$

For alternate notation, suppose $y = f(x)$ , then

$\begin{aligned} \displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2} &\equiv \frac{\mathrm d^2 }{\mathrm dx^2}(y) \equiv \left( \frac{\mathrm d}{\mathrm dx} \right)^2(y) \equiv \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm d}{\mathrm dx}(y)\right) \equiv \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm dy}{\mathrm dx} \right). \end{aligned}$

In turn, we have

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm dy}{\mathrm dx} \right) = \frac{\mathrm d}{\mathrm dx}(f'(x)) = (f')'(x) = f''(x). \end{aligned}$

Theorem 3 (Second Derivative Test). Suppose $f'(c) = 0$ . Then $P(c, f(c))$ is a:
- local minimum if $f''(c) > 0$ ,
- local maximum if $f''(c) < 0$ .
If $f''(c) = 0$ , no conclusion can be made.

Proof Sketch. In the case of a local minimum,

$f'(c^-) < 0,\quad f'(c^+) > 0.$

Since $f''(c) = (f')'(c)$ measures the gradient of $y = f'(c)$ , and $f'$ is increasing from negative to positive, $f''(c) > 0$ . The local maximum case follows similarly. For rigour and detail, see this post.

Example 4. Calculate the stationary points of the graph $y = x + 1/x$ , $x \neq 0$ , and determine their nature.

Solution. To obtain the stationary points of the graph, we use the zero derivative condition

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = 0.$

Evaluating the left-hand side,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} \left( x + \frac 1x \right) \\ &= \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx} \left( \frac 1x \right) \\ &= 1 + \left( -\frac 1{x^2} \right) \\ &= 1 - \frac 1{x^2}. \end{aligned}$

Therefore, we solve the equation

$\begin{aligned} 1 - \frac 1{x^2} &= 0 \\ 1 &= \frac 1{x^2} \\ x^2 &= 1 \\ x &= \pm 1. \end{aligned}$

Now $y(1) = 1+1/1 = 2$ and $y(-1) = -1 + 1/(-1) = -2$ . Therefore, the stationary (not necessarily turning!) points are given by $(1, 2)$ and $(-1, -2)$ .

To determine their nature, we use the second derivative test:

$\begin{aligned} \frac{\mathrm d^2 y}{\mathrm dx^2} &= \frac{\mathrm d}{\mathrm dx} \left( \frac{\mathrm dy}{\mathrm dx} \right) \\ &= \frac{\mathrm d}{\mathrm dx} \left( 1 - \frac 1{x^2} \right) \\ &= \frac{\mathrm d}{\mathrm dx} ( 1 - x^{-2} ) \\ &= \frac{\mathrm d}{\mathrm dx}(1) - \frac{\mathrm d}{\mathrm dx} (x^{-2}) \\ &= 0 - ( -2x^{-2-1} ) \\ &= 2x^{-3} = \frac{2}{x^3}. \end{aligned}$

Using the second derivative test,

$\begin{aligned} \frac{\mathrm d^2 y}{\mathrm dx^2} \Big|_{x = 1}= \frac{2}{1^3} = 2 > 0,\quad \frac{\mathrm d^2 y}{\mathrm dx^2} \Big|_{x = -1}= \frac{2}{(-1)^3} = -2 < 0. \end{aligned}$

Therefore, $(1, 2)$ is a local minimum while $(-1, -2)$ is a local maximum.

Example 5. The diagram below shows the graphs of

$y = x^3,\quad y = x^4,\quad y = -x^4$

respectively.

For each graph, compute $\displaystyle \frac{\mathrm d y}{\mathrm dx}$ and $\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2}$ at $x = 0$ . What do you notice?

Solution. For the first graph,

$\displaystyle y = x^3,\quad \frac{\mathrm dy}{\mathrm dx} = 3x^2,\quad \frac{\mathrm d^2 y}{\mathrm dx^2} = 6x.$

Therefore,

$\displaystyle \frac{\mathrm dy}{\mathrm dx}\Big|_{x = 0} = 3 \cdot 0^2 = 0 ,\quad \frac{\mathrm d^2 y}{\mathrm dx^2}\Big|_{x = 0} = 6 \cdot 0 = 0.$

For the first graph,

$\displaystyle y = x^4,\quad \frac{\mathrm dy}{\mathrm dx} = 4x^3,\quad \frac{\mathrm d^2 y}{\mathrm dx^2} = 12x^2.$

Therefore,

$\displaystyle \frac{\mathrm dy}{\mathrm dx}\Big|_{x = 0} = 4 \cdot 0^3 = 0 ,\quad \frac{\mathrm d^2 y}{\mathrm dx^2}\Big|_{x = 0} = 12 \cdot 0^2 = 0.$

Similarly, in the case $y = -x^4$ ,

$\displaystyle \frac{\mathrm dy}{\mathrm dx}\Big|_{x = 0} = 0 ,\quad \frac{\mathrm d^2 y}{\mathrm dx^2}\Big|_{x = 0} = 0.$

Remark 2. The point of this exercise is to demonstrate that $f'(c) = 0$ tells us that $(c, f(c))$ is a stationary point, but $f''(c) = 0$ does not give us any meaningful information about the nature of $(c, f(c))$ . The latter could occur in all three types of stationary points.

Now, we can talk about constrained optimisation.

Example 6. Determine the smallest perimeter for a rectangle with area $1$ units².

Solution. Sketch the rectangle as follows.

Since $xy = 1$ , we have $y = 1/x$ . Hence the rectangle has a total perimeter of

$\displaystyle P(x) = 2x + 2y = 2x + 2 \cdot \frac 1x = 2 \underbrace{ \left( x + \frac 1x \right) }_{f(x)}.$

Since $x > 0$ , $f(x) = x + 1/x > 0$ . Hence, $(c, P(c))$ is a local minimum if and only if $(c, f(c))$ is a local minimum. By Example 4, $(1, f(1))$ is a local minimum. Therefore, $(1, P(1)) = (1, 4)$ is a local minimum.

In particular, the smallest perimeter achieved is $4$ units², when $x = y = 1$ , i.e. the rectangle is a square with side length $1$ unit.

Remark 3. Strictly speaking, we need to do more work to show that $(1, 4)$ is a global minimum. However, in the context of high school mathematics, most problems uses expressions that do not need this extra technicality.

Example 7. Determine the smallest surface area for a closed cylinder with volume $10$ units³.

Solution. Sketch the cylinder as follows.

Here, $x > 0$ . Recalling the volume of a cylinder,

$\displaystyle \pi x^2 h = 10\quad \Rightarrow \quad h = \frac{10}{\pi x^2}.$

The surface area $A$ is made up of two circles with area $\pi x^2$ each and one curved surface with area $2 \pi x h$ . Hence,

$\begin{aligned} A(x) &= 2 \cdot \pi x^2 + 2 \pi x h \\ &= 2 \pi x^2 + 2 \pi x \cdot \frac{10}{\pi x^2} \\ &= 2 \pi x^2 + \frac{20}{x}. \end{aligned}$

To use the zero derivative condition, we first calculate $\displaystyle \frac{ \mathrm dA }{ \mathrm dx }$ :

$\begin{aligned} \frac{\mathrm dA}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} \left( 2\pi x^2 + \frac{20}{x} \right) \\ &= 2\pi \cdot \frac{\mathrm d}{\mathrm dx} ( x^2 )+ 20 \cdot \frac{\mathrm d}{\mathrm dx}\left( \frac 1x \right) \\ &= 2\pi \cdot 2x + 20 \cdot \left( -\frac 1{ x^2 } \right) \\ &= 4 \left( \pi x - \frac 5{x^2} \right). \end{aligned}$

Hence, we set $\displaystyle \frac{ \mathrm dA }{ \mathrm dx } = 0$ :

$\begin{aligned} 4 \left( \pi x - \frac 5{x^2} \right) &= 0 \\ \pi x - \frac 5{x^2} &= 0 \\ \pi x &= \frac 5{x^2} \\ x^3 &= \frac 5{\pi}. \end{aligned}$

Therefore, $x = \sqrt[3]{5/\pi} \approx 1.17$ units. For the second derivative test, we calculate the second derivative:

$\begin{aligned} \frac{\mathrm d^2 A}{\mathrm dx^2} &= \frac{\mathrm d}{\mathrm dx} \left( \frac{\mathrm dA}{\mathrm dx}\right) \\ &= \frac{\mathrm d}{\mathrm dx} ( 4\pi x - 20 x^{-2} ) \\ &= 4\pi \cdot \frac{\mathrm d}{\mathrm dx} ( x ) - 20 \cdot \frac{\mathrm d}{\mathrm dx} (x^{-2}) \\ &= 4\pi \cdot 1 - 20 \cdot (-2)x^{-3} \\ &= 4\pi + 40 x^{-3} \\ &= 4\pi + \frac{ 40 }{ x^3 } > 0 \end{aligned}$

whenever $x > 0$ . By the second derivative test, $x = \sqrt[3]{5/\pi}$ yields a local maximum for $A$ :

$\begin{aligned} A &= 20 \pi x^2 + \frac{ 20 }{ x } \\ &= \frac 1{x} (2 \pi x^3 + 20 ) \\ &= \frac{ \sqrt[3]{\pi} }{ \sqrt[3]{5} } \left( 2\pi \cdot \frac 5{\pi} + 20\right) \\ &= \frac{ 30 \sqrt[3]{\pi} }{ \sqrt[3]{5} } \approx 25.7\, \text{units}^2. \end{aligned}$

While there are many possible applications of optimisation, especially in profit-maximisation, we want to switch gears and discuss differentiation’s shadow brother—integration. This we will computationally discuss next time.

For now, let’s resolve an interesting generalisation of the questions we solved just now.

Example 8. For positive constants $A, B$ , use calculus to show that

$\displaystyle \frac{A+B}{2} \geq \sqrt{AB}$

with equality if and only if $A = B$ .

Solution. Define the function $f(x) = Ax + B/x$ for $x > 0$ . Taking derivatives, we leave it as an exercise to check that

$\displaystyle f'(x) = A - \frac B{x^2},\quad f''(x) = \frac{2B}{x^3} > 0.$

By the zero derivative condition,

$\displaystyle f'(x) = 0 \quad \iff \quad x = \sqrt{ \frac BA }.$

Since $\sqrt{B/A} > 0$ , $f''(\sqrt{B/A}) > 0$ . By the second derivative test, $x = \sqrt{B/A}$ yields a local minimum at $f(\sqrt{B/A})$ :

$\displaystyle f(\sqrt{B/A}) = A \sqrt{\frac BA } + \frac{B}{\sqrt{B/A}} = 2\sqrt{A}\sqrt{B} = 2\sqrt{AB}.$

In particular, at $x = 1$ ,

$A + B = f(1) \geq f(\sqrt{B/A}) = 2\sqrt{AB}.$

Dividing by $2$ on both sides,

$\displaystyle \frac{A + B}{2} \geq \sqrt{AB}.$

Equality holds if and only if $\sqrt{B/A} = 1 \iff A = B$ .

Remark 4. The left-hand side $(A+B)/2$ is known as the arithmetic mean (AM) of $A,B$ . The right-hand side $\sqrt{AB}$ is known as the geometric mean (GM) of $A,B$ . Therefore, this result is known as the AM-GM inequality.

Remark 5. An alternative proof from just algebra arises from expanding the left-hand side of the not-so-trivial inequality

$(\sqrt A - \sqrt B)^2 \geq 0.$

—Joel Kindiak, 16 Jan 26, 1215H
March 19, 2026
Leftover Angle Properties

We say that two angles $\alpha, \beta$ are supplementary if $\alpha + \beta = 180^\circ$ .

Question 1. In the diagram below, show that $\alpha + \beta = 360^\circ$ . That is, angles at a point sum to $360^\circ$ .

(Click for Solution)

Solution. Extend the straight line as follows.

Then $\beta = \gamma + 180^\circ$ . Since adjacent angles on a straight line are supplementary, $\alpha + \gamma = 180^\circ$ . Therefore

$\begin{aligned} \alpha + \beta &= \alpha + (\gamma + 180^\circ) \\ &= (\alpha + \gamma) + 180^\circ \\ &= 180^\circ + 180^\circ \\ &= 360^\circ. \end{aligned}$

Question 2. In the diagram below, show that $\alpha + \beta = \gamma$ . That is, the external angle equals the sum of opposite interior angles.

(Click for Solution)

Solution. Construct the angle $\theta$ adjacent to $\gamma$ .

Since angles in a triangle are supplementary,

$\alpha + \beta + \theta = 180^\circ.$

Since adjacent angles on a straight line are supplementary,

$\gamma + \theta = 180^\circ.$

Therefore,

$\alpha + \beta + \theta = 180^\circ = \gamma + \theta.$

Canceling $\theta$ from both sides,

$\alpha + \beta = \gamma,$

as required.

—Joel Kindiak, 15 Jan 26, 1808H

March 18, 2026
Applied Trigonometry
Now that we have motivated the definitions of $\sin(\theta)$ and $\cos(\theta)$ for $0^\circ < \theta < 180^\circ$ , can we extend this idea to $0^\circ \leq \theta \leq 360^\circ$ ?

Lemma 1. For any $0^\circ < \theta < 180^\circ$ , $\sin^2( \theta ) + \cos^2( \theta ) = 1$ .

Proof. The case $0^\circ < \theta < 90^\circ$ simply holds due to the vanilla Pythagoras’ theorem. For the case $90^\circ < \theta < 180^\circ$ , write

$\theta = 180^\circ - \alpha \quad \iff \quad \alpha = 180^\circ - \theta,$

so that $\alpha$ is acute. By our definitions of $\sin(\theta)$ and $\cos(\theta)$ ,

$\begin{aligned} \sin^2(\theta) + \cos^2(\theta) &= (\sin(\alpha) )^2 + (-{\cos(\alpha)})^2 \\ &= \sin^2(\alpha) + \cos^2(\alpha) = 1. \end{aligned}$

If you are thinking that the sum-of-squares formula feels like déjà-vu, you are not wrong. The equation $x^2 + y^2 = 1$ is the famous equation of a unit circle (i.e. a circle with centre $(0, 0)$ and radius $1$ ).

Furthermore, by measuring the clockwise angle relative to the positive $x$ -axis and denoting $P_\theta(x_\theta, y_\theta)$ , we observe a remarkable discovery. For acute $\alpha$ ,

$\displaystyle \frac{ x_\alpha }{ 1 } = \cos(\alpha),\quad \frac{ y_\alpha }{ 1 } = \sin(\alpha).$

Hence, we generalise for any $\theta$ :

$\cos(\theta) := x_\theta, \quad \sin(\theta) := y_\theta.$

When $\theta$ is acute, the results hold as per vanilla trigonometry. When $\theta = 180^\circ - \alpha$ is obtuse,

$x_\theta = -{\cos(\alpha)} = \cos(\theta),\quad y = \sin(\alpha) = \sin(\theta).$

Therefore, we can extend the definitions of $\sin(\theta), \cos(\theta)$ to include $0^\circ \leq \theta \leq 360^\circ$ , though baby trigonometry suggests that the expressions $\sin(0^\circ), \cos(0^\circ)$ seemed rather absurd!

Definition 1. For any $0^\circ \leq \theta \leq 360^\circ$ , draw a line segment $OP_\theta$ whose clockwise angle with the positive $x$ -axis is $\theta$ .

Define $(\cos(\theta), \sin(\theta)) := P_\theta$ .

This definition agrees with the usual definitions of $\cos(\theta), \sin(\theta)$ for $0^\circ < \theta < 180^\circ$ .

Example 1. Let $\alpha$ be an acute angle. Use Definition 1 to evaluate $\sin(\theta)$ and $\cos(\theta)$ for

$\theta = 180^\circ + \alpha,\quad \theta = 360^\circ - \alpha,$

in terms of $\alpha$ . Furthermore, evaluate $\sin(\theta)$ and $\cos(\theta)$ for

$\theta = 0^\circ, \quad \theta = 180^\circ,\quad \theta = 270^\circ,\quad \theta = 360^\circ.$

Solution. We annotate on the usual unit circle diagram.

We can then deduce that

$\begin{aligned} (\cos(180^\circ + \alpha), \sin(180^\circ + \alpha)) &= P_{180^\circ + \alpha} = (-{\cos(\alpha)}, -{\sin(\alpha)}), \\ (\cos(360^\circ - \alpha), \sin(360^\circ - \alpha)) &= P_{360^\circ - \alpha} = ({\cos(\alpha)}, -{\sin(\alpha)}). \end{aligned}$

Similarly,

$\begin{aligned} (\cos(0^\circ), \sin(0^\circ)) &= P_{0^\circ} = (1, 0), \\ (\cos(90^\circ), \sin(90^\circ)) &= P_{90^\circ} = (0, 1), \\ (\cos(180^\circ), \sin(180^\circ)) &= P_{180^\circ} = (-1, 0), \\ (\cos(270^\circ), \sin(270^\circ)) &= P_{270^\circ} = (0, -1), \\ (\cos(360^\circ), \sin(360^\circ)) &= P_{360^\circ} = (1, 0). \end{aligned}$

Remark 1. From this point onward, we will switch our discussions to “radian mode”, given by the conversion $360^\circ = 2\pi$ :

$\begin{aligned} (\cos (\pi - \alpha), \sin(\pi - \alpha)) &= (-{\cos(\alpha)}, \sin(\alpha)), \\ (\cos (\pi + \alpha), \sin(\pi + \alpha)) &= (-{\cos(\alpha)}, -{\sin(\alpha)}), \\ (\cos (2\pi - \alpha), \sin(2\pi - \alpha)) &= ({\cos(\alpha)}, -{\sin(\alpha)}). \end{aligned}$

We will soon discuss calculus, which requires angles to be measured in radians.

Example 2. Evaluate $\sin(7\pi/6)$ , $\cos(7\pi/4)$ exactly.

Solution. Using Remark 1,

$\begin{aligned} \sin(7\pi/6) &= \sin(\pi +\pi/6) = -{\sin(\pi/6)} = -\frac{\sqrt 3}{2}, \\ \cos(7\pi/4) &= \cos(2\pi - \pi/4) = {\cos(\pi/4)} = \frac{1}{\sqrt 2}. \end{aligned}$

Example 3. Derive the definition of $\tan(\theta)$ for $\pi/2 \leq \theta \leq 2\pi$ for the different possibilities of $\theta$ .

Solution. Firstly, for $0 < \theta < \pi/2$ , we have the equivalent definition

$\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.$

Since $\cos(\pi/2) = \cos(3\pi/2) = 0$ , $\tan(\pi/2), \tan(3\pi/2)$ would not be well-defined. Therefore, suppose $\theta \neq \pi/2, 3\pi/2$ . For any $0 < \alpha < \pi/2$ ,

$\begin{aligned} \tan(\pi - \alpha) &= \frac{\sin(\pi - \alpha)}{\cos(\pi - \alpha)} = \frac{\sin(\alpha)}{-{\cos(\alpha)}} = -{\tan(\alpha)}, \\ \tan(\pi + \alpha) &= \frac{\sin(\pi + \alpha)}{\cos(\pi + \alpha)} = \frac{-{\sin(\alpha)}}{-{\cos(\alpha)}} = \tan(\alpha), \\ \tan(2\pi - \alpha) &= \frac{\sin(2\pi - \alpha)}{2\cos(\pi - \alpha)} = \frac{-{\sin(\alpha)}}{\cos(\alpha)} = -{\tan(\alpha)}. \end{aligned}$

Furthermore, we remark that $\cos(k\pi) = (-1)^k \neq 0$ for $k = 0, 1, 2$ . Hence,

$\displaystyle \tan(k\pi) = \frac{ \sin(k\pi) }{ \cos(k\pi) } = \frac{ 0 }{ (-1)^k } = 0.$

This post is titled applied trigonometry, but so far, we haven’t applied it in any meaningful sense. Not yet, at least.

The key is that since we have already defined $\sin(\theta)$ for $0 \leq \theta \leq 2\pi$ , we have enough information to sketch (at least approximately) the graph of one cycle of $y = \sin(t)$ . In fact, using the radians version of Example 1,

$\begin{aligned}\sin(0) &= 0,\quad \sin(\pi/2) = 1, \\ \sin(\pi) &= 0,\quad \sin(3\pi/2) = -1,\quad \sin(2\pi) = 0. \end{aligned}$

Hence, we can divvy-up the interval $0 \leq t \leq 2\pi$ into $5-1 = 4$ regions, and these regions correspond to the $5$ special values of $t$ :

Returning to the unit circle, there is no reason to restrict our graph to $0 \leq x \leq 2\pi$ . Since one complete turn corresponds to an angle of $2\pi$ , the number $\theta + 2\pi$ really just corresponds to an angle of $\theta$ . In fact, by parsing Example 1 in radians,

$\sin(0) = 0 = \sin(2\pi),\quad \cos(0) = 1 = \cos(2\pi).$

Hence, we can extend the definitions of $\sin(\theta)$ and $\cos(\theta)$ accordingly.

Definition 2. For any integer $k$ and $0 \leq \theta < 2\pi$ , define

$\sin(\theta + 2k\pi) := \sin(\theta),\quad \cos(\theta + 2k\pi) := \cos(\theta).$

Now our complete graph for $y = \sin(t)$ given real number inputs $t$ looks like this:

If this shape looks familiar to the waves that you see on the seaside, once again, you’re not wrong! These wavy shapes are called sinusoids, or more informally, sine waves. We will call the graph of $y = \sin(t)$ the standard sine wave.

The most general form of a sine wave looks like $f(t) = A \sin(\omega t + \phi)$ , and has very natural visual meanings.

Theorem 1. Define the sine wave $f(t) = A \sin(\omega t + \phi)$ , where $A \neq 0$ , $\omega \neq 0$ , and $\phi$ are real constants.
- For any real $t$ , $-|A| \leq f(t) \leq |A|$ .
- The sine wave first repeats itself after a time interval of $2\pi/\omega$ .
- The roots of the sine wave are given by $(k\pi - \phi)/\omega$ , where $k$ is any integer.
In this case, we give the constants the following names:
- $|A|$ is the amplitude of the sine wave,
- $\omega$ is the angular frequency of the sine wave,
- $\phi/\omega$ is the leftward phase shift of the sine wave,
- $2\pi/\omega$ is the period of the sine wave.
Proof Sketch. One cycle of the graph is obtained by the inequality

$\displaystyle 0 \leq \omega t + \phi \leq 2\pi \quad \Rightarrow \quad -\frac{\phi}{\omega} \leq t \leq \frac{2\pi - \phi}{\omega}.$

We calculated the special points using

$\displaystyle \omega t + \phi = k\pi/2 \quad \iff \quad t = \frac{-\phi + k\pi/2}{\omega},$

where $k = 0,1,2,3,4$ . They match the transformed “four sections” of the standard sine wave.

Example 4. Using Theorem 1, sketch the graph of $y = \cos(t)$ for $0 \leq t \leq 2\pi$ .

Solution. Using the complementary and obtuse angle identities, if $0 < t < \pi/2$ , then

$\begin{aligned} \cos(t) &= \sin(\pi/2 - t) \\ &= \sin( \pi - (\pi/2 - t) ) \\ &= \sin(t + \pi/2). \end{aligned}$

It can be shown that this identity holds for any real $t$ . Therefore, $y = \cos(t)$ is a sine wave with:
- amplitude $1$ ,
- period $2\pi$ ,
- leftward phase shift of $\pi/2$ .
Hence, we sketch $y = \cos(t)$ for $0 \leq t \leq 2\pi$ , and duplicate it for $-2\pi \leq t \leq 0$ :

Therefore, for simplicity, we can just work with sine waves.

What happens when two waves $f(t), g(t)$ meet each other? They combine by adding to give the resultant curve $f(t) + g(t)$ . This result is known as the principle of superposition, and the resulting wave is called an interference of waves.

Theorem 2. Let $f(t), g(t)$ be sine waves with angular frequency $\omega$ . Then their resultant curve $f(t)+g(t)$ is a sine wave with angular frequency $\omega$ .

In particular, given positive constants $A, B$ , for any real $t$ ,

$A \sin(t) \pm B \cos(t) = C \sin(t \pm \alpha), \quad 0 \leq \alpha \leq \pi/2$

where $C = \sqrt{A^2 + B^2}$ and $\tan(\alpha) = B/A$ .

Proof. For the general case, see Problem 3 of this post. We will prove the special case directly. Expand the right-hand side using the addition formula:

$C \sin(t \pm \alpha) = C \sin(t) \cos(\alpha) \pm C \cos(t) \sin(\alpha).$

Setting $A = C \cos(\alpha)$ and $B = C \sin(\alpha)$ ,

$C \sin(t \pm \alpha) = A \sin(t) \pm B \cos(t).$

Using the Pythagorean identity,

$\begin{aligned} A^2 + B^2 &= C^2 \cos^2(\alpha) + C^2 \sin^2(\alpha) \\ &= C^2 (\cos^2(\alpha) + \sin^2(\alpha)) \\ &= C^2 \cdot 1 = C^2. \end{aligned}$

Furthermore,

$\displaystyle \frac{B}{A} = \frac{C \sin(\alpha)}{C \cos(\alpha)} = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha).$

A direct proof is possible but far more cumbersome, and not terribly helpful for our discussions.

Sine waves are responsible for Fourier series that make modern electronics possible in the first place, so if you intend to explore electronics, you will find them helpful.

What we have discussed up to this point covers much of pre-calculus. What lies ahead seems tricky for many but turns out to be one of the most versatile branches of high school mathematics with respect to further studies in college and university—calculus.

We will explore calculus from a computational point of view, rather than explore its rich underlying theory. That exploration can take us down a very, very deep rabbit hole called real analysis, which we shall relegate as an ambitious exercise for a select subset of students.

For now, we shall turn to the first idea of our consideration: differentiation.

—Joel Kindiak, 20 Dec 25, 1240H
March 13, 2026
Baby Trigonometry

Now let’s discuss trigonometry, the bane of high school mathematics. In spite of its rather tragic reputation, the life goal of trigonometry is simple:

What is the relationship between line segments and angles?

An angle, at its heart, deals with circular motion. Angles take considerable effort to construct, but their main purpose is to quantify the “degree” (pun intended) of separation between two line segments.

Trigonometry aims to capture the precise effect angles have on line segments. The name trigono-metry itself suggests our starting point—the triangle. Furthermore, we should start with the simplest triangle—the right-angled triangle—then work our way up to more general triangles.

Definition 1. Consider the following right-angled triangle with acute angle $0^\circ < \theta < 90^\circ$ .

We abbreviate the words opposite, adjacent, and hypotenuse. We define the sine, cosine, and tangent of $\theta$ as follows:

$\begin{aligned} \sin(\theta) := \frac{\text{opp}}{\text{hyp}}, \quad \cos(\theta) := \frac{\text{adj}}{\text{hyp}}, \quad \tan(\theta) := \frac{\text{opp}}{\text{adj}}. \end{aligned}$

These definitions make sense thanks to similar triangles (i.e. right-angled triangles with the same shape will still give the same trigonometric ratios).

Example 1. Using a suitably-drawn triangle, evaluate the trigonometric ratios

$\sin(45^\circ), \quad \cos(45^\circ),\quad \tan(45^\circ).$

Solution. We draw a right-angled isosceles triangle with side length $1$ , whose base angles must be $(180^\circ - 90^\circ) /2 = 45^\circ$ :

Using Pythagoras’ theorem, the triangle has hypotenuse

$\sqrt{1^2 + 1^2} = \sqrt 2.$

By Definition 1,

$\displaystyle \sin(45^\circ) = \frac 1{\sqrt 2},\quad \cos(45^\circ) = \frac 1{\sqrt 2},\quad \tan(45^\circ) = \frac 1{1} = 1.$

Example 2. Using a suitably-drawn triangle, evaluate the trigonometric ratios

$\sin(60^\circ), \quad \cos(60^\circ),\quad \tan(60^\circ).$

Solution. We draw an equilateral triangle divided by its altitude:

We leave it as an exercise to check that the two right-angled triangles are congruent. Hence, relative to $60^\circ$ , the triangle has an adjacent side $2/2 = 1$ . Using Pythagoras’ theorem, the triangle has opposite side

$\sqrt{2^2 - 1^2} = \sqrt 3.$

By Definition 1,

$\displaystyle \sin(60^\circ) = \frac{\sqrt 3}2, \quad \cos(60^\circ) = \frac 12,\quad \tan(60^\circ) = \frac{\sqrt 3}{1} = \sqrt 3.$

Example 3. Use the same diagram in Example 2 to evaluate the trigonometric ratios

$\sin(30^\circ), \quad \cos(30^\circ),\quad \tan(30^\circ).$

Solution. Since angles in a triangle sum to $180^\circ$ , we can draw the $30^\circ$ angle as follows:

Relative to $30^\circ$ , the triangle has opposite side $1$ , adjacent side $\sqrt 3$ , and hypotenuse $2$ . By Definition 1,

$\displaystyle \sin(30^\circ) = \frac{1}2, \quad \cos(30^\circ) = \frac{\sqrt 3}2,\quad \tan(30^\circ) = \frac{1}{\sqrt 3}.$

Now, if we scrutinise these results more carefully, we will notice that

$\cos(60^\circ) = \sin(30^\circ),\quad \cos(30^\circ) = \sin(60^\circ), \quad \tan(60^\circ) \tan(30^\circ) = 1.$

These seeming coincidences are, in fact, not coincidences. We have several results that will always hold for any acute $\theta$ :

Theorem 1. For any acute angle $\theta$ ,

$\cos(\theta) = \sin(90^\circ - \theta),\quad \tan(\theta) \tan(90^\circ - \theta) = 1.$

Proof. Consider the triangle below:

Using Definition 1,

$\displaystyle \cos(\theta) = \frac{a}{c} = \sin(\alpha)$

and

$\displaystyle \tan(\theta) \tan(\alpha) = \frac{b}{a} \cdot \frac ab = 1.$

Since angles in a triangle sum to $180^\circ$ ,

$\alpha + \theta + 90^\circ = 180^\circ \quad \Rightarrow \quad \alpha = 90^\circ - \theta.$

Substituting $\alpha = 90^\circ - \theta$ ,

$\cos(\theta) = \sin(90^\circ - \theta),\quad \tan(\theta) \tan(90^\circ - \theta) = 1.$

We call the acute angles $\alpha, \beta$ complementary if $\alpha + \beta = 90^\circ$ . Hence, $90^\circ - \theta$ is complementary to $\theta$ . In fact, the “co-” in cosine really stands for “complement”, so that $\cos(\theta)$ is the sine of the complement of $\theta$ , i.e.

$\cos(\theta) = \sin(90^\circ - \theta).$

Remark 1. To reduce the clunkiness of the expression, we make the (tragically ambiguous) notations

$\sin^2(\theta):=(\sin(\theta))^2 , \quad \cos^2(\theta):=(\cos(\theta))^2.$

Theorem 2. For any acute angle $\theta$ ,

$\sin^2(\theta) + \cos^2(\theta) = 1.$

Proof. Consider the right-angled triangle below.

Using Definition 1 and Pythagoras’ theorem,

$\displaystyle \sin^2(\theta) + \cos^2(\theta) = \frac{b^2}{c^2} + \frac{a^2}{c^2} = \frac{b^2 + a^2}{c^2} = \frac{c^2}{c^2} = 1.$

This result is simply the Pythagoras’ theorem with the special case hypotenuse equals $1$ .

Strangely enough, together with geometry, we have everything we need to explore high school trigonometry. An alternate formulation of trigonometry, called rational trigonometry, captures more or less the same ideas but simplifies calculations.

Question 1. Given the following triangle with known side lengths $a , b$ and an acute angle $C$ between them, how do we calculate its area?

Solution. Let $h$ denote the height of the triangle.

Then the triangle has area $A = \frac 12 a h$ . By Definition 1,

$\displaystyle \sin(C) = \frac{h}{b} \quad \Rightarrow \quad h = b \sin(C)$ .

Therefore, $A = \frac 12 ab \sin (C)$ .

Remark 2. The proof still holds if $\angle ABC$ is right or obtuse.

Question 2. If instead the angle $C$ is obtuse, how would we calculate the area of the triangle?

Solution. Extend the base of the triangle by $a$ units as follows:

By Question 1, $\Delta AB'C$ has area $\frac 12 ab \sin(\alpha)$ . Since angles on a straight line sum to $180^\circ$ ,

$C + \alpha = 180^\circ \quad \Rightarrow \quad \alpha = 180^\circ - C.$

Since both triangles have the same base and height, they must have the same area $A$ , so that

$A = \frac 12 ab \sin(\alpha) = \frac 12 ab \sin(180^\circ - C).$

If we insist that the formula

$\text{(area of triangle)} = \frac 12 ab \sin(C)$

holds for obtuse $C$ , then we must have

$\sin(C) = \sin(180^\circ - C),$

just as we have explored.

Question 3. Using the same reasoning, how should we define $\sin(90^\circ)$ ?

Solution. Consider the right-angled triangle below.

Assuming the desired area formula holds, we have

$\frac 12 ab \sin(90^\circ) = \frac 12 ab \quad \Rightarrow \quad \sin(90^\circ) = 1.$

This result motivates our definition of $\sin(\theta)$ for $0^\circ < \theta < 180^\circ$ .

Definition 2. Define $\sin(90^\circ) := 1$ . For any acute $\theta$ , define

$\sin(180^\circ - \theta) := \sin(\theta) > 0.$

Using this result, we obtain the beautiful law of sines.

Theorem 3 (Law of Sines). Given the triangle below,

$\displaystyle \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}.$

This result still holds even if some angle is right or obtuse.

Proof. Since the area formula holds in any type of angle (acute, right, obtuse),

$\frac 12 bc \sin(A) = \frac 12 ca \sin(B) = \frac 12 ab \sin(C).$

Dividing by $\frac 12 abc$ on all sides,

$\displaystyle \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}.$

If there is a law of sines, would there be a law of cosines? Yes, and in fact we obtain it via Pythagoras’ theorem.

Lemma 1. Given the triangle below with all acute angles,

$c^2 = a^2 + b^2 - 2ab \cos(C).$

Proof. Sub-divide $a = s+t$ and draw the height of the triangle.

By Pythagoras’ theorem,

$\begin{aligned} c^2 &= t^2 + h^2 \\ &= (a-s)^2 + (b^2 - s^2) \\ &= a^2 - 2as + s^2 + b^2 - s^2 \\ &= a^2 + b^2 - 2as. \end{aligned}$

Using Definition 1,

$\displaystyle \cos(C) = \frac{s}{b} \quad \Rightarrow \quad s = b \cos(C).$

Therefore, $c^2 = a^2 + b^2 - 2ab \cos(C)$ .

Lemma 2. Given the triangle below with obtuse angle $C$ ,

$c^2 = a^2 + b^2 + 2ab \cos(180^\circ - C).$

Proof. Draw the height of the triangle and extend $BC$ as follows:

By Pythagoras’ theorem again,

$\begin{aligned} c^2 &= (a+s)^2 + h^2 \\ &= (a^2 + 2as + s^2) + (b^2-s^2) \\ &= a^2 + b^2 + 2as. \end{aligned}$

Using Definition 1,

$\displaystyle \cos(\alpha) = \frac sb \quad \Rightarrow \quad s = b \cos(\alpha).$

Therefore, $c^2 = a^2 + b^2 + 2ab \cos(\alpha)$ .

Since angles on a straight line sum to $180^\circ$ ,

$C + \alpha = 180^\circ \quad \Rightarrow \quad \alpha = 180^\circ - C.$

Hence,

$c^2 = a^2 + b^2 + 2ab \cos(180^\circ - C).$

Likewise, if we insist that the formula

$c^2 = a^2 + b^2 - 2ab \cos(C)$

holds for obtuse $C$ , then we must have

$\cos(C) = -{\cos(180^\circ - C)}.$

Question 4. Using the same reasoning, how should we define $\cos(90^\circ)$ ?

Solution. We draw a right-angled triangle:

By Pythagoras’ theorem,

$c^2 = a^2 + b^2.$

Therefore,

$c^2 = a^2 + b^2 - 2ab \cos(90^\circ) \quad \iff \quad \cos(90^\circ) = 0.$

Definition 3. Define $\cos(90^\circ) := 0$ . For any acute $\theta$ , define

$\cos(180^\circ - \theta) := -{\cos(\theta)} < 0.$

Using this result, we obtain the corresponding law of cosines.

Theorem 4 (Law of Cosines). Given the triangle below,

$\displaystyle c^2 = a^2 + b^2 - 2ab \cos(C).$

This result still holds even if $C$ is right or obtuse.

Proof. Apply Lemma 1, Lemma 2, and Question 4.

You would have noticed that we have neglected the tangent function. That is not too surprising, thanks to the following observation.

Theorem 5. For any acute $\theta$ , $\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

Proof. We return to the very first triangle in this blog post.

By Definition 1,

$\displaystyle \frac{\sin(\theta)}{\cos(\theta)} = \frac{ \text{opp}/\text{hyp} }{ \text{adj}/\text{hyp} } = \frac{ \text{opp} }{ \text{adj} } = \tan(\theta).$

Since, $\sin(90^\circ) = 1$ and $\cos(90^\circ) = 0$ , superimposing the identity in Theorem 5 will lead to a mathematical error. Nevertheless, $\cos(\theta)$ is well-defined for obtuse $\theta$ . Hence, we define $\tan(\theta)$ according to Theorem 5.

Definition 4. For obtuse $\theta$ , define

$\displaystyle \tan(\theta) := \frac{\sin(\theta)}{\cos(\theta)}.$

In particular, if $\theta$ is acute, then Definitions 2–4 yield.

$\displaystyle \tan(180^\circ - \theta) = \frac{\sin(180^\circ - \theta)}{\cos(180^\circ - \theta)} = \frac{\sin(\theta)}{-{\cos(\theta)}} = -{\tan(\theta)}.$

Corollary 1 (Obtuse Angle Identities). For any $0^\circ < \theta < 180^\circ$ ,

$\begin{aligned} \sin(180^\circ - \theta) &= \sin(\theta), \\ \cos(180^\circ - \theta) &= -{\cos(\theta)}, \\ \tan(180^\circ - \theta) &= -{\tan(\theta)}, \end{aligned}$

whenever the right-hand side is well-defined.

You may explore more trigonometric identities in this exercise post, in which we used the extended definitions of $\sin(\theta)$ and $\cos(\theta)$ for $0^\circ \leq \theta \leq 360^\circ$ , made possible through a “doubling” trick. Here, the only geometric pre-requisite is a special case of the double angle formula.

Nevertheless, in the spirit of geometric reasoning, we will accomplish the same goal by revisiting an old friend—the unit circle.

—Joel Kindiak, 19 Dec 25, 1334H

March 12, 2026