Category: Exercises

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
Divisibility Tricks
Let $n$ be an integer between $0$ and $9$ inclusive. It is obvious that:
- $n$ is even if and only if $n$ is one of the numbers $0, 2, 4, 6, 8$
- $n$ is a multiple of $3$ if and only if $n$ is one of the numbers $0, 3, 6, 9$ .
- $n$ is a multiple of $5$ if and only if $n$ is one of the numbers $0, 5$ .
Let $m$ be number between $1$ and $9$ inclusive. Define the two-digit number

$x(m,n) := 10m + n.$

For example, $x(6, 7) = 10 \times 6 + 7 = 67$ .

Problem 1. Show that $x(m, n)$ is even if and only if $n$ is even (i.e. $n$ is one of $0, 2, 4, 6, 8$ ).

(Click for Solution)

Solution. Write $x(m,n) = 10m + n$ so that

$n = x(m,n) - 10m.$

$(\Rightarrow)$ If $x(m,n)$ is even, then there exists an integer $k$ such that $x(m,n) = 2k$ . Hence,

$\begin{aligned} n &= x(m,n) - 10m \\ &= 2k - 10m \\ &= 2(k-5m). \end{aligned}$

Since $k - 5m$ is still an integer, $n$ must be even.

$(\Leftarrow)$ If $n$ is even, then there exists an integer $\ell$ such that $n = 2 \ell$ . Hence,

$\begin{aligned} x(m,n) &= 10m + n \\ &= 10m + 2\ell \\ &= 2(5m + \ell). \end{aligned}$

Since $5m + \ell$ is still an integer, $x(m,n)$ must be even.

Remark 1. The same proof can be modified to show that $x(m, n)$ is a multiple of $5$ if and only if $n$ is either $0$ or $5$ . Furthermore, $x(m, n)$ is a multiple of $10$ if and only if $n = 0$ . These results work for numbers regardless of the number of digits, for example:
- $123\, 456$ is even,
- $123\, 455$ is a multiple of $5$ ,
- $123\, 450$ is a multiple of $10$ .
We remark that the one-digit multiples of $3$ are

$\begin{aligned} 3 &= 3 \times 1,\\ 6 &= 3 \times 2,\\ 9 &= 3 \times 3. \end{aligned}$

Problem 2. Show that $x(m, n)$ is a multiple of $3$ if and only if $m+n$ is a multiple of $3$ .

(Click for Solution)

Solution. By definition,

$x(m,n) = 10m + n = 9m + (m+n),$

so that $m+n = x(m,n) - 9m$ .

$(\Rightarrow)$ If $x(m,n)$ is a multiple of $3$ , then there exists an integer $k$ such that $x(m,n) = 3k$ . Hence,

$\begin{aligned} m+n &= x(m,n) - 9m \\ &= 3k - 9m \\ &= 3(k-3m). \end{aligned}$

Since $k - 3m$ is still an integer, $m+n$ must be a multiple of $3$ .

$(\Leftarrow)$ If $m+n$ is a multiple of $3$ , then there exists an integer $\ell$ such that $m+n = 3 \ell$ . Hence,

$\begin{aligned} x(m,n) &= 9m + (m+n) \\ &= 9m + 3\ell \\ &= 3(3m + \ell). \end{aligned}$

Since $3m + \ell$ is still an integer, $x(m,n)$ must be a multiple of $3$ .

Example 1. $57$ is a multiple of $3$ , because

$57 \to 5+7 = 12 \to 1+2 = 3.$

This result works for numbers regardless of the number of digits: since

$12\, 357 \to 1+2+3+5+7 = 18 \to 1+8 = 9,$

$12\, 357$ is a multiple of $3$ .

Remark 2. The same proof can be modified to show that $x(m, n)$ is a multiple of $9$ if and only if $m+n$ is a multiple of $9$ . Furthermore, we can use more advanced techniques to reduce all calculations to the set of one-digit numbers $\{1,2,\dots, 9\}$ .

Now let $n_1, n_2, n_3$ be numbers in $0,1,2,\dots, 9$ , and $n_1 > 0$ . Define the three-digit number

$y(n_1,n_2,n_3) := 100n_1 + 10n_2 + n_3.$

For example

$y(4,2,0) = 100 \times 4 + 10 \times 2 + 0 = 420.$

We remark that the first five multiples of $4$ are

$\begin{aligned} 4 &= 4 \times 1,\\ 8 &= 4 \times 2,\\ 12 &= 4 \times 3, \\ 16 &= 4 \times 4, \\ 20 &= 4 \times 5. \end{aligned}$

Furthermore, $0 = 4 \times 0$ is a multiple of $4$ .

Problem 3. Show that:
- $x(n_2,n_3)$ is a multiple of $4$ if and only if there exists a positive integer $k$ such that $x(n_2 - 2k, n_3)$ is a multiple of $4$ ,
- $y(n_1, n_2, n_3)$ is a multiple of $4$ if and only if $x(n_2,n_3)$ is a multiple of $4$ .
(Click for Solution)

Solution. By definition, $x(n_2, n_3) = 10 n_2 + n_3$ and

$x(n_2 - 2k, n_3) = 10 (n_2-2k) + n_3 = 10n_2 + n_3 - 20k = x(n_2, n_3) - 4 \cdot 5k.$

Since $4 \cdot 5k$ is obviously a multiple of $4$ , $x(n_2, n_3)$ is a multiple of $4$ if and only if $x(n_2 - 2k, n_3)$ is.

By definition of $x(n_2, n_3)$ ,

$y(n_1,n_2,n_3) := 100n_1 + 10n_2 + n_3 = 4 \cdot 25n_1 + x(n_2,n_3).$

Therefore $x(n_2, n_3)$ is a multiple of $4$ if and only if

$y(n_1,n_2,n_3) - 4 \cdot 25n_1$

is a multiple of $4$ , which holds if and only if $y(n_1,n_2,n_3)$ is a multiple of $4$ .

Example 2. Since

$924 \to 24 \to 4,$

$924$ is a multiple of $4$ . In fact, this principle works for any number with more than $2$ digits: $123\, 924$ is also a multiple of $4$ since

$123\, 916 \to 16.$

We can reduce all calculations to the set $\{0, 4 , 8, 12, 16\}$ .

—Joel Kindiak, 16 Jan 26, 1432H
March 6, 2026
Return to Circle Geometry

Problem 1. Consider the circle below with equation $x^2 + y^2 = 1$ .

Given $x_0 > 0$ and $y_0 > 0$ , calculate the gradient of the tangent $T$ passing through the point $P(x_0, y_0)$ lying on the circle. Deduce that $OP \perp T$ .

(Click for Solution)

Solution. Since $(x_0, y_0)$ lies on the circle,

$x_0^2 + y_0^2 = 1.$

For any line with gradient $m$ passing through $(x_0, y_0)$ ,

$y = m(x-x_0) + y_0.$

When the line intersects the circle,

$x^2 + (m(x-x_0) + y_0)^2 = 1.$

Expanding the left-hand side, since $x_0^2 + y_0^2 = 1$ ,

$\begin{aligned} x^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 + y_0^2 &= 1 \\ x^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 -(1-y_0^2) &= 0 \\ x^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 -x_0^2 &= 0 \\ x^2 -x_0^2 + m^2(x-x_0)^2 + 2m(x-x_0)y_0 &= 0 \\ (x-x_0)(x+x_0) + m^2(x-x_0)^2 + 2m(x-x_0)y_0 &= 0, \\ (x-x_0)(x+x_0 + m^2(x-x_0) + 2my_0) &= 0.\end{aligned}$

Hence, $x = x_0$ or

$x+x_0 + m^2(x-x_0) + 2my_0 = 0.$

For $T$ to be a tangent with gradient $m_T$ , we require $x_0$ to be the only root. Hence, we must have

$x_0+x_0 + m_T^2(x_0-x_0) + 2m_T \cdot y_0 = 0.$

Solving for $m_T$ ,

$\displaystyle 2x_0 + 2m_T \cdot y_0 = 0 \quad \Rightarrow \quad m_T = -\frac{x_0}{y_0}.$

On the other hand, $m_{OP} = y_0/x_0$ , so that

$\displaystyle m_T \cdot m_{OP} = -\frac{x_0}{y_0} \cdot \frac{y_0}{x_0} = -1.$

Therefore, $OP \perp T$ . We recover the tangent is perpendicular to radius property of circles.

Remark 1. The same argument holds for any general circle, with more careful book-keeping.

Problem 2. Given that $AB$ and $AC$ are tangents to the circle, show that $AB = AC$ and $AO$ bisects $\angle BOC$ .

This result is described by the phrase tangent from external point.

(Click for Solution)

Solution. By Problem 1, $AB \perp OB$ and $AC \perp OC$ . As radii, $OB = OC$ . As a common side, $AO = AO$ . Therefore, by the RHS Criterion,

$\Delta AOB \cong \Delta AOC.$

In particular, $AB = AC$ and $\angle AOB = \angle AOC$ . Hence, $AO$ bisects $\angle BOC$ .

Problem 3. Given that $ABE$ is tangent to the circle, show that $\angle ABC = \angle BDC$ . Deduce that $\angle EBD = \angle BCD$ . This result is known as the alternate segment theorem.

(Click for Solution)

Solution. Make the following constructions, with $O$ denoting the centre of the circle.

We aim to prove that $\alpha = \beta$ . By Problem 1, $\angle ABO = 90^\circ$ . Hence,

$\angle OBC = 90^\circ - \alpha.$

Since $OB = OC$ as radii, $\Delta BOC$ is isosceles. Hence,

$\angle OCB = \angle OBC = 90^\circ - \alpha.$

Since the angle at the centre equals twice the angle at the circumference,

$\angle BOC = 2 \cdot \angle BDC = 2 \beta.$

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

$\begin{aligned} \angle BOC + \angle OBC + \angle OCB &= 180^\circ. \end{aligned}$

Substituting the displays,

$\begin{aligned} 2\beta + (90^\circ - \alpha) + (90^\circ - \alpha) &= 180^\circ \\ 2 \beta + 180^\circ - 2\alpha &= 180^\circ \\ 2\beta - 2\alpha &= 0 \\ 2\beta &= 2\alpha \\ \beta &= \alpha. \end{aligned}$

Therefore, $\angle BDC = \alpha = \beta = \angle ABC$ , as required.

For the second result,

$\begin{aligned} \angle EBD &= 180^\circ - \angle ABC - \angle ABC\quad &(\text{adj}\ \angle\ \text{on st. line}) \\ &= 180^\circ - \angle ABC - \angle BDC \quad &(\angle ABC = \angle BDC) \\ &= \angle BCD.\quad &(\angle\ \text{sum of}\ \Delta) \end{aligned}$

—Joel Kindiak, 9 Jan 26, 1146H

January 9, 2026
Algebraic Calculus

Consider the graph of $y = x^2$ below.

Define the point $P_t$ on the graph by $(t, t^2)$ .

Use this graph to answer Problems 1–4.

Problem 1. Evaluate the gradient of $P_1 P_2$ .

(Click for Solution)

Solution. Since $P_1(1, 1)$ and $P_2(2, 4)$ , the gradient is given by

$\displaystyle m_{P_1P_2} = \frac{4-1}{2-1} = \frac 31 = 3.$

Problem 2. Given $h$ , evaluate the gradient of $P_1 P_{1+h}$ .

Verify your answer in Problem 1.

(Click for Solution)

Solution. Since $P_{1+h}(1+h, (1+h)^2)$ , the gradient is given by

$\begin{aligned} m_{P_1P_{1+h}} &= \frac{(1+h)^2-1}{(1+h)-1} \\ &= \frac{(1+2h+h^2)-1}{(1+h)-1} \\ &= \frac{2h+h^2}{h} \\ &= \frac{(2+h)\cdot h}{h} \\ &= 2+h. \end{aligned}$

This answer agrees with Problem 1, since setting $h = 1$ gives

$\begin{aligned} m_{P_1P_{1+h}} &= 2 + h \\ &= 2 + 1 \\ &= 3 = m_{P_1P_2}. \end{aligned}$

Problem 3. Given $t$ and $h$ , evaluate the gradient of $P_t P_{t+h}$ .

Verify your answer in Problem 2.

(Click for Solution)

Solution. Since $P_t(t,t^2)$ and $P_{t+h}(t+h, (t+h)^2)$ , so that the gradient is given by

$\begin{aligned} m_{P_tP_{t+h}} &= \frac{(t+h)^2-t^2}{(t+h)-t} \\ &= \frac{(t^2+2th+h^2)-t^2}{(t+h)-t} \\ &= \frac{2th+h^2}{h} \\ &= \frac{(2t+h)\cdot h}{h} \\ &= 2t+h. \end{aligned}$

This answer agrees with Problem 2 since setting $t = 1$ gives

$\begin{aligned} m_{P_tP_{t+h}} &= 2t + h \\ &= 2 \cdot 1 + h \\ &= 2 + h \\ &= m_{P_1P_{1+h}}. \end{aligned}$

Problem 4. Let $L_t$ denote the line passing through $P_t$ . Show that $L_t$ is tangent to $y = x^2$ if and only if its gradient is $2t$ .

(Click for Solution)

Solution. Denote the gradient of $L_t$ by $m$ . Then

$L_t : y = m(x - t) + t^2.$

When $L_t$ intersects the curve $y = x^2$ ,

$x^2 = m(x-t) + t^2$

so that solving yields

$\begin{aligned} x^2 - t^2 - m(x-t) &= 0 \\ (x-t)(x+t) - m(x-t) &= 0 \\ (x-t)(x+t-m) &= 0. \end{aligned}$

Then $x = t$ or $x = m - t$ . For $L_t$ to be a tangent to the curve $y = x^2$ at $P_t$ , we must have both roots equal $t$ , so that

$t = m-t \quad \iff \quad m = 2t,$

as required.

Therefore, by setting $h = 0$ in Problem 3, we obtain the answer in Problem 4: the expression $2t$ . Intuitively, this expression describes the gradient of the tangent at $(t, t^2)$ .

Problem 5. Now consider the graph of $y =x^3$ .

Define the point $Q_t$ on the graph by $(t, t^3)$ .

For any $t$ and $h$ , evaluate the gradient of $Q_t Q_{t+h}$ in terms of $h$ .

(Click for Solution)

Solution. Using the same strategy as per Example 1, since $Q_{t+h}(t+h, (t+h)^3)$ , the gradient is given by

$\begin{aligned} m_{P_tP_{t+h}} &= \frac{(t+h)^3-t^3}{(t+h)-t} = \frac{(t+h)^3-t^3}{ h }. \end{aligned}$

We first expand $(t+h)^3$ :

$\begin{aligned} (t+h)^3 &= (t+h)^2 \cdot (t+h) \\ &= (t^2 + 2th + h^2)(t + h) \\ &= t^3 + (2t^2 + t^2)h + (t+2t)h^2 + h^3 \\ &= t^3 + 3t^2h + 3th^2 + h^3.\end{aligned}$

Therefore,

$\begin{aligned} m_{P_tP_{t+h}} & = \frac{(t+h)^3-t^3}{ h } \\ & = \frac{3t^2h + 3th^2 + h^3}{ h } \\ & = \frac{(3t^2 + 3th + h^2)h}{ h } \\ &= 3t^2 + 3th + h^2. \end{aligned}$

If we set $h = 0$ in the final result, we obtain the expression $3t^2$ . Intuitively, this expression describes the gradient of the tangent at $(t, t^3)$ .

Problem 6. Given that $P_t$ lies on $y = x^n$ , evaluate the gradient of the tangent to $P_t$ . You may freely use the factorisation

$x^n - t^n = (x-t)(x^{n-1} + x^{n-2}t + \cdots + xt^{n-2} + t^{n-1}).$

without proof.

(Click for Solution)

Solution. Denote $f(x) := x^n$ and

$\varphi (x) := x^{n-1} + x^{n-2}t + \cdots + xt^{n-2} + t^{n-1}$

so that $f(x)- f(t) = (x-t) \cdot \varphi (x)$ for brevity (and possible generality).

We follow the approach in Problem 4. Denote the gradient of the tangent $L_t$ to $P_t$ by $m$ . Then

$L_t : y = m(x-t) + f(x).$

When $L_t$ intersects the curve $y = f(x)$ ,

$f(x) = m(x-t) + f(t).$

Solving yields

$\begin{aligned} f(x) - f(t) - m(x-t) &= 0 \\ (x-t) \cdot \varphi (x) - m(x-t) &= 0 \\ (x-t)(\varphi (x) - m) &= 0. \end{aligned}$

Hence, $x = t$ or $\varphi (x) = m$ . For $L_t$ to be tangent to $P_t$ , we must have $\varphi (t) = m$ and vice versa. In particular,

$m = \varphi (t) = n \cdot t^{n-1}.$

Remark 1. Problem 6 generalises to other kinds of functions using Carathéodory’s theorem.

—Joel Kindiak, 7 Jan 26, 1542H

January 7, 2026
Trigonometric Identities

In these exercises, you may use the following addition formulae for $\sin$ and $\cos$ : for any $A, B \in \mathbb R$ ,

$\begin{aligned} \sin(A+B) &= \sin(A) \cos(B) + \cos(A) \sin(B), \\ \cos(A+B) &= \cos(A) \cos(B) - \sin(A) \sin(B). \end{aligned}$

Problem 1. Prove the following negative identities: for any $A \in \mathbb R$ , whenever the right-hand side is well-defined,

$\sin(-A) = -\sin(A),\quad \cos(-A) = \cos(A),\quad \tan(-A) = -\tan(A).$

Deduce the subtraction formulae for $\sin$ and $\cos$ : for any $A, B \in \mathbb R$ ,

$\begin{aligned} \sin(A-B) &= \sin(A) \cos(B) - \cos(A) \sin(B), \\ \cos(A-B) &= \cos(A) \cos(B) + \sin(A) \sin(B). \end{aligned}$

(Click for Solution)

Solution. For the first two identities, we use the addition formulae to derive

$\begin{aligned} 0 = \sin(-A + A) &= \sin(-A) \cos(A) + \cos(-A) \sin(A), \\ 1 = \cos(-A + A) &= \cos(-A) \cos(A) - \sin(-A) \sin(A). \end{aligned}$

Solving simultaneous linear equations yields $\sin(-A) = -\sin(A)$ and $\cos(-A) = \cos(A)$ . For the final identity,

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

Replacing $B$ with $-B$ in the addition formulae,

$\begin{aligned} \sin(A-B) &= \sin(A+(-B)) \\ &= \sin(A)\cos(-B) + \cos(A) \sin(-B)\\ &= \sin(A) \cos(B) - \cos(A) \sin(B). \end{aligned}$

A similar calculation yields the cosine subtraction formula.

Problem 2. Prove the following double angle identities: for any $A \in \mathbb R$ ,

$\sin(2A) = 2 \sin(A) \cos(A),\quad \cos(2A) =\cos^2(A) - \sin^2(A).$

(Click for Solution)

Solution. Setting $A = B$ ,

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

Problem 3. Prove that for any $A \in \mathbb R$ ,

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

Deduce that for any $A \in \mathbb R$ ,

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

(Click for Solution)

Solution. For the first identity, it suffices to prove the result for $A \in (0, 2\pi]$ , since

$\sin(A + n \cdot 2\pi) = \sin(A),\quad \cos(A + n \cdot 2\pi) = \cos(A).$

We have already proven the case $A \in (0, \pi/4]$ using the vanilla Pythagorean theorem. For the case $A \in (0, \pi/2]$ , denote $A_2 = A/2$ and use Problem 2 to obtain

$\begin{aligned} \sin^2(A) + \cos^2(A) &= \sin^2(2A_2) + \cos^2(2A_2) \\ &= (2 \sin(A_2) \cos(A_2))^2 + (\cos^2(A_2) - \sin^2(A_2))^2 \\ &= \sin^4(A_2) + 2 \sin^2(A_2) \cos^2(A_2) + \cos^4(A_2) \\ &= (\sin^2(A_2) + \cos^2(A_2))^2 = 1^2 = 1. \end{aligned}$

By repeating the argument, we can “double” the result to hold for $(0, \pi]$ and $(0, 2\pi]$ , as required. For the second identity, use Problem 2 and the first identity to obtain

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

Problem 4. Define $\tan(A) := \sin(A)/{\cos(A)}$ whenever the right-hand side is well-defined. Define the following reciprocal trigonometric functions whenever their right-hand sides are well-defined:

$\displaystyle \sec(A) := \frac{1}{\cos(A)},\quad \csc(A) := \frac{1}{\sin(A)},\quad \cot(A) := \frac{1}{\tan(A)}.$

Prove the following Pythagorean identities, whenever they are well-defined:

$\begin{aligned} \tan^2(A) + 1 &= \sec^2(A), \\ 1 + \cot^2(A) &= \csc^2(A). \end{aligned}$

(Click for Solution)

Solution. Using Problem 3,

$\begin{aligned} \tan^2(A) + 1 &= \frac{\sin^2(A)}{\cos^2(A)} + 1 \\ &= \frac{\sin^2(A) + \cos^2(A)}{\cos^2(A)} \\ &= \frac 1{\cos^2(A)} = \sec^2(A). \end{aligned}$

Similarly,

$\begin{aligned} 1 + \cot^2(A) &= 1 + \frac{\cos^2(A)}{\sin^2(A)} + 1 \\ &= \frac{\sin^2(A) + \cos^2(A)}{\sin^2(A)} \\ &= \frac 1{\sin^2(A)} = \csc^2(A). \end{aligned}$

Problem 5. Prove the following half angle identities: for any $A \in \mathbb R$ ,

$\displaystyle \sin^2(A/2) =\frac {1 - \cos(A)}2, \quad \cos^2(A/2) =\frac {1 + \cos(A)}2.$

(Click for Solution)

Solution. Using Problem 3,

$\cos(A) = 2 \cos^2(A/2) - 1 = 1 - 2\sin^2(A/2).$

Algebruh yields the desired result.

Problem 6. Prove the following triple angle identities: for any $A \in \mathbb R$ ,

$\begin{aligned} \sin(3A) &=3 \sin(A) - 4 \sin^3(A), \\ \cos(3A) &= 4 \cos^3(A) - 3 \cos(A).\end{aligned}$

(Click for Solution)

Solution. Using the addition formulae and Problem 3,

$\begin{aligned} \sin(3A) &= \sin(A + 2A) \\ &= \sin(A) \cos(2A) + \cos(A) \sin(2A) \\ &= \sin(A) (1 - 2\sin^2(A)) + \cos(A) \cdot 2 \sin(A)\cos(A) \\ &= \sin(A) (1 - 2\sin^2(A)) + 2 \sin(A) (1 - \sin^2(A))\\ &= \sin(A) - 2\sin^3(A) + 2 \sin(A) - 2 \sin^3(A) \\ &= 3 \sin(A) - 4 \sin^3(A). \end{aligned}$

Similarly,

$\begin{aligned} \cos(3A) &= \cos(A + 2A) \\ &= \cos(A) \cos(2A) - \sin(A) \sin(2A) \\ &= \cos(A) (2\cos^2(A)-1) - \sin(A) \cdot 2 \sin(A)\cos(A) \\ &= \cos(A) (2\cos^2(A)-1) - 2 \cos(A) (1 - \cos^2(A))\\ &= 2\cos^3(A) - \cos(A) - 2 \cos(A) + 2 \cos^3(A) \\ &= 4 \cos^3(A) - 3 \cos(A). \end{aligned}$

Problem 7. Prove the following tangent identities: for any $A,B \in \mathbb R$ , whenever the right-hand side is well-defined,

$\begin{aligned} \tan(A + B) &= \frac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}, \\ \tan(3A) &= \frac{3 \tan(A) - \tan^3(A) }{ 1 - 3 \tan^2(A) }.\end{aligned}$

(Click for Solution)

Solution. Using the addition formulae,

$\begin{aligned} \tan(A + B) &= \frac{\sin(A+B)}{\cos(A+B)} \\ &= \frac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A)\sin(B)} \\ &= \frac{\frac{\sin(A)}{\cos(A)} \cdot 1 + 1 \cdot \frac{\sin(B)}{\cos(B)} }{ 1 - \frac{\sin(A)}{\cos(A)} \cdot \frac{\sin(B)}{\cos(B)} } \\ &= \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}. \end{aligned}$

Using Problems 4 and 6,

$\begin{aligned} \tan(3A) &= \frac{\sin(3A)}{\cos(3A)} \\ &= \frac{3 \sin(A) - 4 \sin^3(A)}{4 \cos^3(A) - 3\cos(A)} \\ &= \frac{3 \tan(A) \sec^2(A) - 4 \tan^3(A)}{4 - 3 \sec^2(A)} \\ &= \frac{3 \tan(A) (1 + \tan^2(A)) - 4 \tan^3(A)}{4 - 3 (1 + \tan^2(A))} \\ &= \frac{ (3 \tan(A) + 3 \tan^3(A)) - 4 \tan^3(A)}{4 - (3 + 3\tan^2(A))} \\ &= \frac{ 3 \tan(A) - \tan^3(A) }{1 - 3\tan^2(A)}. \end{aligned}$

—Joel Kindiak, 4 Jun 25, 1918H

November 14, 2025
Several Differentiation Drills

For any integer $n$ , you may freely use the results

$\begin{aligned} (x^n)' &= nx^{n-1},\\ \sin'(x) &= \cos(x),\\ \cos'(x) &= -{\sin(x)},\\ \tan'(x) &= \sec^2(x), \\ (e^x)' &= e^x, \\ \ln'(x) &= \frac 1x, \end{aligned}$

from either previous posts or from real analysis. You may also apply the chain rule, product rule, and quotient rule however you wish.

Problem 1. For any $a > 0$ , prove that

$\displaystyle \begin{aligned} (a^x)' = a^x \ln(a), \quad \log_a'(x) = \frac{1}{x \ln(a)}. \end{aligned}$

(Click for Solution)

Solution. By exponential properties,

$a^x = (e^{\ln(a)})^x = e^{x \ln(a)}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (a^x)' &= (e^{x \ln(a)})'\\ &= e^{x \ln(a)} \cdot \frac{\mathrm d}{\mathrm dx}(x \ln a) \\ &= e^{x \ln(a)} \cdot \ln(a) \\ &= a^x \ln(a). \end{aligned}$

By logarithm properties and linearity,

$\displaystyle \begin{aligned} \log_a'(x) &= \left( \frac{\ln(x)}{\ln(a)} \right)'\\ &= \frac{1}{\ln(a)} \ln'(x) \\ &= \frac{1}{\ln(a)} \cdot \frac 1x = \frac 1{x \ln (a)}. \end{aligned}$

Up till now, we have proven that $(x^n)' = nx^{n-1}$ only for integer values of $n$ . We extend this result to all real values of $n$ .

Problem 2. For any real $r$ , use Problem 1 to prove that $(x^r)' = rx^{r-1}$ .

(Click for Solution)

Solution. By exponential properties,

$x^r = (e^{\ln x})^r = e^{r \ln x}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (e^{r \ln x})'\\ &= e^{r \ln x} \cdot \frac{\mathrm d}{\mathrm dx}(r \ln x) \\ &= x^r \cdot r \cdot \frac 1x = rx^{r-1}. \end{aligned}$

Example 1. By Problem 2,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{\sqrt 2}) = \sqrt 2 \cdot x^{\sqrt 2 - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^\pi) = \pi \cdot x^{\pi - 1}.$

In particular,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}\neq e^x = \frac{\mathrm d}{\mathrm dx}(e^x).$

Problem 3. Prove that

$\displaystyle \sec'(x) = \sec(x)\tan(x),\quad \csc'(x) = -{\csc(x)\cot(x)}.$

(Click for Solution)

Solution. Writing $\displaystyle \sec(x) = \frac 1{\cos(x)}$ , apply the quotient rule to obtain

$\displaystyle \begin{aligned} \sec'(x) &= \frac{ (1)' \cdot \cos(x) - 1 \cdot \cos'(x) }{ \cos^2(x) } \\ &= \frac{ 0 \cdot \cos(x) - 1 \cdot (-\sin(x)) }{ \cos^2(x) } \\ &= \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x) \tan(x). \end{aligned}$

We recall that $\csc(x) = \sec(\pi/2-x)$ so that

$\displaystyle \begin{aligned}\csc'(x) &= (\sec(\pi/2-x))' \\ &= \sec'(\pi/2-x)) \cdot (-1) \\ &= \sec(\pi/2-x) \tan(\pi/2-x) \cdot (-1) \\ &= \csc(x) \cot(x) \cdot (-1) \\ &= - \csc(x) \cot(x). \end{aligned}$

Problem 4. Given that $t > 0$ , $x \equiv x(t) > 0$ , and $y \equiv y(t) > 0$ , evaluate $\displaystyle \frac{\mathrm d}{\mathrm dt}(x^y)$ .

(Click for Solution)

Solution. Writing $x^y = e^{y \ln(x)}$ and using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dt}(x^y) &= \frac{\mathrm d}{\mathrm dt} (e^{y \ln(x)}) \\ &= e^{y \ln(x)} \left( \frac{\mathrm d}{\mathrm dt}(y \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac{\mathrm d}{\mathrm dt} ( \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac 1x \cdot \frac{\mathrm dx}{\mathrm dt} \right) \\ &= x^y \ln(x) \cdot \frac{\mathrm dy}{\mathrm dt} + yx^{y-1} \cdot \frac{\mathrm dx}{\mathrm dt}. \end{aligned}$

Remark 1. If $x = t$ and $y$ is constant, we recover the power rule. If $x$ is constant and $y = t$ , we recover Problem 1. Furthermore, if $y = x = t$ , we obtain

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x) = x^x (\ln(x) + 1).$

In the language of partial derivatives, denoting $z = x^y \equiv { x(t) }^{ y(t) }$ ,

$\displaystyle \frac{\mathrm d z}{\mathrm dt} = \frac{\partial z}{\partial y} \cdot \frac{\mathrm dy}{\mathrm dt} + \frac{\partial z}{\partial x} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Problem 5. Suppose $f, f^{-1}$ are differentiable on $\mathbb R$ . Evaluate $(f^{-1})'(x)$ .

(Click for Solution)

Solution. By definition of the inverse function,

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

—Joel Kindiak, 9 Feb 25, 2126H

February 9, 2025

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