Category Archives: Algebra

October 15, 2025 · 6:19 am

Hard Equation Solving Question

Find the value(s) of $k$ such that the equation below has two numerically equal but opposite sign solutions (e.g. $5$ and $-5$ ).

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{k-1}{k+1}\end{equation}$

$\begin{equation*}(x^2-2x)(k+1)=(k-1)(4x-1)\end{equation}$

$\begin{equation*}(k+1)x^2-2kx-2x=4kx-k-4x+1\end{equation}$

$\begin{equation*}(k+1)x^2-2kx-4kx-2x+4x-1=0\end{equation}$

$\begin{equation*}(k+1)^2x^2-(6k-2)x-1=0\end{equation}$

For there to be two numerically equal but opposite sign solutions, the $b$ term of the quadratic equation must be $0$ .

$\begin{equation*}6k-2=0\end{equation}$

Hence $k=\frac{1}{3}$ .

When $k=\frac{1}{3}$ the equation becomes

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{\frac{-2}{3}}{\frac{4}{3}}\end{equation}$

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{-1}{2}\end{equation}$

$\begin{equation*}2x^2-4x=-4x+1\end{equation}$

$\begin{equation*}2x^2-1=0\end{equation}$

$\begin{equation*}x^2=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\pm \frac{1}{\sqrt{2}}\end{equation}$

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Filed under Algebra, Polynomials, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as hard algebra question, quadratic, solving quadratic equations

October 9, 2025 · 6:37 am

Trigonometric Exact Values

Find exactly $sin(18^\circ)$

We must be able to find an arithmetic combination of the exact values we knew to find $18$ .

$\begin{equation*}90=5\times 18\end{equation}$

$\begin{equation*}90-3(18)=2(18)\end{equation}$

I re-arranged as above, so I could take advantage of $cos(90)=0$ and $sin(90)=1$

cos(2x)=cos^2(x)-sin^2(x)=1-2sin^2(x)=2cos^2(x)-1

$\begin{equation*}sin(90-3(18))=sin(2(18))\end{equation}$

$\begin{equation*}sin(90)cos(3(18))-sin(3(18))cos(90)=2sin(18)cos(18)\end{equation}$

$\begin{equation*}cos(3(18))=2sin(18)cos(18)\end{equation}$

$\begin{equation*}4cos^3(18)-3cos(18)=2sin(18)cos(18)\end{equation}$

$\begin{equation*}4cos^3(18)-3cos(18)-2sin(18)cos(18)=0\end{equation}$

$\begin{equation*}cos(18)(4cos^2(18)-3-2sin(18))=0\end{equation}$

Hence,

$\begin{equation*}4cos^2(18)-2sin(18)-3=0\end{equation}$

$\begin{equation*}4-4sin^2(18)-2sin(18)-3=0\end{equation}$

$\begin{equation*}-4sin^2(18)-2sin(18)+1=0\end{equation}$

Use the quadratic equation formula

$\begin{equation*}sin(18)=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}sin(18)=\frac{2 \pm \sqrt{4-4(-4)(1)}}{-8}\end{equation}$

$\begin{equation*}sin(18)=\frac{2 \pm \sqrt{20}}{-8}\end{equation}$

$\begin{equation*}sin(18)=\frac{-2 \mp 2\sqrt{5}}{8}\end{equation}$

$\begin{equation*}sin(18)=\frac{-1 \mp \sqrt{5}}{4}\end{equation}$

As $sin(18)>0$ , $sin(18)=\frac{-1+\sqrt{5}}{4}$

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Filed under Addition and Subtraction Identities, Algebra, Identities, Quadratic, Quadratics, Solving, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as exact values, sin(18), trig exact values, trigonometric identities

September 30, 2025 · 3:54 am

Geometry Problem

*Geometry Snacks* by Ed Southall and Vincent Pantaloni

I started by trisecting another side of the triangle

This makes it clearer that the two lines are parallel

Which means the two angles labelled above are corresponding and therefore congruent.

Let the side length be $x$ .

The area of the equilateral triangle is

(1) $\begin{equation*}A=\frac{1}{2}x^2 sin(60)=\frac{\sqrt{3}x^2}{4}\end{equation*}$

$\begin{equation*}cos(60)=\frac{y}{\frac{2x}{3}}\end{equation}$

$\begin{equation*}y=\frac{x}{3}\end{equation}$

Area of right triangle

(2) $\begin{equation*}A=(\frac{1}{2})(\frac{2x}{3})(\frac{x}{3})sin(60)=\frac{\sqrt{3}x^2}{18}\end{equation*}$

The fraction of the area is

$\begin{equation*}=\frac{\frac{\sqrt{3}x^2}{18}}{\frac{\sqrt{3}x^2}{4}}=\frac{2}{9}\end{equation}$

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Filed under Algebra, Area, Area of Triangles (Sine), Finding an area, Geometry, Puzzles, Right Trigonometry, Simplifying fractions, Trigonometry

Tagged as geometry problem, geometry snacks

September 17, 2025 · 7:03 am

Divisibility Rule for 11

I was working on a question and involved 11 and I wondered what the divisibility rule was?

So then I had a bit of a think about it.

Let $N$ be a number divisible by $11$ . The $N (mod11)=0$

$\begin{equation*}N=a_n\times10^n+a_{n-1}\times10^{n-1}+a_{n-2}\times10^{n-2}+...+a_0\times10^0\end{equation}$

$\begin{equation*}0=a_n\times10^n (mod11)+a_{n-1}\times10^{n-1}(mod11)+a_{n-2}\times10^{n-2}(mod11)+...+a_0\times10^0(mod11)\end{equation}$

Now $10 (mod11)=10$ which is congruent to $-1$ because $10-(-1)=11$ , which is a multiple of 11.

Thus

$\begin{equation*}0=a_n(-1)^n+a_{n-1}(-1)^{n-1}+a_{n-2}(-1)^{n-2}+...+a_0(-1)^0\end{equation}$

Odd powers will be negative and even positive.

So if we start at one end of the number and add every second digit (i.e. first digit plus third digit plus fifth digit etc.) and then subtract the other digits (i.e. second digit, fourth digit, six digit, etc.), if that equals zero then the number is divisible by 11.

For example, is $1756238$ divisible by $11$ ?

$1+5+2+8-7-6-3=16-16=0$

Hence $1756238$ is divisible by $11$

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Filed under Algebra, Arithmetic, Divisibility, Index Laws, Interesting Mathematics, Number Bases

Tagged as Divisibility rules, divisible by 11

September 15, 2025 · 3:12 am

Deriving the sum formula for an Arithmetic Progression

Arithmetic progressions (or arithmetic sequences) are sequences with a common difference (i.e. the same number is added or subtracted to get the next number in the sequence).

For example,

$2, 5, 8, 11, 14, ...$

$12,10, 8, 6, 4, ...$

The $n^{th}$ term of an arithmetic progression is $T_n=a+(n-1)d$ where $a$ is the first term and $d$ is the common difference.

i.e. For the sequence above, $T_4=12+(4-1)(-2)=6$

An arithmetic series is the sum of the arithmetic progression.

For example, if the sequence is

$3, 7, 11, 15, ...$

then $S_1=3, S_2=3+7=10, S_3=3+7+11=21$

The series is also a sequence and we are going to find the general term, $S_n$ .

$\begin{equation*}S_n=T_1+T_2+T_3+...+T_n\end{equation}$

which we can write as

$\begin{equation*}S_n=a+a+d+a+2d+...+a+(n-1)d\end{equation}$

Now, I am going to write that in reverse order (to make the next bit more obvious)

$\begin{equation*}S_n=a+(n-1)d+a+(n-2)d+a+(n-3)d+ ... +a\end{equation}$

I am going to add the two versions of $S_n$ together

Each term has an $a$ and there are $n$ terms, so we now have $2na$ . The $d$ terms, we going to group together

$d+(n-1)d+2d+(n-2)d+3d+(n-3)d+ ... +(n-1)d+d$

Which simplifies to $nd+nd+nd+ ...+nd$ and we have $(n-1)$ $d$ terms. So this part of the sum is $(n-10nd$

Thus we have

$\begin{equation*}2S_n=2an+(n-1)nd\end{equation}$

Which simplifies to

(1) $\begin{equation*}S_n=\frac{n}{2}(2a+(n-1)d)\end{equation*}$

Let’s test it, remember the sequence $3, 7, 11, 15, ...$ . We know $S_3=21$

$\begin{equation*}S_3=\frac{3}{2}(2(3)+(3-1)(4))=\frac{3}{2}(14)=21\end{equation}$

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Filed under Algebra, Arithmetic, Sequences, Year 11 Mathematical Methods

Tagged as arithmetic progressions, arithmetic sequences, arithmetic series, sequences

September 1, 2025 · 3:50 am

Area Problem

Two rectangular garden beds have a combined area of $40m^2$ . The larger bed has twice the perimeter of the smaller and the larger side of the smaller bed is equal to the smaller side of the larger bed. If the two beds are not similar, and if all edges are a whole number of metres, what is the length, in metres, of the longer side of the larger bed?
AMC 2007 S.14

Let’s draw a diagram

From the information in the question, we know

(1) $\begin{equation*}xy+xz=40\end{equation*}$

and

$\begin{equation*}2x+2y=4x+4z\end{equation}$

$\begin{equation*}x+y=2x+2z\end{equation}$

(2) $\begin{equation*}y=x+2z\end{equation*}$

Equation $1$ becomes

$\begin{equation*}x(x+3z)=40\end{equation}$

As the sides are whole numbers, consider the factors of 40.

$1, 2, 4, 5, 8, 10, 20, 40$

Remember $z<x<y$

$x$	$x+3z$	$z$	$y$	Perimeter Large	Perimeter Small	Comment
$2$	$20$	$6$				$x$ must be greater than $z$
$4$	$10$	$2$	$8$	$2(4+8)=24$	$2(2+4)=12$	This one works
$5$	$8$	$1$	$7$	$2(5+7)=24$	$2(5+1)=12$	This one also works
$8$	$10$	$\frac{2}{3}$				$z$ not a whole number
$10$	$4$	$z<0$				Not possible
$20$	$2$	$z<0$				Not possible
$40$	$1$	$z<0$				Not possible

There are two possibilities

The large garden bed could be $4$ by $8$ and the smaller $4$ by $2$ (Area $40$ Perimeters $24$ and $12$ )

The large garden bed could be $5$ by $7$ and the smaller $5$ by $1$ (Area $40$ Perimeters $24$ and $12$ )

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Filed under Area, Interesting Mathematics, Measurement, Puzzles, Solving Equations, Year 8 Mathematics

Tagged as area problem

August 14, 2025 · 6:39 am

Trig Identities and Exact Values

My Year 11 Specialist Mathematics students are working on Trig identities. We came across this question

Without the use of a calculator, evaluate
(a) $cos20^\circ\times cos40^\circ\times cos80^\circ$

(b) $cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$

OT Lee Year 11 Specialist Mathematics textbook

I spent a bit of time thinking about the question. Can you use a product to sum identity twice? But I was always being left with an angle that doesn’t have a nice exact value.

I tried a few things, had a chat to Meta AI, and finally stumbled upon this method.

Remember

$\begin{equation*}sin(2x)=2sin(x)cos(x)\end{equation}$

Which can be rearranged to

$\begin{equation*}cos(x)=\frac{sin(2x)}{sin(x)}\end{equation}$

(a) $cos20^\circ\times cos40^\circ\times cos80^\circ=\frac{sin(40)}{2sin(20)}\frac{sin(80)}{2sin(40)}\frac{sin(160)}{2sin(80)}$

Which simplifies to

$\begin{equation*}\frac{sin(160)}{8sin(20)}\end{equation}$

Now $sin(160)=sin(20)$

Hence $cos20^\circ\times cos40^\circ\times cos80^\circ=\frac{1}{8}$

And we will do the same for part (b)

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{sin(\frac{2\pi}{7})}{2sin(\frac{\pi}{7})}\frac{sin(\frac{4\pi}{7})}{2sin(\frac{2\pi}{7})}\frac{sin(\frac{8\pi}{7})}{2sin(\frac{4\pi}{7})}$

Which simplifies to

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{sin(\frac{8\pi}{7})}{8sin(\frac{\pi}{7})}$

Now $sin(\frac{8\pi}{7})=-sin(\frac{\pi}{7})$

Hence $cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{-1}{8}$

And then I had to test them on my Classpad.

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Filed under Classpad Skills, Identities, Simplifying fractions, Trigonometry, Year 11 Specialist Mathematics

Tagged as exact values, trig identities

August 2, 2025 · 9:30 am

Trigonometric Exact Value

Using an appropriate double angle identity, find the exact value of
$cos(\frac{\pi}{12})$

The double angle identity for sine is

(1) $\begin{equation*}cos(2A)=cos^2A-sin^2A=2cos^2A-1=1-2sin^2A\end{equation*}$

That means $\frac{\pi}{12}$ is either $2A$ or $A$ .

It must be $A$ as $2\times\frac{\pi}{12}=\frac{\pi}{6}$ as there are exact values for $\frac{\pi}{6}$

Hence,

$\begin{equation*}cos{\frac{\pi}{6}}=2cos^2{\frac{\pi}{12}}-1\end{equation}$

$\begin{equation*}\frac{\sqrt{3}}{2}=2cos^2{\frac{\pi}{12}}-1\end{equation}$

$\begin{equation*}\frac{\sqrt{3}}{2}+1=2cos^2{\frac{\pi}{12}}\end{equation}$

$\begin{equation*}\frac{\frac{\sqrt{3}}{2}+1}{2}=cos^2{\frac{\pi}{12}}\end{equation}$

$\begin{equation*}\frac{\sqrt{3}+2}{4}=cos^2{\frac{\pi}{12}}\end{equation}$

$\begin{equation*}\sqrt{\frac{\sqrt{3}+2}{4}}=cos{\frac{\pi}{12}}\end{equation}$

As $\frac{\pi}{12}$ is in the first quadrant, we don’t need to consider the negative version.

$\begin{equation*}cos(\frac{\pi}{12})=\frac{\sqrt{3}+2}{2}\end{equation}$

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Filed under Algebra, Identities, Trigonometry, Year 11 Mathematical Methods

Tagged as trig double angle identities, trig exact values

July 28, 2025 · 8:09 am

Trig Identities – Addition and Subtraction

Deriving the addition and subtraction trigonometric identities.

We will start with cosine, and use the result to derive the remaining identities.

Proving $cos(A-B)=cos(A)cos(B)+sin(A)(sin(B)$ .

$A$ and $B$ are represented in the unit circle below.

Remember $P(x_1,y_1)=(cos(A), sin(A))$ and $Q(x_2, y_2)=(cos(B), sin(B))$

Using the cosine rule and triangle $OPQ$ , find $PQ$

$\begin{equation*}(PQ)^2=1^2+1^2-2(1)(1)cos(A-B)\end{equation}$

$\begin{equation*}(PQ)^2=2-2cos(A-B)\end{equation}$

Using the distance between points, find $PQ$

$\begin{equation*}(PQ)^2=(x_1-x_2)^2+(y_1-y_2)^2\end{equation}$

$\begin{equation*}(PQ)^2=(cosA-cosB)^2+(sinA-sinB)^2\end{equation}$

$(PQ)^2=cos^2A-2cosAcosB+cos^2B+sin^2A-2sinAsinB+sin^2B$

Remember the Pythagorean identity

$\begin{equation*}cos^2\theta+sin^2\theta=1\end{equation}$

$\begin{equation*}(PQ)^2=2-2cosAcosB-2sinAsinB\end{equation}$

Hence

$\begin{equation*}2-2cos(A-B)=2-2cosAcosB-2sinAsinB\end{equation}$

(1) $\begin{equation*}cos(A-B)=cosAcosB+sinAsinB\end{equation*}$

We can then use this identity to find $cos(A+B)$ .

$\begin{equation*}cos(A+B)=cos(A-(-B))\end{equation}$

$\begin{equation*}cos(A-(-B))=cos(A)cos(-B)+sinAsin(-B)\end{equation}$

Remember $cos(-B)=cos(B)$ and $sin(-B)=-sin(B)$

$\begin{equation*}cos(A-(-B))=cosAcosB-sinAsinB\end{equation}$

(2) $\begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation*}$

We can also find $sin(A+B)$

Remember, $sin\theta=cos(\frac{\pi}{2}-\theta)$

$\begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-(A+B))\end{equation}$

$\begin{equation*}sin(A+B)=cos((\frac{\pi}{2}-A)-B)\end{equation}$

$\begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-A)cosB+sin(\frac{\pi}{2}-A)sinB\end{equation}$

(3) $\begin{equation*}sin(A+B)=sinAcosB+cosAsinB\end{equation*}$

We can use equation $2$ to find $sin(A-B)$

$\begin{equation*}sin(A-B)=sin(A+(-B))\end{equation}$

$\begin{equation*}sin(A+(-B))=sinAcos(-B)+cosAsin(-B)\end{equation}$

$\begin{equation*}sin(A+(-B))=sinAcos(B)-cosAsin(B)\end{equation}$

(4) $\begin{equation*}sin(A-B)=sinAcos(B)-cosAsin(B)\end{equation*}$

And we can use both the sine and cosine identities to find $tan(A+B)$

Remember $tan\theta=\frac{sin\theta}{cos\theta}$

$\begin{equation*}tan(A+B)=\frac{sin(A+B)}{cos(A+B)}\end{equation}$

$\begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\end{equation}$

$\begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\times \frac{cosAcosB}{cosAcosB}\end{equation}$

$\begin{equation*}=\frac{\frac{sinAcosB}{cosAcosB}+\frac{cosAsinB}{cosAcosB}}{\frac{cosAcosB}{cosAcosB}+\frac{sinAsinB}{cosAcosB}}\end{equation}$

$\begin{equation*}=\frac{tanA+tanB}{1-tanAtanB}\end{equation}$

(5) $\begin{equation*}tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\end{equation*}$

and

(6) $\begin{equation*}tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}\end{equation*}$

$\begin{equation*}cos(A\pm B)=cosAcosB\mp sinAsinB\end{equation}$

$\begin{equation*}sin(A\pm B)=sinAcosB\pm cosAsinB\end{equation}$

$\begin{equation*}tan(A \pm B)=\frac{tan A \pm tanB}{1 \mp tanAtanB}\end{equation}$

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Filed under Addition and Subtraction Identities, Identities, Non-Right Trigonometry, Simplifying fractions, Trigonometry, Year 11 Mathematical Methods, Year 11 Specialist Mathematics

Tagged as addition and subtraction trig identities, trig identities

July 26, 2025 · 8:45 am

General Solutions to Trigonometric Equations

Solve $sinx=\frac{1}{2}$ for $0\le x \le 2\pi$

Sine is positive in the first and second quadrants.

$\begin{equation*}sinx=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6} \text{ and } x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\end{equation}$

But what if we aren’t given a domain for the $x$ values?

Then we need to give general solutions.

For example,

Solve $sinx=\frac{1}{2}$

As you can see from the sketch above, there are infinite solutions.

The sine function has a period of $360^\circ$ , and so if $\frac{\pi}{6}$ is a solution then $\2pi+\frac{\pi}{6}$ is also a solution. This means $\frac{\pi}{6}+2\pi n, n\in\mathbb{Z}$ is a general solution. And we can do the same for the second solution $\frac{5\pi}{6}+2\pi n$ .

In general

$\begin{equation*}sinx=y\end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi-arcsin(y)+2\pi n \end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi(2n+1)-arcsin(y), n \in \mathbb{Z}\end{equation}$

We can turn this into one equation
$\begin{equation*}x=(-1)^n arcsin(y)+n\pi, n \in \mathbb{Z}\end{equation}$

What about cosine?

Solve $cosx=\frac{1}{2}$

Cosine is positive in the first and fourth quadrants (it also has a period of $2\pi$ . The first two (positive) solutions are $\frac{\pi}{3}$ and $2\pi-\frac{\pi}{3}$ .