Category Archives: Year 11 Mathematical Methods

November 22, 2025 · 6:10 am

Polynomial Long Division

I usually choose to use synthetic division when factorising polynomials, but I know some teachers are unhappy when their students do this. So for completeness, here is my PDF for Polynomial Long Division.

Polynomial Long Division Download

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Filed under Algebra, Cubics, Factorisation, Factorising, Factorising, Polynomials, Quadratics, Solving, Solving, Solving Equations, Year 11 Mathematical Methods

Tagged as polynomial long division, solving cubic equations

November 5, 2025 · 6:57 am

Equation of a Circle and Geometry Question

A circle has equation $x^2+y^2+4x-6y=36$
(a) Find the centre and radius of the circle.
Points $S$ and $T$ lie on the circle such that the origin is the midpoint of $ST$ .
(b) Show that $ST$ has a length of 12.

(a)We need to put the circle equation into completed square form

$\begin{equation*}(x+2)^2-4+(y-3)^2-9=36\end{equation}$

$\begin{equation*}(x+2)^2+(y-3)^2=49\end{equation}$

The centre is $(-2, 3)$ and the radius is $7$ .

(b)Draw a diagram

We know $SO$ and $TO$ are radii of the circle. Hence $\Delta{SOT}$ is isosceles and the line segment from $O$ to the origin is perpendicular to $ST$ .

$OT=7$ and the distance from $O$ to the origin is

$\begin{equation*}\sqrt{(-2-0)^2+(3-0)^2}=\sqrt{13}\end{equation}$

We can use Pythagoras to find the distance from the origin to $T$ .

$\begin{equation*}x=\sqrt{7^2-(\sqrt{13})^2}=\sqrt{49-13}=\sqrt{36}=6\end{equation}$

Hence $ST=2\times6=12$

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Filed under Co-ordinate Geometry, Geometry, Pythagoras, Year 11 Mathematical Methods

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

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

June 4, 2025 · 5:30 am

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

At 1pm, object $H$ travelling with constant velocity $\begin{pmatrix}200\\10\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}-90\\-100\end{pmatrix}$ km. At 2pm object $J$ travelling with constant velocity $\begin{pmatrix}100\\-100\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}20\\-120\end{pmatrix}$ km. Determine the minimum distance between $H$ and $J$ and when this occurs.

OT Lee Mathematics Specialist Year 11 Unit 1 and 2 Exercise 10.1 Question 6.

(1) $\begin{equation*}\mathbf{r_H}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}\end{equation*}$

(2) $\begin{equation*}\mathbf{r_J}=\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix}\end{equation*}$

$\begin{pmatrix}-80\\-20\end{pmatrix}$ is the position vector of $J$ at 1pm.

Find the relative displacement of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\mathbf{r_H}-\mathbf{r_J}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}-(\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix})\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

Find the relative velocity of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{v_J}=\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

The relative displacement is perpendicular to the relative velocity at the closest approach.

That is

(3) $\begin{equation*}\mathbf{_H}\mathbf{r_J}\cdot\mathbf{_H}\mathbf{v_J}=0\end{equation*}$

$\begin{equation*}(\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix})\cdot(\begin{pmatrix}100\\110\end{pmatrix})=0\end{equation}$

$\begin{equation*}(-10+100t)(100)+(-80+110t)(110)=0\end{equation}$

$\begin{equation*}-1000+10 000t-8800+12100t=0\end{equation}$

$\begin{equation*}22100t=9800\end{equation}$

$\begin{equation*}t=\frac{98}{221}\end{equation}$

Substitute $t=\frac{98}{221}$ into the relative displacement and find the magnitude.

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+\frac{98}{221}\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\end{equation}$

$\begin{equation*}\|\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\|=46.4\end{equation}$

The closest objects $H$ and $J$ get to each other is $46.4$ km at 1:27pm.

I have made an e-activity for this.

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Filed under Classpad Skills, Closest Approach, Vectors, Year 11 Mathematical Methods

Tagged as casio classpad, eactivity, vectors closest approach

May 6, 2025 · 7:27 am

Combinations Proof

Prove $\begin{pmatrix}n+1\\r\end{pmatrix}=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}$

Let’s start with the right hand side.

$RHS=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}$

$RHS=\frac{n!}{(n-(r-1))!(r-1)!}+\frac{n!}{(n-r)!r!}$

Simplify

$RHS=\frac{n!}{(n+1-r)!}+\frac{n!}{(n-r)!r!}$

There is a common denominator of $(n+1-r)!r!$

$RHS=\frac{n!}{(n+1-r)!}\times \frac{r}{r}+\frac{n!}{(n-r)!r!}\times \frac{n+1-r}{n+1-r}$

$RHS=\frac{n!r+n!(n+1-r)}{(n+1-r)r!}$

$RHS=\frac{n!r+(n+1)n!-n!r}{(n+1-r)!r!}$

$RHS=\frac{(n+1)!}{(n+1-r)!r!}$

$RHS=\begin{pmatrix}n+1\\r\end{pmatrix}$

$RHS=LHS$

We know this intuitively from Pascal’s triangle

Where each entry is the sum of the two entries above it – for example,

$6+4=10$

Remember, Pascal’s triangle can be written as combinations,

so $\begin{pmatrix}5\\3\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}4\\3\end{pmatrix}$

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Filed under Algebra, Counting Techniques, Year 11 Mathematical Methods

April 3, 2025 · 7:28 am

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an $n-$ sided regular polygon if you know the side length, $l$ .

Find the $h$ of the triangle in terms of $l$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{h}$

$h=\frac{l}{2tan(\theta)}$

Remember the area of a triangle is $A=\frac{1}{2}bh$

Hence, $A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}$

And $\theta=\frac{360}{2n}=\frac{180}{n}$

Therefore $A=\frac{l^2}{4tan(\frac{180}{n})}$

There are $n$ triangles in an $n-$ sided polygon

(1) $\begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}$

A=\frac{6\times10^2}{4tan(\frac{180}{6})}

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of $a$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{a}$

$l=2atan(\theta)$

$A=\frac{1}{2}bh$

$A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)$

And $\theta=\frac{180}{n}$

Hence for an $n-$ sided polygon

(2) $\begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}$

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle ( $R$ in the diagram above)

Remember the area of triangle formula

$A=\frac{1}{2}absin(\theta)$

$A=\frac{1}{2}R^2sin(\theta)$

$\theta=\frac{360}{n}$

Hence, $A=\frac{1}{2}R^2sin(\frac{360}{n})$

Hence, for an $n-$ sided polygon

(3) $\begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}$

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Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Tagged as area of regular polygons, area of triangles, trig

March 5, 2025 · 2:00 am

Binomial Expansion Theorem

My Year 11 Mathematics Methods students are working on the Binomial Expansion Theorem.

But before we get onto that, remember Pascal’s triangle

Now we can use combinations to find the numbers in each row. For example, $1 4 6 4 1$ is $\begin{pmatrix}4\\0\end{pmatrix}=1, \begin{pmatrix}4\\1\end{pmatrix}=4, \begin{pmatrix}4\\2\end{pmatrix}=6, \begin{pmatrix}4\\3\end{pmatrix}=4, \begin{pmatrix}4\\4\end{pmatrix}=1$

As you can see, the coefficients are the row of pascal’s triangle corresponding to the power. So $(x+y)^6$ would have co-efficients from the sixth row of the table $1, 6, 15, 20, 15, 6, 1$ .

To generalise

$(x+y)^n=\begin{pmatrix}n\\0\end{pmatrix}x^ny^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}y^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}y^2+ ...+\begin{pmatrix}n\\n-1\end{pmatrix}x^1{y^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^0y^n$

Which we can condense to

$(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i$

Worked Examples

$(1)$ Expand $(2x-3)^4$

$(2x-3)^4=\begin{pmatrix}4\\0\end{pmatrix}(2x)^4(-3)^0+\begin{pmatrix}4\\1\end{pmatrix}(2x)^3(-3)^1+\begin{pmatrix}4\\2\end{pmatrix}(2x)^2(-3)^2+\begin{pmatrix}4\\3\end{pmatrix}(2x)^1(-3)^3+\begin{pmatrix}4\\4\end{pmatrix}(2x)^0(-3)^4$
$(2x-3)^4=16x^4-96x^3+216x^2-216x+81$

$(2)$ Find the co-efficient of the $x^3$ term in the expansion of $(2-5x)^5$ .

Remember $(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i$ , the $x^3$ is when $i=3$
$\begin{pmatrix}5\\3\end{pmatrix}(2)^2(-5)^3=10\times 2\times -125=-5000$

$(3)$ Find the constant term in the expansion of $(x^2+\frac{3}{x^4})^6$

We need to find the term where the $x$ ‘s cancel out. Each term is $\begin{pmatrix}6\\i\end{pmatrix}(x^2)^{6-i}(\frac{3}{x^4})^i$ .
$\begin{pmatrix}6\\i\end{pmatrix}(x^{12-2i})(3^ix^{-4i})$ .
We need $12-2i-4i=0$ , hence $i=2$
Therefore, the co-efficient is $\begin{pmatrix}6\\2\end{pmatrix}\times3^2=135$

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Filed under Algebra, Binomial Expansion Theorem, Counting Techniques, Year 11 Mathematical Methods

October 7, 2024 · 7:43 am

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If $T_m=n$ and $T_n=m$ , then prove that $T_{m+n}=0$ . Here where $T_n$ and $T_m$ are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

\begin{equation*}n=a+(m-1)d\end{equation*}

As you can see from equation $(6), T_{m+n}=0$

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

Tagged as Arthimetic, sequences, year 11 mathematical methods

Expression	Expansion	Co-efficients
$(x+y)^2$	$x^2+2xy+y^2$	$1, 2, 1$
$(x+y)^3$	$x^3+3x^2y+3xy^2+y^3$	$1, 3, 3, 1$
$(x+y)^4$	$x^4+4x^3y+6x^2y^2+4xy^3+y^4$	$1, 4, 6, 4, 1$

Category Archives: Year 11 Mathematical Methods

Polynomial Long Division

Equation of a Circle and Geometry Question

Deriving the sum formula for an Arithmetic Progression

Trigonometric Exact Value

Trig Identities – Addition and Subtraction

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

Combinations Proof

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Finding the area of a polygon if you know the inradius or the apothem

Finding the area of a regular polygon given the circumradius

Binomial Expansion Theorem

Worked Examples

Arithmetic Sequence

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