Category Archives: Simplifying fractions

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

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

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 1, 2025 · 10:35 am

Synthetic Division (for factorising and/or solving polynomials)

I use synthetic division to factorise polynomials with a degree greater than 2. For example, $f(x)=x^3+2x^2-5x-6$

It works best with monic polynomials but can be adapted to non-monic ones (see example below).

The only problem is that you need to find a root to start.

Try the factors of $-6$ i.e. $(-1, 1, 2, -2, 3, -3, 6, -6)$

$f(-1)=(-1)^3+2(-1)^2-5(-1)-6=0$

Hence, $x=-1$ is a root and $(x+1)$ is a factor of the polynomial.

Set up as follows

Bring the first number down

Multiply by the root and place under the second co-efficient

Add down

Repeat the process

The numbers at the bottom $(1, 1, -6)$ are the coefficients of the polynomial factor.

We now know $x^3+2x^2-5x-6=(x+1)(x^2+x-6)$ .

We can factorise the quadratic in the usual way.

$x^2+x-6=(x+3)(x-2)$

Hence $x^3+2x^2-5x-6=(x+1)(x+3)(x-2)$ .

Let’s try a non-monic example

Factorise $6x^4+39x^3+91x^2+89x+30$

I know $-2$ is a root. Otherwise I would try the factors of 30.

Use synthetic division

Because this was non-monic we need to divide our new co-efficients $(6, 27, 37, 15)$ by 6 (the co-efficient of the $x^4$ term)

$x^3+\frac{9}{2}x^2+\frac{37}{6}+\frac{5}{2}$

We now need to go again. I know that $\frac{-3}{2}$ is a root and $(2x+3)$ is a factor.

Our quadratic factor is $x^2+3x+5/3$ , which is $3x^2+9x+5$ .

The quadratic factor doesn’t have integer factors so,

$6x^4+39x^3+91x^2+89x+30=(x+2)(2x+3)(3x^2+9x+5)$

I think this is much quicker than polynomial long division.

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Filed under Algebra, Factorising, Factorising, Fractions, Polynomials, Quadratic, Simplifying fractions, Solving Equations

Tagged as non monic synthetic division, Synthetic division

May 3, 2025 · 3:34 am

Geometry Puzzle

Another puzzle from this book

Two ladders are propped up vertically in a narrow passageway between two vertical buildings. The ends of the ladders are 8 metres and 4 metres above the pavement.
Find the height above the ground, $T$ ,

$\Delta ABC \sim \Delta TEC$ and $\Delta DCB \sim \Delta TEB$ (Angle Angle Similarity)

Hence,

(1) $\begin{equation*}\frac{h}{8}=\frac{x}{x+y}\end{equation*}$

(2) $\begin{equation*}\frac{h}{4}=\frac{y}{x+y}\end{equation*}$

From equation $1$ $h=\frac{8x}{x+y}$ and from $2$ $h=\frac{4y}{x+y}$

Hence, $\frac{8x}{x+y}=\frac{4y}{x+y}$

Therefore, $8x=4y$ and $y=2x$

From equation $1$

$\begin{equation*}h=\frac{8x}{3x}=\frac{8}{3}\end{equation}$

Hence $T$ is $2 \frac{2}{3}$ m above the ground.

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Filed under Geometry, Puzzles, Similarity, Simplifying fractions, Solving Equations

Tagged as Puzzle

September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let $x$ be the number of cats and $y$ be the number of dogs.
Then $0.9x+0.1y$ think they are cats.
But we also know 20% of the total think they are cats.
$0.2(x+y)$
Therefore, $0.9x+0.1y=0.2(x+y)$
$0.9x+0.1y=0.2x+0.2y$
$0.7x=0.1y$
$7x=y$
Percentage of cats is $\frac{x}{x+y}\times100$
Substitute $7x$ for $y$
$\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%$
$\therefore 12.5$ % of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

Tagged as Fractions, percent, uk maths challenge

Category Archives: Simplifying fractions

Geometry Problem

Trig Identities and Exact Values

Trig Identities – Addition and Subtraction

Synthetic Division (for factorising and/or solving polynomials)

Geometry Puzzle

Cats and Dogs

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