Monthly Archives: August 2025

August 25, 2025 · 3:29 am

Matrices -Linear Transformations (Rotation)

We are going to find a matrix to rotate a point about the origin a number of degrees (or radians).

$P$ and $P'$ are equidistant from the origin. I.e. $\sqrt{c^2+d^2}=\sqrt{a^2+b^2}$

Remember, anti-clockwise angles are positive.

$\begin{equation*}cos(\theta+\alpha)=\frac{c}{\sqrt{c^2+d^2}}\end{equation}$

$\begin{equation*}c=\sqrt{a^2+b^2}cos(\theta+\alpha)\end{equation}$

Use the cosine addition identity.

$\begin{equation*}c=\sqrt{a^2+b^2}(cos(\theta)cos(\alpha)-sin(\theta)sin(\alpha))\end{equation}$

$\begin{equation*}c=\sqrt{a^2+b^2}(cos(\theta)\frac{a}{\sqrt{a^2+b^2}}-sin(\theta)\frac{b}{\sqrt{a^2+b^2}})\end{equation}$

(1) $\begin{equation*}c=acos(\theta)-bsin(\theta)\end{equation*}$

We will do the same for $d$

$\begin{equation*}sin(\theta+\alpha)=\frac{d}{\sqrt{a^2+b^2}}\end{equation}$

$\begin{equation*}d=\sqrt{a^2+b^2}sin(\theta+\alpha)\end{equation}$

Use the sine addition identity.

$\begin{equation*}d=\sqrt{a^2+b^2}(sin(\theta)cos(\alpha)+cos(\theta)sin(\alpha)\end{equation}$

$\begin{equation*}d=\sqrt{a^2+b^2}(sin(\theta)\frac{a}{\sqrt{a^2+b^2}}+cos(\theta)\frac{b}{\sqrt{a^2+b^2}})\end{equation}$

(2) $\begin{equation*}d=asin(\theta)+bcos(\theta)\end{equation*}$

Let $R$ be the rotation matrix, then

$\begin{equation*}R\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}acos(\theta)-bsin(\theta)\\asin(\theta)+bcos(\theta)\end{bmatrix}\end{equation}$

Hence $R$ must be

(3) $\begin{equation*}R=\begin{bmatrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{bmatrix}\end{equation*}$

Example

Find the image of the line $y=x+1$ after it is rotated $60^\circ$ about the origin.

I am going to select two points on the line and transform them.

$\begin{equation*}\begin{bmatrix}cos(60)&-sin(60)\\sin(60)&cos(60)\end{bmatrix}\begin{bmatrix}0&4\\1&5\end{bmatrix}=\begin{bmatrix}x'_1&x'_2\\y'_1&y'_2\end{bmatrix}\end{equation}$

$\begin{equation*}\begin{bmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}\begin{bmatrix}0&4\\1&5\end{bmatrix}=\begin{bmatrix}x'_1&x'_2\\y'_1&y'_2\end{bmatrix}\end{equation}$

$\begin{equation*}\begin{bmatrix}-\frac{\sqrt{3}}{2}&\frac{4-5\sqrt{3}}{2}\\\frac{1}{2}&2\sqrt{3}+\frac{5}{2}\end{bmatrix}=\begin{bmatrix}x'_1&x'_2 \\y'_1&y'_2\end{bmatrix}\end{equation}$

We can then find the equation of the line.

$\begin{equation*}m=\frac{2\sqrt{3}+\frac{5}{2}-\frac{1}{2}}{\frac{4-5\sqrt{3}+\sqrt{3}}{2}}\end{equation}$

$\begin{equation*}m=-2-\sqrt{3}\end{equation}$

$\begin{equation*}y-\frac{1}{2}=(-2-\sqrt{3})(x+\frac{\sqrt{3}}{2})\end{equation}$

$\begin{equation*}y=(-2-\sqrt{3})x-1-\sqrt{3}\end{equation}$

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Filed under Addition and Subtraction Identities, Identities, Matrices, Transformations, Trigonometry, Year 11 Specialist Mathematics

Tagged as matrix transformations, rotation matrix

August 21, 2025 · 8:31 am

Matrices – Linear Transformations (reflection in y=mx)

We are going to derive the transformation matrix for a reflection across a line $y=mx$ .

Things to remember about a reflection:

The distance between $P'$ and the line is same as the distance between $P$ and the line. Hence $M$ is the midpoint of $P$ and $P'$ .
The line segment joining $P$ and $P'$ is perpendicular to the line.

Our aim is to find a general identity for $P'$ and then use that to derive a transformation matrix.

Let’s start by finding the equation of the line joining $P$ and $P'$ .

We know the gradient of this line is perpendicular to the gradient of $y=mx$ , hence the gradient is $-\frac{1}{m}$ .

$\begin{equation*}y-y_0=-\frac{1}{m}(x-x_0)\end{equation}$

And $(a,b)$ is a point on the line.

$\begin{equation*}y-b=-\frac{1}{m}(x-a)\end{equation}$

Which simplifies to

(1) $\begin{equation*}y=-\frac{1}{m}x+\frac{a+bm}{m}\end{equation*}$

We are going to find the co-ordinates of $M$ in two ways; as the midpoint of $P$ and $P'$ , and as the point of intersection of $y=mx$ and $y=-\frac{1}{m}x+\frac{a+bm}{m}$

As the midpoint of $P$ and $P'$

(2) $\begin{equation*}M=(\frac{a+c}{2}, \frac{b+d}{2})\end{equation*}$

As the point of intersection of $y=mx$ and $y=-\frac{1}{m}x+\frac{a+bm}{m}$

$\begin{equation*}mx=-\frac{1}{m}x+\frac{a+bm}{m}\end{equation}$

$\begin{equation*}mx+\frac{x}{m}=\frac{a+bm}{m}\end{equation}$

$\begin{equation*}\frac{m^2x+x}{m}=\frac{a+bm}{m}\end{equation}$

$\begin{equation*}x(m^2+1)=a+bm\end{equation}$

$\begin{equation*}x=\frac{a+bm}{m^2+1}\end{equation}$

Substitute $x$ into $y=mx$

$\begin{equation*}y=\frac{am+bm^2}{m^2+1}\end{equation}$

(3) $\begin{equation*}M=(\frac{a+bm}{m^2+1},\frac{am+bm^2}{m^2+1})\end{equation*}$

$M$ must equal $M$ , hence we can find $(c, d)$ in terms of $(a, b)$

Equate equation $2$ and $3$

$\begin{equation*}\frac{a+c}{2}=\frac{a+bm}{m^2+1}\end{equation}$

$\begin{equation*}a+c=2(\frac{a+bm}{m^2+1})\end{equation}$

$\begin{equation*}c=2(\frac{a+bm}{m^2+1})-a\end{equation}$

$\begin{equation*}c=\frac{2a+2bm-am^2-a}{m^2+1}\end{equation}$

$\begin{equation*}c=\frac{a+2bm-am^2}{m^2+1}\end{equation}$

(4) $\begin{equation*}c=\frac{1-m^2}{m^2+1}a+\frac{2m}{m^2+1}b\end{equation*}$

And

$\begin{equation*}\frac{b+d}{2}=\frac{am+bm^2}{m^2+1}\end{equation}$

$\begin{equation*}b+d=2(\frac{am+bm^2}{m^2+1})\end{equation}$

$\begin{equation*}d=2(\frac{am+bm^2}{m^2+1})-b\end{equation}$

$\begin{equation*}d=\frac{2am+2bm^2-bm^2-b}{m^2+1}\end{equation}$

$\begin{equation*}d=\frac{2m}{m^2+1}a+\frac{m^2-1}{m^2+1}b\end{equation}$

$\begin{equation*}d=\frac{2m}{m^2+1}a-\frac{-m^2+1}{m^2+1}b\end{equation}$

(5) $\begin{equation*}d=\frac{2m}{m^2+1}a-\frac{1-m^2}{m^2+1}b\end{equation*}$

Hence $P'=(\frac{1-m^2}{m^2+1}a+\frac{2m}{m^2+1}b, \frac{2m}{m^2+1}a-\frac{1-m^2}{m^2+1}b)$

For ease of writing, let $p=\frac{1-m^2}{m^2+1}$ and $q=\frac{2m}{m^2+1}$

Then

$P'=(pa+qb, qa-px)$ , which we can generalise to $(px+qy, qx-py)$

Now let’s think about a transformation matrix, we want to transform $(x, y)$ to $(px+qy, qx-py)$

$\begin{equation*}\begin{bmatrix}p&q\\q &-p\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}px+qy\\qx-py\end{bmatrix}\end{equation}$

Remember the gradient of a line is the tangent of the angle of inclination, $tan(\theta)=m$

$\begin{equation*}p=\frac{1-m^2}{m^2+1}=\frac{1-tan^2(\theta)}{tan^2(\theta)+1}\end{equation}$

Remember the identity

$\begin{equation*}tan^2(\theta)+1=sec^2(\theta)\end{equation}$

$\begin{equation*}p=\frac{1-(sec^2(\theta)-1)}{sec^2(\theta)}\end{equation}$

$\begin{equation*}p=\frac{2-sec^2(\theta)}{sec^2(\theta)}\end{equation}$

$\begin{equation*}p=\frac{2}{sec^2(\theta)}-\frac{sec^2(\theta)}{sec^2(\theta)}\end{equation}$

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

(6) $\begin{equation*}p=cos(2\theta)\end{equation*}$

And we will do the same for $q$ .

$\begin{equation*}q=\frac{2m}{m^2+1}=\frac{2tan(\theta)}{tan^2(\theta)+1}\end{equation}$

$\begin{equation*}q=\frac{\frac{2sin(\theta)}{cos(\theta)}}{sec^2(\theta)}\end{equation}$

$\begin{equation*}q=\frac{2sin(\theta)}{cos(\theta)} \times cos^2(\theta)\end{equation}$

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

(7) $\begin{equation*}q=sin(2\theta)\end{equation*}$

Hence our transformation matrix is

(8) $\begin{equation*}\begin{bmatrix}cos(2\theta)&sin(2\theta)\\sin(2\theta)&-cos(2\theta)\end{bmatrix}\end{equation*}$

Example

The vertices of a triangle $T$ are $A(-3, 1), B(6, -4)$ and $C(1, 5)$ . $T'$ is a reflection of $T$ in the line $y-x=0$

The gradient of the line $y=x$ is $1$ .

$\begin{equation*}tan(\theta)=1\end{equation}$

$\begin{equation*}\theta=\frac{\pi}{4}\end{equation}$

Our transformation matrix is
$\begin{equation*}\begin{bmatrix}cos(\frac{\pi}{2})&sin\frac{\pi}{2})\\sin\frac{\pi}{2})&-cos(\frac{\pi}{2})\end{bmatrix}\end{equation}$

Which is

$\begin{equation*}\begin{bmatrix}0&1\\1&0\end{bmatrix}\end{equation}$

So

$\begin{equation*}T'=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}-3&6&1\\1&-4&5\end{bmatrix}\end{equation}$

$\begin{equation*}T'=\begin{bmatrix}1&-4&5\\-3&6&1\end{bmatrix}\end{equation}$

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Filed under Matrices, Transformations

Tagged as reflection across line y=mx, Transformations

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 10, 2025 · 4:54 am

Trigonometric Identities – Product to Sum

Let’s think about the sine and cosine addition and subtraction trig identities.

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

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

If we add equation $1$ and $2$ , we get

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

Hence, $sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))$

If we subtract equation $2$ from equation $1$ , we get

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

Hence, $cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B)$

What about the cosine addition and subtraction idenities?

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

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

If we add equation $3$ and $4$ , we get

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

Hence, $cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))$

If we subtract $3$ from $4$ , we get

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

Hence, $sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

These are the product to sum identities.

$\begin{equation*}sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))\end{equation}$

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

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

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

Examples

(1) Solve $sin(5x)-sin(x)=0$ for $0\le x \le 2\pi$

Remember,

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

$\begin{equation*}A+B=5\end{equation}$

$\begin{equation*}A-B=1\end{equation}$

Therefore, $A=3$ and $B=2$

$\begin{equation*}sin(5x)-sin(x)=2cos(3x)sin(2x)=0\end{equation}$

$\begin{equation*}cos(3x)=0\end{equation}$

$\begin{equation*}3x=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{\9\pi}{2}, \frac{11\pi}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\end{equation}$

$\begin{equation*}sin(2x)=0\end{equation}$

$\begin{equation*}2x=0, \pi, 2\pi, 3\pi, 4\pi\end{equation}$

$\begin{equation*}x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{equation}$

Hence $x=0, \frac{\pi}{6}. \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, 2\pi$

(2)Solve $sin(7\theta)-sin(\theta)=sin(3\theta)$ for $0 \le \theta \le\2\pi$

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

Therefore, $A+B=7$ and $A-B=1$

$A=4, B=3$

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

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

$\begin{equation*}sin(3\theta)(2cos(4\theta)-1)=0\end{equation}$

$sin(3\theta)=0$ and $cos(4\theta)=\frac{1}{2}$

$3\theta=0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi$

$\theta=0, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}. 2\pi$

$cos(4\theta)=\frac{1}{2}$

$4\theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}, \frac{19\pi}{3}, \frac{23\pi}{3}$

$\theta=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}$

Hence $\theta=0, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{5\pi}{3}, \frac{23\pi}{12}, 2\pi$