Racquel Sanderson

September 9, 2025 · 11:19 am

Linear Transformation (Rotation) Question

The unit square is rotated about the origin by $45^\circ$ anti-clockwise.
(a) Find the matrix of this transformation.
(b) Draw the unit square and its image on the same set of axes.
(c) Find the area of the over lapping region.

Remember the general rotation matrix is

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

Hence

$\begin{equation*}\begin{bmatrix}cos(45)&-sin(45)\\sin(45)&cos(45)\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}\end{equation}$

The unit square has co-ordinates

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

Transform the unit square

$\begin{bmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}0 & 1 & 1& 0\\0&0&1&1\end{bmatrix}=\begin{bmatrix}0&\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\0&\frac{1}{\sqrt{2}}&\frac{2}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$

The overlapping area is the area of $\Delta ADC$ – the area of $\Delta EFC$

We know $\angle{FCE}=45^\circ$ because the diagonal of a square bisects the angle.

We know $\angle{EFC}$ is a right angle as it’s on a straight line with the vertex of a square.

Hence $\Delta EFC$ is isosceles.

$\overline{AC}=\sqrt{1^2+1^2}=\sqrt{2}$ and $\overline{AF}=1$ , hence $\overline{FC}=\sqrt{2}-1$

$A_{\Delta ADC}=\frac{1}{2}(1)(1)=\frac{1}{2}$

$A_{\Delta ECF}=\frac{1}{2}(\sqrt{2}-1)(\sqrt{2}-1)$

Area of shaded region = $\frac{1}{2}-\frac{1}{2}(3-2\sqrt{2})=\sqrt{2}-1$

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Filed under Area, Co-ordinate Geometry, Finding an area, Geometry, Matrices, Transformations, Year 11 Specialist Mathematics

Tagged as matrix tranformations, rotations

September 6, 2025 · 9:50 am

Clock (Geometry Question)

At what time after 2pm will the minute hand over take the hour hand?

Let the distance the minute hand moves be $x^\circ$ . Then the distance the hour hand moves is $x-60^\circ$ (The angle between the hands at 2pm is $60^\circ$ )

The rate the hour hand moves is $R_H=\frac{30}{60}=\frac{1}{2}^\circ$ per minute, and the rate the minute hand moves is $R_M=\frac{360}{60}=6^\circ$ per minute.

The time is the distance divided by the rate,

$\begin{equation*}\frac{x}{6}=\frac{x-60}{\frac{1}{2}}\end{equation}$

$\begin{equation*}\frac{x}{2}=6x-360\end{equation}$

$\begin{equation*}360=\frac{11x}{2}\end{equation}$

$\begin{equation*}x=65\frac{5}{11}^\circ\end{equation}$

As the minute hand moves $65\frac{5}{11}^\circ$ , the change in time is $65\frac{5}{11}^\circ \div 6=10$ minutes and $54$ seconds. Hence the time is $2:11$ pm

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Filed under Finding an angle, Geometry, Puzzles

Tagged as geometry clock question

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

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Filed under Addition and Subtraction Identities, Identities, Product to Sum idenitites, Trigonometry

Tagged as product to sum trig identities, solving trig equations, 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 $2\pi-\frac{\pi}{3}$ .