Category Archives: Trigonometry

September 13, 2026 · 3:15 am

Simple Harmonic Motion Question

An object in simple harmonic motion passes through the central point $x=0$ at time $t=0$ and every second thereafter.

Find an equation for the motion if $v(0)=-4.$

Because the particle passes through the central point $x=0$ when $t=0$ , it is natural to use a sine function: $x=A\sin(\omega t).$

The particle passes through the central point every second.

However, passing through the centre occurs twice during each complete cycle — once moving in one direction and once moving in the other.

Therefore, the time from one central crossing to the next is half a period: $\frac{T}{2}=1.$

Hence, $T=2\text{ seconds}.$

For simple harmonic motion, $\omega=\frac{2\pi}{T}.$

Since $T=2$ , $\omega=\frac{2\pi}{2}=\pi.$

Our equation is therefore $x=A\sin(\pi t).$

We still need to determine $A$ .

For simple harmonic motion,

$v^2=k^2(A^2-x^2)$

When $t-0, x=0$ and $v=-4$

$16=\pi^2(A^2-0)$

$A=\frac{4}{\pi}$

Hence,

$x(t)=\frac{4}{\pi}sin(\pi t)$

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Filed under Rectilinear Motion, Simple Harmonic Motion, Trigonometry, Year 12 Specialist Mathematics

Tagged as mathematics specialist, rectilinear motion, simple harmonic motion, Year 12 Specialist Mathematics

July 30, 2026 · 11:06 am

Finding the Area formula of an Ellipse

Ellipses appear in both complex loci and vector calculus.

We are going to use calculus (integration) to derive the formula for the area of an ellipse. The equation of an ellipse is

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. $$

Our aim is to find the area under the upper half of the ellipse and then double the result to obtain the total area.

Rearrange the equation to make $y$ the subject.

$$ \frac{y^2}{b^2}=1-\frac{x^2}{a^2} $$ $$ y^2=b^2\left(1-\frac{x^2}{a^2}\right) $$ $$ y=\pm b\sqrt{1-\frac{x^2}{a^2}}. $$

We only need to consider the positive branch.

$$ y=b\sqrt{1-\frac{x^2}{a^2}}. $$

The area of the upper half of the ellipse is

$$ A=\int_0^a b\sqrt{1-\frac{x^2}{a^2}}\,dx. $$

Let

$$ x=a\sin\theta. $$

Then

$$ dx=a\cos\theta\,d\theta. $$

When $x=0$, $\theta=0$, and when $x=a$, $\theta=\frac{\pi}{2}$.

The integral becomes

$$ A =ab\int_0^{\pi/2}\cos^2\theta\,d\theta. $$

Remember the trigonometric identity

$$ \cos^2\theta=\frac{1+\cos(2\theta)}{2}. $$

Hence,

$$ A =\frac{ab}{2}\int_0^{\pi/2}\left(1+\cos(2\theta)\right)\,d\theta. $$ $$ =\frac{ab}{2}\left[\theta+\frac{1}{2}\sin(2\theta)\right]_0^{\pi/2} $$ $$ =\frac{ab}{2}\left(\frac{\pi}{2}\right) $$ $$ =\frac{\pi ab}{4}. $$

This is the area of one quarter of the ellipse. Therefore, the total area of the ellipse is

$$ 4\times\frac{\pi ab}{4}=\boxed{\pi ab}. $$

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Filed under Area, Definite, Ellipse, Identities, Integration, Trigonometric Substitution, Year 12 Specialist Mathematics

February 11, 2026 · 3:28 am

Complex Numbers and Trig Idenities

My Year 12 Specialist Students are using complex numbers to prove trigonometric identities.

Things like

$\begin{equation*}sin(5\theta)=16sin^5(\theta) -20sin^3(\theta)+5sin(\theta) \end{equation}$

Method 2 might be a little bit easier depending upon how your brain works.

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Filed under Algebra, Binomial Expansion Theorem, Complex Numbers, Identities, Trig Identities, Trigonometry, Year 12 Specialist Mathematics

Tagged as complex numbers, de moivre's theorem, trigonometric identities

January 22, 2026 · 9:27 am

Complex Loci Question

A sketch of the locus of a complex number $z$ is shown above, determine the maximum value of $arg(z)$ correct to two decimal places where $0\le z \le 2\pi$

Draw tangent lines from the origin to the circle.

Remember tangent lines are perpendicular to the radii

The maximum argument is this angle

I am going to find the angle in two sections

From the diagram the radius of the circle is $2$ and the centre is $(4, 3)$ . Hence the distance from the origin to the centre is $5$ .

$\begin{equation*}sin(\theta_1)=\frac{2}{5}\end{equation}$

$\begin{equation*}\theta_1=0.412\end{equation}$

$\begin{equation*}sin(\theta_2)=\frac{3}{5}\end{equation}$

$\begin{equation*}\theta_2=0.644\end{equation}$

Hence maximum $arg(z)=1.06$

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Filed under Complex Numbers, Right Trigonometry, Sketching Complex Regions, Trigonometry, Year 12 Specialist Mathematics

Tagged as argument of a complex number, complex locus, right trig

December 10, 2025 · 7:54 am

Geometry Problem

Problem from *50 Maths Problems with Solutions* – Ajmal KP

The blue shaded area is the area of triangles $APO$ and $AQO$ subtract the sector $POQ$ .

We can use Heron’s law to find the area of the triangle $\Delta{ABC}$

$\begin{equation*}A=\sqrt{s(s-a)(s-b)(s-c)}\end{equation}$

where $s=\frac{a+b+c}{2}$

$\begin{equation*}A=\sqrt{20(20-16)(20-10)(20-14)}=40\sqrt{3}\end{equation}$

We also know the area of triangle $\Delta{ABC}=sr$ where $r$ is the radius of the inscribed circle.

Hence, $40\sqrt{3}=20r$ and $r=2\sqrt{3}$

We know $AP=AQ, CQ=CR$ , and $BP=BR$ – tangents to a circle are congruent.

$\begin{equation*}14-x=6+x\end{equation}$

(1) $\begin{equation*}8=2x\end{equation*}$

(2) $\begin{equation*}x=4\end{equation*}$

Area $\Delta{AQO}=\frac{1}{2}10\times 2\sqrt{3}=10\sqrt{3}$

Area $\Delta{APO}=$ Area $\Delta{AQO}$

$\begin{equation*}tan(\theta)=\frac{10}{2\sqrt{3}}\end{equation}$

$\begin{equation*}\theta=70.9^{\circ}\end{equation}$

Area of sector $OPQ=\frac{2\times70.9}{360}\pi (2\sqrt{3})^2=14.8$

Blue area = $20\sqrt{3}-14.8=19.8cm^2$

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Filed under Algebra, Area, Finding an angle, Finding an area, Geometry, Heron's Law, Interesting Mathematics, Puzzles, Radius and Semi-Perimeter, Right Trigonometry, Solving Equations, Trigonometry

Tagged as geometry puzzle

November 26, 2025 · 10:57 am

Trigonometric Equation

Solve $cos(4\theta)+cos(2\theta)+cos(\theta)=0$ for $0\le \theta \le\pi$

Remember the identity

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

Hence

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

Now I have

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

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

$cos(\theta)=0$ or $cos(3\theta)=\frac{-1}{2}$

$\theta=\frac{\pi}{2}$

$cos(3\theta)=-\frac{1}{2}$ for $0 \le \theta \le 3\pi$

$3\theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}$

$\theta=\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}$

Hence $\theta =\frac{\pi}{2},\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}$

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Filed under Identities, Quadratic, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as solving trig equations, sum to product trig identity

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

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 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

Category Archives: Trigonometry

Simple Harmonic Motion Question

Finding the Area formula of an Ellipse

Complex Numbers and Trig Idenities

Complex Loci Question

Geometry Problem

Trigonometric Equation

Trigonometric Exact Values

Geometry Problem

Matrices -Linear Transformations (Rotation)

Example

Trig Identities and Exact Values

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