Category Archives: Classpad Skills

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

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

October 22, 2024 · 6:55 am

Year 12 Mathematics Applications – Finance Question

What happens with an investment or a loan when the compounding periods are not the same as the payment periods?

For example,

Gerald invests $450 000 at a rate of 6.4% p.a. compounding monthly. At the end of every quarter he receives $35 000 from his investment.

(a) Write a recursive rule that will enable you to find the balance of the account, $T_n$ after $n$ payments.

(b) Find the balance of the account after $3$ years.

(d) What will be the amount of Gerald’s last payment?

T_{n+1}=T_n(1+\frac{6.4}{100\times12})^{12\div4}-35 000, T_0=450 000

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Filed under Classpad Skills, Finance, Finance, Uncategorized, Year 12 Mathematics Applications

Tagged as compounding rate different from payment rate, finance, Year 12 Mathematics Applications

June 12, 2024 · 7:37 am

Year 12 Mathematics Applications Finance

Ming, a former high school student and now a successful business owner, wishes to set up a perpetuity of $6000 per year to be paid to a deserving student from her school. The perpetuity is to be paid at the start of the year in one single payment.

(a) A financial institution has agreed to maintain an account for the perpetuity paying a fixed rate of 5.9% p.a. compounded monthly. Show that an amount of $98 974, to the nearest dollar, is required to maintain this perpetuity.

(b) Ming allows herself five years to accumulate the required $98 974 by making regular quarterly payments into an account paying 5.4% p.a. compounded monthly. Determine the quarterly payment needed to reach the required amount after five years if Ming starts the account with an initial deposit of $1000.
SCSA 2017 CA 8

(a) For a perpetuity, we want the interest to equal the payment.

Remember the compound interest formula is

$A=P(1+\frac{r}{100n})^{nt}$