### Trigonometry Formulae & Applications (Part 2)

I meant to share more on factor Formulae today. However, a few students are not so sure how to get the R-formulae correctly during their preliminary exams recently. So I thought that I’ll share how they can derive the R-Formulae from the MF26.

The following is the R-Formulae which students should have memorised. It is under assumed knowledge, just saying…

$a \text{cos} \theta \pm b \text{sin} \theta = R \text{cos} (\theta \mp \alpha)$

$a \text{sin} \theta \pm b \text{cos} \theta = R \text{sin} (\theta \pm \alpha)$

where $R = \sqrt{a^2 + b^2}$ and $\text{tan} \alpha = \frac{b}{a}$ for $a > 0, b > 0$ and $\alpha$ is acute.

So here, I’ll write the addition formulae that’s found in MF26.

$\text{sin}(A \pm B) \equiv \text{sin}A \text{cos} B \pm \text{cos} A \text{sin} B$

$\text{cos}(A \pm B) \equiv \text{cos}A \text{cos} B \mp \text{sin} A \text{sin} B$

I’ll use an example I discussed previously.

$f(x) = 3 \text{cos}t - 2 \text{sin}t$

Write $f(x)$ as a single trigonometric function exactly.

Lets consider the formulae from MF26.

$\text{cos}(A \pm B) \equiv \text{cos}A \text{cos} B \mp \text{sin} A \text{sin} B$

$R\text{cos}(A \pm B) \equiv R \text{cos}A \text{cos} B \mp R \text{sin} A \text{sin} B$

We can let

$3 = R \text{cos} B ---(1)$

$2 = R \text{sin} B ---(2)$

$\Rightarrow \sqrt{ 3^2 + 2^2 } = \sqrt{ R^2 \text{cos}^2 B + R^2 \text{sin}^2 B}$

$\Rightarrow \sqrt{13} = R$

$\Rightarrow \frac{R \text{sin} B}{R \text{cos} B} = \frac{2}{3}$

$\Rightarrow \text{tan} B = \frac{2}{3}$

Putting things together, we have that

$f(x) = \sqrt{13} \text{cos} ( t + \text{tan}^{\text{-1}} (\frac{2}{3}))$

### Trigonometry Formulae & Applications (Part 1)

Upon request by some students, I’ll discuss a few trigonometry formulae here and also some of their uses in A’levels. I’ve previously discussed the use of factor formulae here under integration.

I’ll start with the R-Formulae. It should require no introduction as it is from secondary Add Math. This formulae is not given in MF26, although students can derive it out using existing formulae in MF26.

$a \text{cos} \theta \pm b \text{sin} \theta = R \text{cos} (\theta \mp \alpha)$

$a \text{sin} \theta \pm b \text{cos} \theta = R \text{sin} (\theta \pm \alpha)$

where $R = \sqrt{a^2 + b^2}$ and $\text{tan} \alpha = \frac{b}{a}$ for $a > 0, b > 0$ and $\alpha$ is acute.

Here is a quick example,

$f(x) = 3 \text{cos}t - 2 \text{sin}t$

Write $f(x)$ as a single trigonometric function exactly.

Here, we observe, we have to use the R-Formulae where

$R = \sqrt{3^2 + 2^2} = \sqrt{13}$

$\alpha = \text{tan}^{\text{-1}} (\frac{2}{3})$

We have that

$f(x) = \sqrt{13} \text{cos} ( t + \text{tan}^{\text{-1}} (\frac{2}{3}))$.

I’ll end with a question from HCI Midyear 2017 that uses R-Formulae in one part of the question.

A curve D has parametric equations

$x = e^{t} \text{sin}t, y = e^{t} \text{cos}t, \text{~for~} 0 \le t \le \frac{\pi}{2}$

(i) Prove that $\frac{dy}{dx} = \text{tan} (\frac{\pi}{4} - t)$.

I’ll discuss about Factor Formulae soon.  And then the difference and application between this two formulae.

### Random Questions from 2017 Prelims #2

Today I’ll share a question that came out of CJC Prelim 2017 Paper 1 for H2 Mathematics 9758. I think some of my student would have seen this question before and we discussed it in class before. Very technical question. This is question 11, I’ll share only the first part which is on the application of ratio theorem or mid point theorem. The second part is on application: Ray Tracing which I’ll discuss in class.

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. For the triangle show below, O, A and B are vertices, where O is the origin. $\vec{OA} = a$ and $\vec{OB} = b$. The midpoints of OB, OA and AB are M, N and T respectively.

It is given that X is the point of intersection between the medians of triangle OAB from vertices A and B.

(i) Show that $\vec{OX} = \frac{1}{3} (a +b)$

(ii) Prove that X also lies on OT, the median of triangle OAB from vertex O.

### Random Sec 4 Differentiations

B6

$y = 3e^x + \frac{4}{e^x}$

$\frac{dy}{dx} = 3e^x - \frac{4}{e^x}$

$\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}$

let $\frac{dy}{dx} = 0$

$3e^x - \frac{4}{e^x} = 0$

$3e^{2x} = 4$

$2x = \mathrm{ln} \frac{4}{3}$

$x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$

Sub $x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$ to $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} > 0$ Thus, it is a min point.

C7

$y = \mathrm{ln} \frac{5-4x}{3+2x}$

$y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)$

$\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}$

let $\frac{dy}{dx} = 0$

$\frac{-4}{5-4x} - \frac{2}{3+2x} = 0$

$\frac{-4}{5-4x} = \frac{2}{3+2x}$

$-4(3+2x) = 2(5-4x)$

$-12 - 8x = 10 - 8x$

$-12 = 10$ (NA).

There are no stationary points for this curve.

C8

$x = \frac{1}{3}e^{y(2x+5)}$

$\mathrm{ln}(3x) = y(2x+5)$

$\frac{\mathrm{ln}(3x)}{2x+5} = y$

$y = \frac{\mathrm{ln}(3x)}{2x+5}$

$\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}$

Let $x = e^2$

$\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}$

Evaluate with a calculator…

### Reimagining mathematical notations

Maths notation is often overly complex. Let’s see how this can be better:

### Question of the Day #17

This is a pretty cool and interesting question from AMC.

There are four lifts in a building. Each makes three stops, which do not have to be on consecutive floors on include the ground floor. For any two floors, there is at least one lift which stops on both of them. What is the maximum number of floors that this building can have?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 12

### Question of the Day #16

The positive integers $x$ and $y$ satisfy

$3x^2 -8y^2 +3x^2 y^2 = 2008$

What is the value of $xy$?

### Arithmetic Problem #7

I came across this interesting algebra/ arithmetic problem the other day.

$(1 \times 2 \times 3) + 2 = 8 = 2^3$
$(2 \times 3 \times 4) + 3 = 27 = 3^3$
$(3 \times 4 \times 5) + 4 = 64 = 4^3$

Do you think this will always work?

Hint: We can use algebra to prove this cool sequence easily.

### O’levels Results 2016!

All of us wish the students receiving the O’levels Results 2016 the best! And do not let grades define you. 🙂

If you’re keen to meet up with us for the JC Talk, you may contact Newtonapple Learning Hub at +65 9222 3423 for more details.

This talk will be opened freely to the public and existing students. There will be discussion about Subject Combinations, discussion and introduction with different subjects. Come down to find out more 🙂

### 30 Free Awesome Resources for Students

New year, new you. Be ready to tackle on the new academic year with these awesome resources curated by Donut.sg

You’re welcome.

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