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

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

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• […] as it is from secondary Add Math. This formulae is not given in MF26, although students can derive it out using existing formulae in […]