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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