The R-Formula: Solving and Finding a Maximum

Not examined in 0606. Expressing a sinθ + b cosθ in the form R sin(θ ± α) is not in the Cambridge IGCSE Additional Mathematics 0606 (2025–2027) syllabus. It is A Level (9709) and SPM material. This page is extension practice for students bridging upward. For 0606-examined trig, see the trigonometry topic.

The R-formula turns a two-term expression into a single sine (or cosine), which unlocks two things at once: you can solve equations, and you can read off the maximum and minimum directly. This question uses both.

(i) Express $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. (ii) Hence solve $3\sin\theta + 4\cos\theta = 2$ for $0^\circ \le \theta \le 360^\circ$. (iii) State the greatest value of $3\sin\theta + 4\cos\theta$.

The working

(i) For $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$, match coefficients: $R\cos\alpha = 3, \qquad R\sin\alpha = 4$

Then $R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$, and $\tan\alpha = \dfrac{4}{3} \Rightarrow \alpha = 53.13^\circ$: $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13^\circ)$

(ii) Replace the left side using part (i): $5\sin(\theta + 53.13^\circ) = 2 \;\Rightarrow\; \sin(\theta + 53.13^\circ) = 0.4$

Let $\phi = \theta + 53.13^\circ$. As $\theta$ runs $0^\circ$ to $360^\circ$, $\phi$ runs $53.13^\circ$ to $413.13^\circ$. Solve $\sin\phi = 0.4$ in that window. The principal value is $23.58^\circ$ (out of range), so use $180^\circ - 23.58^\circ$ and $360^\circ + 23.58^\circ$: $\phi = 156.42^\circ \quad \text{and} \quad \phi = 383.58^\circ$

Subtract $53.13^\circ$: $\theta = 103.3^\circ \quad \text{and} \quad \theta = 330.5^\circ \quad \text{(1 d.p.)}$

(iii) The greatest value of $5\sin(\theta + 53.13^\circ)$ occurs when the sine equals $1$: $\text{greatest value} = R = 5$

Where the marks are won and lost

• Shift the range with the angle. Solving $\sin\phi = 0.4$ over $0^\circ$ to $360^\circ$ instead of $53.13^\circ$ to $413.13^\circ$ gives the wrong solutions or misses one. Convert the interval, solve, then convert back.
• The greatest value is $R$, not $R$ plus anything. For $R\sin(\theta + \alpha)$ the range is $-R$ to $R$.
• Keep $\alpha$ to enough accuracy ($53.13^\circ$) throughout, or the final angles drift.

Common mistakes

• Forgetting to adjust the interval for $\phi = \theta + \alpha$.
• Giving only one solution in (ii).
• Stating the maximum as $5 +$ something, or using the minimum by mistake.

Full method: The R-Formula notes. Topic home: Trigonometry pillar.