The R-Formula (a sinθ ± b cosθ)

Not in the 0606 syllabus. Expressing $a\sin\theta + b\cos\theta$ in the form $R\sin(\theta \pm \alpha)$ is not part of the Cambridge IGCSE Additional Mathematics 0606 (2025–2027) subject content, so it will not be set as a 0606 exam question. It is A Level (9709) and SPM Additional Mathematics material. This page is kept for students bridging into A Level Maths or sitting SPM alongside 0606. If you are preparing only for 0606, you can skip it and spend the time on the trig identities and equations that are examined.

Any combination $a\sin\theta + b\cos\theta$ collapses into a single wave:

$a\sin\theta + b\cos\theta =$ $R\sin(\theta + \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \dfrac{b}{a}$

One function instead of two, which makes equations solvable and maxima readable. Where it is examined (A Level, SPM), the question almost always comes in three parts.

Part 1. Express

Write $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$, $0^\circ < \alpha < 90^\circ$. $R = \sqrt{9 + 16} =$ $5$ $\tan\alpha = \frac{4}{3}$ $\to$ $\alpha = 53.1^\circ$ $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.1^\circ)$

Derivation logic (worth knowing, occasionally demanded): expand $R\sin(\theta + \alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta$ and match coefficients. $R\cos\alpha = a$, $R\sin\alpha = b$. Dividing gives $\tan\alpha = \dfrac{b}{a}$; squaring and adding gives $R^2 = a^2 + b^2$. The matching lines are method marks when “show that” appears.

Variants: $R\cos(\theta - \alpha)$ and the minus-sign forms each match differently, expand the target form and compare rather than memorising four cases. The question’s stated range for $\alpha$ tells you which quadrant it lives in; check your $\alpha$ lands there.

Part 2. Solve the equation

Hence solve $3\sin\theta + 4\cos\theta = 2$ for $0^\circ \le \theta \le 360^\circ$. $5\sin(\theta + 53.1^\circ) = 2$ $\to$ $\sin(\theta + 53.1^\circ) = 0.4$ $\theta + 53.1^\circ$ runs over $53.1^\circ$ to $413.1^\circ$, solve over that range: $\theta + 53.1^\circ = 156.4^\circ, 383.6^\circ$ $\theta = 103.3^\circ, 330.5^\circ$

The compound-angle range expansion is the standard equation discipline; a hence would mean the R-form is the intended route.

Part 3. Max, min, and where

The collapsed form answers instantly: maximum $R$ ($= 5$) where $\sin(\theta + \alpha) = 1$, i.e. $\theta = 90^\circ - \alpha = 36.9^\circ$; minimum $-R$ at $\theta = 270^\circ - \alpha$. Related: the maximum of $\dfrac{1}{3\sin\theta + 4\cos\theta + 7}$ occurs at the minimum of the denominator, a favourite twist.

Common mistakes

• $\tan\alpha = \frac{a}{b}$ (inverted)
• $\alpha$‘s quadrant unchecked against the stated range
• The equation solved without expanding the range for $\theta + \alpha$, solutions missing
• Max stated as $R +$ something when no constant exists (or the constant ignored when it does)
• “Hence” ignored in part 2

Full topic context: Trigonometry notes.