R-form: Rsin(θ+α) / Rcos(θ−α)

Expressing $a\sin\theta + b\cos\theta$ as a single sinusoid — finding $R$ and $\alpha$
9709 Paper 3 · §3.3 Trigonometry II · Week 4 Lesson 1
💡 Lead In: Reversing a Compound Angle

The Scenario

You can expand a compound angle. For example, using $\cos(A-B)=\cos A\cos B+\sin A\sin B$ with $A=\theta$, $B=60^{\circ}$:

$$\cos(\theta-60^{\circ})=\cos\theta\cos60^{\circ}+\sin\theta\sin60^{\circ}=\tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta.$$

Now run it backwards: the messy sum $\tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta$ is secretly the single cosine $\cos(\theta-60^{\circ})$. What if the coefficients are not friendly like $\tfrac12$ and $\tfrac{\sqrt3}{2}$, but plain numbers such as $3$ and $4$?

Review · Activating Prior Knowledge

Quick check — pick the correct statement in each question before we build the R-form.

Review 1 — Which is the correct expansion of $\cos(A-B)$?
$\cos(A-B)=\cos A\cos B-\sin A\sin B$
$\cos(A-B)=\sin A\cos B+\cos A\sin B$
$\cos(A-B)=\sin A\cos B-\cos A\sin B$
$\cos(A-B)=\cos A\cos B+\sin A\sin B$

$\cos(A-B)=\cos A\cos B+\sin A\sin B$. The R-form works by reading this expansion backwards: $\cos A\cos B+\sin A\sin B$ collapses into a single cosine.

Review 2 — Which set of exact values is correct?
$\cos 60^{\circ}=\dfrac{1}{2}, \; \sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$
$\cos 60^{\circ}=\dfrac{\sqrt{3}}{2}, \; \sin 60^{\circ}=\dfrac{1}{2}$
$\cos 60^{\circ}=\dfrac{1}{\sqrt{2}}, \; \sin 60^{\circ}=\dfrac{1}{2}$
$\cos 60^{\circ}=1, \; \sin 60^{\circ}=0$

$\cos 60^{\circ}=\dfrac{1}{2}$ and $\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$. These exact values will let us check the angle $\alpha$ when we match coefficients.

The Question

Can $3\cos\theta+4\sin\theta$ be written as a single $R\cos(\theta-\alpha)$? If we "factor out" $R$, how big must $R$ be so that the remaining coefficients become a cosine and a sine of $\alpha$ that actually fit on the unit circle?

💡 The Bridge (click to reveal)

We need $R\cos\alpha=3$ and $R\sin\alpha=4$. Squaring and adding: $R^2=3^2+4^2=25$, so $R=5$, and $\tan\alpha=\dfrac{4}{3}$, giving $\alpha\approx53.1^{\circ}$:

$$3\cos\theta+4\sin\theta=5\cos(\theta-53.1^{\circ}).$$

This is the R-form $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$ with $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\dfrac{b}{a}$.

📖 Key Concepts: The R-form (Auxiliary Angle)

Any expression of the form $a\cos\theta + b\sin\theta$ can be written as a single sinusoid $R\cos(\theta - \alpha)$ or $R\sin(\theta + \alpha)$. The trick is to expand the target form using the compound angle formulae and then match coefficients of $\cos\theta$ and $\sin\theta$.

Key relation 1 — the $R\cos(\theta - \alpha)$ form: $$R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$$

Matching with $a\cos\theta + b\sin\theta$:

$$a = R\cos\alpha, \qquad b = R\sin\alpha$$

$$\boxed{\;R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \dfrac{b}{a}\;}$$

Key relation 2 — the $R\sin(\theta + \alpha)$ form: $$R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$$

Matching with $a\sin\theta + b\cos\theta$:

$$a = R\cos\alpha, \qquad b = R\sin\alpha$$

$$\boxed{\;R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \dfrac{b}{a}\;}$$

The formula for $R$ and $\tan\alpha$ is identical. What changes is which coefficient ($\sin\theta$ or $\cos\theta$) plays the role of $a$ and $b$.

Key relation 3 — geometric interpretation:

As $\theta$ varies, the value of $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$ traces a single sinusoid. Geometrically, the point $(a\cos\theta,\; b\sin\theta)$ is the sum of two perpendicular oscillations, and it lies on a circle of radius $R$. The combined wave is just a cosine wave shifted left by $\alpha$, with amplitude $R$.

120°
Drag to scan the waves — at the cursor, green height + indigo height = amber height.

Sign of $\alpha$: When both $a$ and $b$ are positive, $\alpha$ lies in the first quadrant ($0 < \alpha < 90^{\circ}$). When the signs differ, use the signs of $R\cos\alpha = a$ and $R\sin\alpha = b$ to place $\alpha$ in the correct quadrant (since $R > 0$ always, $\cos\alpha$ has the sign of $a$ and $\sin\alpha$ has the sign of $b$).

Derivation · Whole Class Activity 1: Deriving the R-form — Where Do $R$ and $\alpha$ Come From?

Follow the algebra: $a\cos\theta + b\sin\theta$ collapses into a single wave $R\cos(\theta-\alpha)$.

Step 1 — Assume the target form

$$a\cos\theta + b\sin\theta = R\cos(\theta - \alpha), \qquad R > 0$$

Step 2 — Expand (compound angle)

$$R\cos(\theta - \alpha) = R(\cos\theta\cos\alpha + \sin\theta\sin\alpha)$$

Step 3 — Regroup

$$R\cos(\theta - \alpha) = (R\cos\alpha)\cos\theta + (R\sin\alpha)\sin\theta$$

Step 4 — Match coefficients

$$R\cos\alpha = a, \qquad R\sin\alpha = b$$

Step 5 — Square & add

$$(R\cos\alpha)^2 + (R\sin\alpha)^2 = a^2 + b^2 \;\Rightarrow\; R^2 = a^2 + b^2$$

$$\boxed{\;R = \sqrt{a^2 + b^2}\;}$$

Step 6 — Divide

$$\frac{R\sin\alpha}{R\cos\alpha} = \frac{b}{a} \;\Rightarrow\; \tan\alpha = \frac{b}{a}$$

$$\boxed{\;\alpha = \tan^{-1}\!\left(\frac{b}{a}\right)\;}$$

Step 7 — Check: $a = 3$, $b = 4$

$$R = \sqrt{3^2 + 4^2} = 5, \qquad \tan\alpha = \frac{4}{3} \Rightarrow \alpha \approx 53.1^{\circ}$$

$$\boxed{\;3\cos\theta + 4\sin\theta = 5\cos(\theta - 53.1^{\circ})\;}$$

Result: $a\cos\theta + b\sin\theta = R\cos(\theta-\alpha)$ with $\;R = \sqrt{a^2+b^2}\;$ and $\;\tan\alpha = \dfrac{b}{a}\;$ (place $\alpha$ by the signs of $a$, $b$).
Investigate · Pairs Activity 2: Rule Recall — Turn $(a, b)$ into $R$ and $\alpha$

From Activity 1's derivation: which formula turns $(a, b)$ into $R$? And which turns $(a, b)$ into $\alpha$? Complete each checkpoint to lock in the rule.

Checkpoint 1 — Which expression gives $R$ from $a$ and $b$?
Checkpoint 2 — Which expression gives $\tan\alpha$ from $a$ and $b$?
Group · Guided Worked Example 1: The $R\cos(\theta - \alpha)$ Form Exercise 3D Q1

Express each of the following in the form $r \cos( \theta - \alpha)$, where $r > 0$ and $0^{\circ} < \alpha < 90^{\circ}$.

(i) $\cos \theta + \sin \theta$   (ii) $20 \cos \theta + 21 \sin \theta$   (iii) $\cos \theta + \sqrt{3} \sin \theta$   (iv) $\sqrt{5} \cos \theta + 2 \sin \theta$

👁 Reveal Solution
Part (i): $\cos\theta + \sin\theta$

$a = 1$, $b = 1$. $r = \sqrt{1^2 + 1^2} = \sqrt{2}$. $\tan\alpha = \dfrac{1}{1} = 1 \Rightarrow \alpha = 45^{\circ}$.

$\boxed{\sqrt{2}\cos(\theta - 45^{\circ})}$

Part (ii): $20\cos\theta + 21\sin\theta$

$a = 20$, $b = 21$. $r = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29$. $\tan\alpha = \dfrac{21}{20} = 1.05 \Rightarrow \alpha = 46.4^{\circ}$ (1 d.p.).

$\boxed{29\cos(\theta - 46.4^{\circ})}$

Part (iii): $\cos\theta + \sqrt{3}\sin\theta$

$a = 1$, $b = \sqrt{3}$. $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. $\tan\alpha = \dfrac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \alpha = 60^{\circ}$.

$\boxed{2\cos(\theta - 60^{\circ})}$

Part (iv): $\sqrt{5}\cos\theta + 2\sin\theta$

$a = \sqrt{5}$, $b = 2$. $r = \sqrt{(\sqrt{5})^2 + 2^2} = \sqrt{5 + 4} = \sqrt{9} = 3$. $\tan\alpha = \dfrac{2}{\sqrt{5}} = 0.8944 \Rightarrow \alpha = 41.8^{\circ}$ (1 d.p.).

$\boxed{3\cos(\theta - 41.8^{\circ})}$

Individual · Guided Worked Example 2: The $R\sin(\theta + \alpha)$ Form Exercise 3D Q3

Express each of the following in the form $r \sin( \theta + \alpha)$, where $r > 0$ and $0^{\circ} < \alpha < 90^{\circ}$.

(i) $\sin \theta + 2 \cos \theta$   (ii) $2 \sin \theta + \sqrt{5} \cos \theta$

👁 Reveal Solution
Part (i): $\sin\theta + 2\cos\theta$

Match with $r\sin(\theta+\alpha) = r\sin\theta\cos\alpha + r\cos\theta\sin\alpha$:

$a = 1$ (coeff of $\sin\theta$), $b = 2$ (coeff of $\cos\theta$). So $r\cos\alpha = 1$, $r\sin\alpha = 2$.

$r = \sqrt{1^2 + 2^2} = \sqrt{5}$. $\tan\alpha = \dfrac{2}{1} = 2 \Rightarrow \alpha = 63.4^{\circ}$ (1 d.p.).

$\boxed{\sqrt{5}\sin(\theta + 63.4^{\circ})}$

Part (ii): $2\sin\theta + \sqrt{5}\cos\theta$

$a = 2$ (coeff of $\sin\theta$), $b = \sqrt{5}$ (coeff of $\cos\theta$). So $r\cos\alpha = 2$, $r\sin\alpha = \sqrt{5}$.

$r = \sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4 + 5} = \sqrt{9} = 3$. $\tan\alpha = \dfrac{\sqrt{5}}{2} = 1.118 \Rightarrow \alpha = 48.2^{\circ}$ (1 d.p.).

$\boxed{3\sin(\theta + 48.2^{\circ})}$

✏️ Concept Quiz
Score: 0 / 4
1. Expressing $a\cos\theta + b\sin\theta$ as $R\cos(\theta - \alpha)$, what is $R$?
$a + b$
$\sqrt{a^2 + b^2}$
$|a - b|$
$a \cdot b$
From $R\cos\alpha = a$ and $R\sin\alpha = b$, squaring and adding gives $R^2(\cos^2\alpha + \sin^2\alpha) = a^2 + b^2$, so $R^2 = a^2 + b^2$ and $R = \sqrt{a^2 + b^2}$ (taking the positive root since $R > 0$). The amplitudes combine by Pythagoras, not by addition.
2. For $3\cos\theta + 4\sin\theta$ written as $R\cos(\theta - \alpha)$, what is $\alpha$ (to 1 d.p.)?
$36.9^{\circ}$
$53.1^{\circ}$
$30.0^{\circ}$
$45.0^{\circ}$
$a = 3$, $b = 4$. $\tan\alpha = \dfrac{b}{a} = \dfrac{4}{3} \approx 1.333$, so $\alpha = \arctan(1.333) \approx 53.1^{\circ}$. The trap $36.9^{\circ}$ comes from computing $\arctan(3/4)$ — that would be the angle if $a$ and $b$ were swapped. Always use $\tan\alpha = b/a$ where $a$ is the $\cos\theta$ coefficient and $b$ is the $\sin\theta$ coefficient.
3. To write $a\sin\theta + b\cos\theta$ as $R\sin(\theta + \alpha)$, which is correct?
$R\cos\alpha = b$, $R\sin\alpha = a$
$R\cos\alpha = a$, $R\sin\alpha = b$
$R\cos\alpha = a$, $R\sin\alpha = -b$
$R = a + b$
Expanding: $R\sin(\theta+\alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$. Matching with $a\sin\theta + b\cos\theta$: the $\sin\theta$ coefficient gives $R\cos\alpha = a$, and the $\cos\theta$ coefficient gives $R\sin\alpha = b$. The same rule $R = \sqrt{a^2+b^2}$, $\tan\alpha = b/a$ follows — only the identification of which coefficient is $a$ vs $b$ differs from the cos form.
4. As $\theta$ varies, the value of $a\cos\theta + b\sin\theta$ (with $a, b$ not both zero) traces a wave whose amplitude is $R = \sqrt{a^2+b^2}$. Why must the maximum be $R$ rather than $a + b$?
The two terms peak at different $\theta$ values, so they never reach $a$ and $b$ simultaneously.
Sine and cosine are always negative, so the sum is smaller.
The maximum of any trig expression is always $1$.
$R = \sqrt{a^2+b^2}$ equals $a + b$ only when one of them is zero.
$\cos\theta = 1$ requires $\theta = 0^{\circ}$, but $\sin\theta = 1$ requires $\theta = 90^{\circ}$. The two peaks never coincide, so the combined maximum is strictly less than $a + b$ (unless one of $a, b$ is zero). Writing the expression as $R\cos(\theta-\alpha)$ makes the maximum obvious: $\cos(\theta-\alpha) \leq 1$, so the max is exactly $R$, attained when $\theta = \alpha$.
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📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Exercise 3D Q2 Standard

Express each of the following in the form $r \cos( \theta + \alpha)$, where $r > 0$ and $0 < \alpha < \dfrac{\pi}{2}$.

(i) $\cos \theta - \sin \theta$   (ii) $\sqrt{3} \cos \theta - \sin \theta$

💡 Hint

Expand $r\cos(\theta+\alpha) = r\cos\theta\cos\alpha - r\sin\theta\sin\alpha$.

Match with $a\cos\theta + b\sin\theta$: here $b$ is negative, so $r\sin\alpha = -b > 0$, keeping $\alpha$ in the first quadrant.

👁 Reveal Solution
Part (i): $\cos\theta - \sin\theta$

$r\cos(\theta+\alpha) = r\cos\theta\cos\alpha - r\sin\theta\sin\alpha$. Matching with $\cos\theta - \sin\theta$:

$r\cos\alpha = 1$, $r\sin\alpha = 1$ (the minus sign is absorbed by the formula's own minus).

$r = \sqrt{1^2 + 1^2} = \sqrt{2}$. $\tan\alpha = \dfrac{1}{1} = 1 \Rightarrow \alpha = \dfrac{\pi}{4}$.

$\boxed{\sqrt{2}\cos\!\left(\theta + \dfrac{\pi}{4}\right)}$

Part (ii): $\sqrt{3}\cos\theta - \sin\theta$

$r\cos\alpha = \sqrt{3}$, $r\sin\alpha = 1$.

$r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$. $\tan\alpha = \dfrac{1}{\sqrt{3}} \Rightarrow \alpha = \dfrac{\pi}{6}$.

$\boxed{2\cos\!\left(\theta + \dfrac{\pi}{6}\right)}$

Exercise 3D Q4 Standard

Express each of the following in the form $r \sin( \theta - \alpha)$, where $r > 0$ and $0 < \alpha < \dfrac{\pi}{2}$.

(i) $\sin \theta - \cos \theta$   (ii) $\sqrt{7} \sin \theta - \sqrt{2} \cos \theta$

💡 Hint

Expand $r\sin(\theta-\alpha) = r\sin\theta\cos\alpha - r\cos\theta\sin\alpha$.

Match with $a\sin\theta + b\cos\theta$: $r\cos\alpha = a$, $r\sin\alpha = -b$ (so the negative $b$ gives a positive $r\sin\alpha$).

👁 Reveal Solution
Part (i): $\sin\theta - \cos\theta$

$r\sin(\theta-\alpha) = r\sin\theta\cos\alpha - r\cos\theta\sin\alpha$. Matching with $\sin\theta - \cos\theta$:

$r\cos\alpha = 1$, $r\sin\alpha = 1$.

$r = \sqrt{1^2 + 1^2} = \sqrt{2}$. $\tan\alpha = \dfrac{1}{1} = 1 \Rightarrow \alpha = \dfrac{\pi}{4}$.

$\boxed{\sqrt{2}\sin\!\left(\theta - \dfrac{\pi}{4}\right)}$

Part (ii): $\sqrt{7}\sin\theta - \sqrt{2}\cos\theta$

$r\cos\alpha = \sqrt{7}$, $r\sin\alpha = \sqrt{2}$.

$r = \sqrt{(\sqrt{7})^2 + (\sqrt{2})^2} = \sqrt{7 + 2} = \sqrt{9} = 3$. $\tan\alpha = \dfrac{\sqrt{2}}{\sqrt{7}} = \sqrt{\dfrac{2}{7}} \approx 0.5345 \Rightarrow \alpha \approx 0.491\,\text{rad}$ (3 s.f.).

$\boxed{3\sin\!\left(\theta - \arctan\!\sqrt{\tfrac{2}{7}}\right) \approx 3\sin(\theta - 0.491)}$

Exercise 3D Q5(i)(ii) Medium

Express each of the following in the form $r \cos( \theta - \alpha)$, where $r > 0$ and $-180^{\circ} < \alpha < 180^{\circ}$.

(i) $\cos \theta - \sqrt{3} \sin \theta$   (ii) $2\sqrt{2} \cos \theta - 2\sqrt{2} \sin \theta$

💡 Hint

Here $b$ (the $\sin\theta$ coefficient) is negative, so $r\sin\alpha = b < 0$ while $r\cos\alpha = a > 0$. This places $\alpha$ in quadrant IV — a negative angle between $-90^{\circ}$ and $0^{\circ}$.

👁 Reveal Solution
Part (i): $\cos\theta - \sqrt{3}\sin\theta$

$a = 1$, $b = -\sqrt{3}$. So $r\cos\alpha = 1 > 0$ and $r\sin\alpha = -\sqrt{3} < 0$ — $\alpha$ is in Q4.

$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. $\tan\alpha = \dfrac{-\sqrt{3}}{1} = -\sqrt{3}$.

Reference angle: $\arctan(\sqrt{3}) = 60^{\circ}$. In Q4: $\alpha = -60^{\circ}$.

$\boxed{2\cos(\theta - (-60^{\circ})) = 2\cos(\theta + 60^{\circ})}$

Part (ii): $2\sqrt{2}\cos\theta - 2\sqrt{2}\sin\theta$

$a = 2\sqrt{2}$, $b = -2\sqrt{2}$. So $r\cos\alpha = 2\sqrt{2} > 0$ and $r\sin\alpha = -2\sqrt{2} < 0$ — $\alpha$ is in Q4.

$r = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$. $\tan\alpha = \dfrac{-2\sqrt{2}}{2\sqrt{2}} = -1$.

Reference angle: $\arctan(1) = 45^{\circ}$. In Q4: $\alpha = -45^{\circ}$.

$\boxed{4\cos(\theta - (-45^{\circ})) = 4\cos(\theta + 45^{\circ})}$

Exercise 3D Q5(iii)(iv) Medium

Express each of the following in the form $r \cos( \theta - \alpha)$, where $r > 0$ and $-180^{\circ} < \alpha < 180^{\circ}$.

(iii) $\sin \theta + \sqrt{3} \cos \theta$   (iv) $5 \sin \theta + 12 \cos \theta$

💡 Hint

Rewrite each as (coeff of $\cos\theta$)$\cos\theta$ + (coeff of $\sin\theta$)$\sin\theta$ first, so you can read off $a$ (the $\cos\theta$ coefficient) and $b$ (the $\sin\theta$ coefficient).

For (iii): $\sin\theta + \sqrt{3}\cos\theta = \sqrt{3}\cos\theta + \sin\theta$, so $a = \sqrt{3}$, $b = 1$. Both positive — $\alpha$ is in Q1.

For (iv): $5\sin\theta + 12\cos\theta = 12\cos\theta + 5\sin\theta$, so $a = 12$, $b = 5$. Both positive — $\alpha$ is in Q1.

👁 Reveal Solution
Part (iii): $\sin\theta + \sqrt{3}\cos\theta$

Rewrite as $\sqrt{3}\cos\theta + \sin\theta$. Then $a = \sqrt{3}$, $b = 1$.

$r\cos\alpha = \sqrt{3} > 0$, $r\sin\alpha = 1 > 0$ — $\alpha$ is in Q1.

$r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$. $\tan\alpha = \dfrac{1}{\sqrt{3}} \Rightarrow \alpha = 30^{\circ}$.

$\boxed{2\cos(\theta - 30^{\circ})}$

Part (iv): $5\sin\theta + 12\cos\theta$

Rewrite as $12\cos\theta + 5\sin\theta$. Then $a = 12$, $b = 5$.

$r\cos\alpha = 12 > 0$, $r\sin\alpha = 5 > 0$ — $\alpha$ is in Q1.

$r = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$. $\tan\alpha = \dfrac{5}{12} \approx 0.4167 \Rightarrow \alpha = 22.6^{\circ}$ (1 d.p.).

$\boxed{13\cos(\theta - 22.6^{\circ})}$

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