The Scenario
By now the R-form is your own four-step machine:
$$a\cos\theta+b\sin\theta \;\xrightarrow{\;1\;}\; R\cos(\theta-\alpha) \;\xrightarrow{\;2\;}\; R=\sqrt{a^2+b^2},\ \tan\alpha=\tfrac{b}{a} \;\xrightarrow{\;3\;}\; \text{solve } R\cos(\theta-\alpha)=k \;\xrightarrow{\;4\;}\; \text{extrema }\pm R.$$
Today the machine is put to work on fuller, examination-style problems. Before we start, let's check the toolbox is fully loaded.
Review ยท Activating Prior Knowledge
Quick check โ pick the correct statement in each question before we run the full toolbox.
$R=\sqrt{3^2+4^2}=\sqrt{25}=5$. This is the amplitude of the single cosine curve that replaces $3\cos x+4\sin x$.
$\tan\alpha=\dfrac{b}{a}$. To place $\alpha$ in the correct quadrant, use the signs of $\cos\alpha=\dfrac{a}{R}$ and $\sin\alpha=\dfrac{b}{R}$ โ a common source of half-marks in exams.
Since $-1\leqslant\cos(\theta-\alpha)\leqslant1$, the expression $R\cos(\theta-\alpha)$ oscillates between $R$ and $-R$. Solve the equation over the given interval as a standard cosine equation.
The Question
For $3\cos x+4\sin x$, run the toolbox: what is $R$, what is $\alpha$, and how would you then solve $3\cos x+4\sin x=4.5$ on $0^{\circ}<x<360^{\circ}$?
๐ก The Bridge (click to reveal)
$R=\sqrt{3^2+4^2}=5$ and $\tan\alpha=\tfrac43$ so $\alpha\approx53.13^{\circ}$, giving $3\cos x+4\sin x=5\cos(x-53.13^{\circ})$. Then $5\cos(x-53.13^{\circ})=4.5$ is a standard cosine equation. Today we apply this full pipeline to past-paper-style questions.
Past paper R-form questions follow a predictable pattern. Master this four-step system and you can handle any problem of the form โexpress $a\cos\theta + b\sin\theta$ in R-form, hence solve โฆโ.
Write the expression as $a\cos\theta + b\sin\theta$ (or $a\sin\theta + b\cos\theta$). Note the signs of $a$ and $b$ โ they determine the quadrant of $\alpha$.
$$R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \dfrac{b}{a} \text{ (or } \dfrac{a}{b} \text{ for sin form)}$$
Always state the exact value of $R$ when possible (e.g. $R = 5$, not $R \approx 5$). Give $\alpha$ to the required decimal places (usually 2 d.p.).
Replace $a\cos\theta + b\sin\theta$ with $R\cos(\theta - \alpha)$ (or the appropriate form) in the equation. Divide both sides by $R$ to get a standard trig equation.
Solve the resulting $\cos(\theta \pm \alpha) = k$ or $\sin(\theta \pm \alpha) = k$ equation. Always check that your solutions lie in the required interval, and use the periodicity of $\cos$/$\sin$ to find all solutions.
The class works together to derive the constraint equation for the table-in-corridor problem. The teacher poses questions and collects multiple responses before confirming the answer โ wrong answers are resources, not failures.
Work through this short problem individually. The teacher circulates to check the first few steps โ catching errors early is the key to fluency. Self-check your answer using the interactive tool.
(i) Express $3\cos x + 4\sin x$ in the form $R\cos(x - \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$, stating the exact value of $R$ and giving the value of $\alpha$ correct to 2 decimal places.
(ii) Hence solve the equation $3\cos x + 4\sin x = 4.5$, giving all solutions in the interval $0^\circ < x < 360^\circ$.
๐ Reveal Solution
Expand $R\cos(x - \alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha$.
Match coefficients with $3\cos x + 4\sin x$:
$R\cos\alpha = 3$ โ and โ $R\sin\alpha = 4$
Square and add: $R^2(\cos^2\alpha + \sin^2\alpha) = 9 + 16 = 25$, so $R^2 = 25$, $R = 5$ (exact).
Divide: $\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha} = \dfrac{4}{3}$, so $\alpha = \arctan\!\left(\dfrac{4}{3}\right) \approx 53.13^\circ$.
Both $R\cos\alpha = 3 > 0$ and $R\sin\alpha = 4 > 0$, so $\alpha$ is in the first quadrant, consistent with $0^\circ < \alpha < 90^\circ$.
$\boxed{3\cos x + 4\sin x = 5\cos(x - 53.13^\circ)}$
Substitute the R-form: $5\cos(x - 53.13^\circ) = 4.5$
$\cos(x - 53.13^\circ) = \dfrac{4.5}{5} = 0.9$
Let $\beta = x - 53.13^\circ$. Since $0^\circ < x < 360^\circ$, we have $-53.13^\circ < \beta < 306.87^\circ$.
Base angle: $\beta_0 = \arccos(0.9) \approx 25.84^\circ$.
Solutions for $\beta$: $\beta = 25.84^\circ$ or $\beta = -25.84^\circ$ (both in range).
Solution 1: $x - 53.13^\circ = 25.84^\circ \Rightarrow x = 78.97^\circ \approx 79.0^\circ$
Solution 2: $x - 53.13^\circ = -25.84^\circ \Rightarrow x = 27.29^\circ \approx 27.3^\circ$
Both solutions lie in $0^\circ < x < 360^\circ$.
$\boxed{x \approx 27.3^\circ, \quad 79.0^\circ}$ (to 1 d.p.)
(i) Express $7\cos\theta + 24\sin\theta$ in the form $R\cos(\theta - \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$, giving the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.
(ii) Hence solve the equation $7\cos\theta + 24\sin\theta = 15$, giving all solutions in the interval $0^\circ \leqslant \theta \leqslant 360^\circ$.
๐ Reveal Solution
Expand $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Match coefficients with $7\cos\theta + 24\sin\theta$:
$R\cos\alpha = 7$ โ and โ $R\sin\alpha = 24$
Square and add: $R^2 = 49 + 576 = 625$, so $R = 25$ (exact).
Divide: $\tan\alpha = \dfrac{24}{7}$, so $\alpha = \arctan\!\left(\dfrac{24}{7}\right) \approx 73.74^\circ$.
Both $R\cos\alpha > 0$ and $R\sin\alpha > 0$, so $\alpha$ is in the first quadrant, consistent with $0^\circ < \alpha < 90^\circ$.
$\boxed{7\cos\theta + 24\sin\theta = 25\cos(\theta - 73.74^\circ)}$
Substitute the R-form: $25\cos(\theta - 73.74^\circ) = 15$
$\cos(\theta - 73.74^\circ) = \dfrac{15}{25} = 0.6$
Let $\beta = \theta - 73.74^\circ$. Since $0^\circ \leqslant \theta \leqslant 360^\circ$, we have $-73.74^\circ \leqslant \beta \leqslant 286.26^\circ$.
Base angle: $\beta_0 = \arccos(0.6) \approx 53.13^\circ$.
Solutions for $\beta$: $\beta = 53.13^\circ$ or $\beta = -53.13^\circ$ (both in range).
Solution 1: $\theta - 73.74^\circ = 53.13^\circ \Rightarrow \theta = 126.87^\circ \approx 126.9^\circ$
Solution 2: $\theta - 73.74^\circ = -53.13^\circ \Rightarrow \theta = 20.61^\circ \approx 20.6^\circ$
Both solutions lie in $0^\circ \leqslant \theta \leqslant 360^\circ$.
$\boxed{\theta \approx 20.6^\circ, \quad 126.9^\circ}$ (to 1 d.p.)
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest. All questions are from past CIE 9709 papers.
(i) Express $5\cos\theta - \sin\theta$ in the form $R\cos(\theta + \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$, giving the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.
(ii) Hence solve the equation $5\cos\theta - \sin\theta = 4$, giving all solutions in the interval $0^\circ \leqslant \theta \leqslant 360^\circ$.
๐ก Hint
(i) Use $R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$. Match: $R\cos\alpha = 5$, $R\sin\alpha = 1$. Then $R = \sqrt{25 + 1} = \sqrt{26}$ and $\tan\alpha = 1/5$.
(ii) Substitute: $\sqrt{26}\cos(\theta + 11.31^\circ) = 4$, so $\cos(\theta + 11.31^\circ) = 4/\sqrt{26} \approx 0.784$. Let $\beta = \theta + 11.31^\circ$. Solve $\cos\beta = 0.784$ for $\beta \in [11.31^\circ, 371.31^\circ]$.
๐ Reveal Solution
Expand $R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$.
Match with $5\cos\theta - \sin\theta$:
$R\cos\alpha = 5$ โ and โ $R\sin\alpha = 1$
Square and add: $R^2 = 25 + 1 = 26$, so $R = \sqrt{26}$ (exact).
Divide: $\tan\alpha = \dfrac{1}{5} = 0.2$, so $\alpha = \arctan(0.2) \approx 11.31^\circ$.
Both $R\cos\alpha > 0$ and $R\sin\alpha > 0$, so $\alpha$ is in the first quadrant.
$\boxed{5\cos\theta - \sin\theta = \sqrt{26}\cos(\theta + 11.31^\circ)}$
Substitute: $\sqrt{26}\cos(\theta + 11.31^\circ) = 4$
$\cos(\theta + 11.31^\circ) = \dfrac{4}{\sqrt{26}} \approx 0.7845$
Let $\beta = \theta + 11.31^\circ$. Since $0^\circ \leqslant \theta \leqslant 360^\circ$, we have $11.31^\circ \leqslant \beta \leqslant 371.31^\circ$.
Base angle: $\beta_0 = \arccos(0.7845) \approx 38.34^\circ$.
Solutions for $\beta$: $\beta = 38.34^\circ$ and $\beta = 360^\circ - 38.34^\circ = 321.66^\circ$ (both in range).
Solution 1: $\theta = 38.34^\circ - 11.31^\circ = 27.03^\circ \approx 27.0^\circ$
Solution 2: $\theta = 321.66^\circ - 11.31^\circ = 310.35^\circ \approx 310.4^\circ$
$\boxed{\theta \approx 27.0^\circ, \quad 310.4^\circ}$ (to 1 d.p.)
Solve the equation $\cos\theta + 3\sin\theta = 2$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.
๐ก Hint
First express $\cos\theta + 3\sin\theta$ in the form $R\cos(\theta - \alpha)$. Match: $R\cos\alpha = 1$, $R\sin\alpha = 3$. Then $R = \sqrt{1 + 9} = \sqrt{10}$ and $\tan\alpha = 3$.
Substitute: $\sqrt{10}\cos(\theta - \alpha) = 2$, so $\cos(\theta - \alpha) = 2/\sqrt{10} \approx 0.632$. Solve for $\theta$.
๐ Reveal Solution
Expand $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Match with $\cos\theta + 3\sin\theta$:
$R\cos\alpha = 1$ โ and โ $R\sin\alpha = 3$
$R = \sqrt{1 + 9} = \sqrt{10}$ (exact), โ $\tan\alpha = 3$, โ $\alpha = \arctan(3) \approx 71.57^\circ$
$\cos\theta + 3\sin\theta = \sqrt{10}\cos(\theta - 71.57^\circ)$
Substitute: $\sqrt{10}\cos(\theta - 71.57^\circ) = 2$
$\cos(\theta - 71.57^\circ) = \dfrac{2}{\sqrt{10}} \approx 0.6325$
Let $\beta = \theta - 71.57^\circ$. Since $0^\circ \leqslant \theta \leqslant 360^\circ$, we have $-71.57^\circ \leqslant \beta \leqslant 288.43^\circ$.
Base angle: $\beta_0 = \arccos(0.6325) \approx 50.77^\circ$.
Solutions for $\beta$: $\beta = 50.77^\circ$ or $\beta = -50.77^\circ$ (both in range).
Solution 1: $\theta = 50.77^\circ + 71.57^\circ = 122.34^\circ \approx 122.3^\circ$
Solution 2: $\theta = -50.77^\circ + 71.57^\circ = 20.80^\circ \approx 20.8^\circ$
$\boxed{\theta \approx 20.8^\circ, \quad 122.3^\circ}$ (to 1 d.p.)
By expressing $8\sin\theta - 6\cos\theta$ in the form $R\sin(\theta - \alpha)$, solve the equation $8\sin\theta - 6\cos\theta = 7$, for $0^\circ \leqslant \theta \leqslant 360^\circ$.
๐ก Hint
Expand $R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$. Match with $8\sin\theta - 6\cos\theta$: $R\cos\alpha = 8$, $R\sin\alpha = 6$. Then $R = \sqrt{64 + 36} = 10$ and $\tan\alpha = 6/8 = 3/4$.
Substitute: $10\sin(\theta - \alpha) = 7$, so $\sin(\theta - \alpha) = 0.7$. Solve for $\theta$ in $[0^\circ, 360^\circ]$.
๐ Reveal Solution
Expand $R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$.
Match with $8\sin\theta - 6\cos\theta$:
$R\cos\alpha = 8$ โ and โ $R\sin\alpha = 6$
$R = \sqrt{64 + 36} = \sqrt{100} = 10$ (exact), โ $\tan\alpha = \dfrac{6}{8} = \dfrac{3}{4}$, โ $\alpha = \arctan(0.75) \approx 36.87^\circ$
$8\sin\theta - 6\cos\theta = 10\sin(\theta - 36.87^\circ)$
Substitute: $10\sin(\theta - 36.87^\circ) = 7$
$\sin(\theta - 36.87^\circ) = \dfrac{7}{10} = 0.7$
Let $\beta = \theta - 36.87^\circ$. Since $0^\circ \leqslant \theta \leqslant 360^\circ$, we have $-36.87^\circ \leqslant \beta \leqslant 323.13^\circ$.
Base angle: $\beta_0 = \arcsin(0.7) \approx 44.43^\circ$.
Solutions for $\beta$: $\beta = 44.43^\circ$ or $\beta = 180^\circ - 44.43^\circ = 135.57^\circ$ (both in range).
Solution 1: $\theta = 44.43^\circ + 36.87^\circ = 81.30^\circ \approx 81.3^\circ$
Solution 2: $\theta = 135.57^\circ + 36.87^\circ = 172.44^\circ \approx 172.4^\circ$
$\boxed{\theta \approx 81.3^\circ, \quad 172.4^\circ}$ (to 1 d.p.)
The diagram shows a table stuck in a corridor. The table is 120 cm long and 80 cm wide, and the width of the corridor is 130 cm.
(i) Show that $12\sin\theta + 8\cos\theta = 13$.
(ii) Hence find the angle $\theta$. (There are two answers.)
๐ก Hint
(i) When the table tilts at angle $\theta$, the horizontal projection is $120\sin\theta + 80\cos\theta$. At the critical angle, this equals the corridor width 130 cm. Divide by 10.
(ii) Express $12\sin\theta + 8\cos\theta$ in the form $R\sin(\theta + \alpha)$. Match: $R\cos\alpha = 12$, $R\sin\alpha = 8$. Then $R = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}$ and $\tan\alpha = 8/12 = 2/3$. Solve $\sin(\theta + \alpha) = 13/(4\sqrt{13})$.
๐ Reveal Solution
When the table tilts at angle $\theta$ to the corridor direction:
โข The length (120 cm) contributes $120\sin\theta$ to the horizontal width (perpendicular to the corridor).
โข The width (80 cm) contributes $80\cos\theta$ to the horizontal width (along the corridor).
The table jams when the total horizontal projection equals the corridor width:
$$120\sin\theta + 80\cos\theta = 130$$
Dividing both sides by 10:
$$12\sin\theta + 8\cos\theta = 13 \quad \checkmark$$
Express $12\sin\theta + 8\cos\theta$ in the form $R\sin(\theta + \alpha)$:
$R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$
Match: $R\cos\alpha = 12$, โ $R\sin\alpha = 8$
$R = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}$, โ $\tan\alpha = \dfrac{8}{12} = \dfrac{2}{3}$, โ $\alpha \approx 33.69^\circ$
So $12\sin\theta + 8\cos\theta = 4\sqrt{13}\sin(\theta + 33.69^\circ)$.
Substitute: $4\sqrt{13}\sin(\theta + 33.69^\circ) = 13$
$\sin(\theta + 33.69^\circ) = \dfrac{13}{4\sqrt{13}} = \dfrac{13}{4\sqrt{13}} \cdot \dfrac{\sqrt{13}}{\sqrt{13}} = \dfrac{13\sqrt{13}}{4 \times 13} = \dfrac{\sqrt{13}}{4} \approx 0.9014$
Let $\beta = \theta + 33.69^\circ$. Solve $\sin\beta = 0.9014$.
Base angle: $\beta_0 = \arcsin(0.9014) \approx 64.34^\circ$.
Solutions: $\beta = 64.34^\circ$ or $\beta = 180^\circ - 64.34^\circ = 115.66^\circ$.
Solution 1: $\theta = 64.34^\circ - 33.69^\circ = 30.65^\circ \approx 30.7^\circ$
Solution 2: $\theta = 115.66^\circ - 33.69^\circ = 81.97^\circ \approx 82.0^\circ$
$\boxed{\theta \approx 30.7^\circ, \quad 82.0^\circ}$ (to 1 d.p.)
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest โ this guides next lesson's review.