R-form Applications I

Expressing $a\cos\theta + b\sin\theta$ as a single trig function — solving equations, finding max/min, sketching graphs
9709 Paper 3 · §3.3 Trigonometry · Lesson 2
💡 Lead In: From Form to Solutions

The Scenario

Yesterday you learned to compress $a\cos\theta+b\sin\theta$ into a single cosine. For example,

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

Now the payoff: that single form lets us solve equations and read off maximum and minimum values immediately.

Review · Activating Prior Knowledge

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

Review 1 — For $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$, which pair is correct?
$R=\sqrt{a^2+b^2}, \; \tan\alpha=\dfrac{b}{a}$
$R=a+b, \; \tan\alpha=\dfrac{a}{b}$
$R=\sqrt{a^2+b^2}, \; \tan\alpha=\dfrac{a}{b}$
$R=a^2+b^2, \; \tan\alpha=\dfrac{b}{a}$

$a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$ with $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\dfrac{b}{a}$ (taking care to choose the correct quadrant for $\alpha$).

Review 2 — After writing the equation as $R\cos(\theta-\alpha)=k$, what is the next step?
Add $\alpha$ to both sides immediately, then solve
Divide by $\theta$
Solve $\cos(\theta-\alpha)=\dfrac{k}{R}$ for $\theta-\alpha$, then add $\alpha$
Square both sides and take square roots

Divide through by $R$ to get $\cos(\theta-\alpha)=\dfrac{k}{R}$, solve this standard cosine equation for $\theta-\alpha$ (remembering all solutions in the interval), then add $\alpha$ back.

Review 3 — What are the maximum and minimum values of $R\cos(\theta-\alpha)$?
Max $1$, min $-1$
Max $R^2$, min $-R^2$
Max $R+1$, min $R-1$
Max $R$, min $-R$

Since $-1\leqslant\cos(\theta-\alpha)\leqslant1$, multiplying by $R>0$ gives $-R\leqslant R\cos(\theta-\alpha)\leqslant R$. The amplitude of the R-form is exactly $R$.

The Question

How does converting to the R-form turn "solve $\cos\theta+\sqrt3\sin\theta=1$" into a routine cosine equation, and how does it let you state the maximum and minimum of $3\cos\theta+4\sin\theta$ without any graphing?

💡 The Bridge (click to reveal)

Expressing $\cos\theta+\sqrt3\sin\theta=2\cos(\theta-60^{\circ})$ turns the equation into $2\cos(\theta-60^{\circ})=1$, i.e. $\cos(\theta-60^{\circ})=\tfrac12$, which is routine. For extrema, $3\cos\theta+4\sin\theta=5\cos(\theta-\alpha)$ has maximum $5$ and minimum $-5$. Today we practise both applications.

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

Any expression of the form $a\cos\theta + b\sin\theta$ can be written as a single trigonometric function. This is called the R-form or auxiliary angle method. The key idea is to run the compound angle formula in reverse.

Form 1: $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$

Expand $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$ and match coefficients:

$$R\cos\alpha = a, \quad R\sin\alpha = b$$ $$R = \sqrt{a^2 + b^2}, \quad \tan\alpha = \dfrac{b}{a}$$

Use the signs of $a$ and $b$ to determine the correct quadrant for $\alpha$.

Form 2: $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$

Expand $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$ and match coefficients:

$$R\cos\alpha = a, \quad R\sin\alpha = b$$ $$R = \sqrt{a^2 + b^2}, \quad \tan\alpha = \dfrac{b}{a}$$

Form 3: $a\cos\theta - b\sin\theta = R\cos(\theta + \alpha)$

Expand $R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$ and match coefficients:

$$R\cos\alpha = a, \quad R\sin\alpha = b$$ $$R = \sqrt{a^2 + b^2}, \quad \tan\alpha = \dfrac{b}{a}$$

Form 4: $a\sin\theta - b\cos\theta = R\sin(\theta - \alpha)$

Expand $R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$ and match coefficients:

$$R\cos\alpha = a, \quad R\sin\alpha = b$$ $$R = \sqrt{a^2 + b^2}, \quad \tan\alpha = \dfrac{b}{a}$$

Maximum and minimum values:

Since $-1 \leqslant \cos(\theta \pm \alpha) \leqslant 1$ and $-1 \leqslant \sin(\theta \pm \alpha) \leqslant 1$:

$$\text{Maximum of } a\cos\theta + b\sin\theta = R = \sqrt{a^2 + b^2}$$ $$\text{Minimum of } a\cos\theta + b\sin\theta = -R = -\sqrt{a^2 + b^2}$$

Group · Guided Worked Example 1: R-form with Degrees — Max, Min, Graph, Equation Exercise 3D Q6

(i) Express $5\cos\theta - 12\sin\theta$ in the form $r\cos(\theta + \alpha)$, where $r > 0$ and $0^\circ < \alpha < 90^\circ$.

(ii) State the maximum and minimum values of $5\cos\theta - 12\sin\theta$.

(iii) Sketch the graph of $y = 5\cos\theta - 12\sin\theta$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.

(iv) Solve the equation $5\cos\theta - 12\sin\theta = 4$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.

👁 Reveal Solution
Part (i): Express in the form $r\cos(\theta + \alpha)$

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

Match coefficients with $5\cos\theta - 12\sin\theta$:

$r\cos\alpha = 5$   and   $r\sin\alpha = 12$

Square and add: $r^2(\cos^2\alpha + \sin^2\alpha) = 25 + 144 = 169$, so $r^2 = 169$, $r = 13$.

Divide: $\tan\alpha = \dfrac{r\sin\alpha}{r\cos\alpha} = \dfrac{12}{5}$, so $\alpha = \arctan(2.4) \approx 67.38^\circ$.

Both $r\cos\alpha = 5 > 0$ and $r\sin\alpha = 12 > 0$, so $\alpha$ is in the first quadrant, consistent with $0^\circ < \alpha < 90^\circ$.

$\boxed{5\cos\theta - 12\sin\theta = 13\cos(\theta + 67.38^\circ)}$

Part (ii): Maximum and minimum values

Since $-1 \leqslant \cos(\theta + 67.38^\circ) \leqslant 1$:

Maximum $= 13 \times 1 = \boxed{13}$

Minimum $= 13 \times (-1) = \boxed{-13}$

Part (iii): Sketch the graph

The graph of $y = 13\cos(\theta + 67.38^\circ)$ is a cosine wave with:

• Amplitude $= 13$

• Period $= 360^\circ$

• Phase shift: the curve reaches its maximum when $\theta + 67.38^\circ = 0^\circ$, i.e. at $\theta = -67.38^\circ$ (equivalently, the maximum in $[0^\circ, 360^\circ]$ occurs at $\theta = 360^\circ - 67.38^\circ = 292.62^\circ$).

• Crosses the $\theta$-axis when $\cos(\theta + 67.38^\circ) = 0$, i.e. $\theta + 67.38^\circ = 90^\circ, 270^\circ$, giving $\theta = 22.62^\circ, 202.62^\circ$.

• At $\theta = 0^\circ$: $y = 5$. At $\theta = 90^\circ$: $y = -12$.

Part (iv): Solve $5\cos\theta - 12\sin\theta = 4$

Substitute the R-form: $13\cos(\theta + 67.38^\circ) = 4$

$\cos(\theta + 67.38^\circ) = \dfrac{4}{13} \approx 0.3077$

Let $\beta = \theta + 67.38^\circ$. Then $\cos\beta = 0.3077$.

$\beta = \arccos(0.3077) \approx 72.08^\circ$   or   $\beta = 360^\circ - 72.08^\circ = 287.92^\circ$

Solution 1: $\theta + 67.38^\circ = 72.08^\circ \Rightarrow \theta = 4.70^\circ$

Solution 2: $\theta + 67.38^\circ = 287.92^\circ \Rightarrow \theta = 220.54^\circ$

Both solutions lie in $0^\circ \leqslant \theta \leqslant 360^\circ$.

$\boxed{\theta \approx 4.7^\circ, \quad 220.5^\circ}$ (to 1 d.p.)

Individual · Guided Worked Example 2: R-form with Radians — Max, Min, Graph, Equation Exercise 3D Q7

(i) Express $\sqrt{3}\sin\theta - \cos\theta$ in the form $r\sin(\theta - \alpha)$, where $r > 0$ and $0 < \alpha < \dfrac{\pi}{2}$.

(ii) State the maximum and minimum values of $\sqrt{3}\sin\theta - \cos\theta$.

(iii) Sketch the graph of $y = \sqrt{3}\sin\theta - \cos\theta$ for $0 \leqslant \theta \leqslant 2\pi$.

(iv) Solve the equation $\sqrt{3}\sin\theta - \cos\theta = 1$ for $0 \leqslant \theta \leqslant 2\pi$.

👁 Reveal Solution
Part (i): Express in the form $r\sin(\theta - \alpha)$

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

Match coefficients with $\sqrt{3}\sin\theta - \cos\theta$:

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

Square and add: $r^2 = 3 + 1 = 4$, so $r = 2$.

Divide: $\tan\alpha = \dfrac{1}{\sqrt{3}}$, so $\alpha = \dfrac{\pi}{6}$.

Both $r\cos\alpha > 0$ and $r\sin\alpha > 0$, so $\alpha$ is in the first quadrant, consistent with $0 < \alpha < \dfrac{\pi}{2}$.

$\boxed{\sqrt{3}\sin\theta - \cos\theta = 2\sin\!\left(\theta - \dfrac{\pi}{6}\right)}$

Part (ii): Maximum and minimum values

Since $-1 \leqslant \sin\!\left(\theta - \dfrac{\pi}{6}\right) \leqslant 1$:

Maximum $= 2 \times 1 = \boxed{2}$

Minimum $= 2 \times (-1) = \boxed{-2}$

Part (iii): Sketch the graph

The graph of $y = 2\sin\!\left(\theta - \dfrac{\pi}{6}\right)$ is a sine wave with:

• Amplitude $= 2$

• Period $= 2\pi$

• Phase shift: the curve starts at $y = 0$ when $\theta - \dfrac{\pi}{6} = 0$, i.e. $\theta = \dfrac{\pi}{6}$. It reaches its maximum at $\theta - \dfrac{\pi}{6} = \dfrac{\pi}{2}$, i.e. $\theta = \dfrac{2\pi}{3}$.

• Crosses the $\theta$-axis at $\theta = \dfrac{\pi}{6}, \dfrac{7\pi}{6}$.

• At $\theta = 0$: $y = 2\sin\!\left(-\dfrac{\pi}{6}\right) = -1$. At $\theta = 2\pi$: $y = 2\sin\!\left(\dfrac{11\pi}{6}\right) = -1$.

Part (iv): Solve $\sqrt{3}\sin\theta - \cos\theta = 1$

Substitute the R-form: $2\sin\!\left(\theta - \dfrac{\pi}{6}\right) = 1$

$\sin\!\left(\theta - \dfrac{\pi}{6}\right) = \dfrac{1}{2}$

Let $\beta = \theta - \dfrac{\pi}{6}$. Then $\sin\beta = \dfrac{1}{2}$.

In the range $0 \leqslant \theta \leqslant 2\pi$, we have $-\dfrac{\pi}{6} \leqslant \beta \leqslant \dfrac{11\pi}{6}$.

$\beta = \dfrac{\pi}{6}$   or   $\beta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$

Solution 1: $\theta - \dfrac{\pi}{6} = \dfrac{\pi}{6} \Rightarrow \theta = \dfrac{\pi}{3}$

Solution 2: $\theta - \dfrac{\pi}{6} = \dfrac{5\pi}{6} \Rightarrow \theta = \pi$

Both solutions lie in $0 \leqslant \theta \leqslant 2\pi$.

$\boxed{\theta = \dfrac{\pi}{3}, \quad \pi}$

✏️ Concept Quiz
Score: 0 / 4
1. Express $3\cos\theta + 4\sin\theta$ in the form $R\cos(\theta - \alpha)$. What are $R$ and $\alpha$?
$R = 5, \quad \alpha = \arctan\!\left(\dfrac{4}{3}\right) \approx 53.1^\circ$
$R = 7, \quad \alpha = \arctan\!\left(\dfrac{4}{3}\right) \approx 53.1^\circ$
$R = 5, \quad \alpha = \arctan\!\left(\dfrac{3}{4}\right) \approx 36.9^\circ$
$R = 25, \quad \alpha = \arctan\!\left(\dfrac{4}{3}\right) \approx 53.1^\circ$
Match $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$ with $3\cos\theta + 4\sin\theta$: $R\cos\alpha = 3$, $R\sin\alpha = 4$. Then $R = \sqrt{9 + 16} = 5$ and $\tan\alpha = \dfrac{4}{3}$, so $\alpha \approx 53.1^\circ$. The common error is to swap $a$ and $b$ in $\tan\alpha$.
2. The maximum value of $7\cos\theta + 24\sin\theta$ is:
$25$
$31$
$17$
$7$
$R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$. The maximum of $R\cos(\theta - \alpha)$ is $R = 25$. The values 31, 17, and 7 are distractors: 31 = 7 + 24, 17 = 24 - 7, and 7 is just the cosine coefficient.
3. Given $5\cos\theta - 12\sin\theta = 13\cos(\theta + \alpha)$ where $\tan\alpha = \dfrac{12}{5}$, the equation $5\cos\theta - 12\sin\theta = 6.5$ is equivalent to:
$\cos(\theta + \alpha) = \dfrac{1}{2}$
$\cos(\theta + \alpha) = 6.5$
$\cos(\theta + \alpha) = 13$
$\sin(\theta + \alpha) = \dfrac{1}{2}$
Substitute: $13\cos(\theta + \alpha) = 6.5$, so $\cos(\theta + \alpha) = \dfrac{6.5}{13} = \dfrac{1}{2}$. The value 6.5 on the right must be divided by the amplitude 13, not used directly. Since $\dfrac{1}{2} \leqslant 1$, a solution exists.
4. The graph of $y = 13\cos(\theta + 67^\circ)$ has:
Amplitude $13$, period $360^\circ$, shifted left by $67^\circ$
Amplitude $67$, period $13^\circ$, shifted right by $13^\circ$
Amplitude $13$, period $180^\circ$, shifted right by $67^\circ$
Amplitude $67$, period $360^\circ$, shifted left by $13^\circ$
For $y = R\cos(\theta + \alpha)$: the amplitude is $R = 13$ (the coefficient), the period of $\cos\theta$ is $360^\circ$, and the $+\alpha$ inside the bracket shifts the graph left by $\alpha = 67^\circ$. A common error is to confuse the amplitude with the phase shift, or to think $+\alpha$ shifts right.
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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 Q5(v)(vi) Standard

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

(v) $\sin\theta - \cos\theta$   (vi) $\sqrt{2}\sin\theta - \sqrt{2}\cos\theta$

💡 Hint

Expand $r\cos(\theta - \alpha) = r\cos\theta\cos\alpha + r\sin\theta\sin\alpha$. Match the coefficient of $\cos\theta$ (which is negative here) with $r\cos\alpha$, and the coefficient of $\sin\theta$ with $r\sin\alpha$. Since $r\cos\alpha < 0$ and $r\sin\alpha > 0$, $\alpha$ is in the second quadrant.

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

Rewrite as $-\cos\theta + \sin\theta$ to identify coefficients.

Match with $r\cos\theta\cos\alpha + r\sin\theta\sin\alpha$:

$r\cos\alpha = -1$   (coefficient of $\cos\theta$)

$r\sin\alpha = 1$   (coefficient of $\sin\theta$)

$r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$

$\tan\alpha = \dfrac{1}{-1} = -1$

Since $r\cos\alpha < 0$ and $r\sin\alpha > 0$, $\alpha$ is in the second quadrant.

$\alpha = 180^\circ - 45^\circ = 135^\circ$

$\boxed{\sin\theta - \cos\theta = \sqrt{2}\cos(\theta - 135^\circ)}$

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

Rewrite as $-\sqrt{2}\cos\theta + \sqrt{2}\sin\theta$.

Match: $r\cos\alpha = -\sqrt{2}$, $r\sin\alpha = \sqrt{2}$

$r = \sqrt{2 + 2} = \sqrt{4} = 2$

$\tan\alpha = \dfrac{\sqrt{2}}{-\sqrt{2}} = -1$

Since $r\cos\alpha < 0$ and $r\sin\alpha > 0$, $\alpha$ is in the second quadrant.

$\alpha = 180^\circ - 45^\circ = 135^\circ$

$\boxed{\sqrt{2}\sin\theta - \sqrt{2}\cos\theta = 2\cos(\theta - 135^\circ)}$

Exercise 3D Q9 Medium

(i) Express $\cos\theta + \sqrt{2}\sin\theta$ in the form $r\cos(\theta - \alpha)$, where $r > 0$ and $0^\circ < \alpha < 90^\circ$.

(ii) State the maximum and minimum values of $\cos\theta + \sqrt{2}\sin\theta$ and the smallest positive values of $\theta$ for which they occur.

(iii) Sketch the graph of $y = \cos\theta + \sqrt{2}\sin\theta$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.

(iv) State the maximum and minimum values of $3 + \cos\theta + \sqrt{2}\sin\theta$ and the smallest positive values of $\theta$ for which they occur.

💡 Hint

(i) Match $r\cos\alpha = 1$ (coefficient of $\cos\theta$), $r\sin\alpha = \sqrt{2}$ (coefficient of $\sin\theta$).

(ii) Max $= r$ when $\cos(\theta - \alpha) = 1$; Min $= -r$ when $\cos(\theta - \alpha) = -1$.

(iv) The $+3$ shifts the whole graph up by 3. The max/min of $3 + f(\theta)$ are $3 + \text{max}(f)$ and $3 + \text{min}(f)$, occurring at the same $\theta$ values.

👁 Reveal Solution
Part (i): Express in the form $r\cos(\theta - \alpha)$

Match $r\cos(\theta - \alpha) = r\cos\theta\cos\alpha + r\sin\theta\sin\alpha$ with $\cos\theta + \sqrt{2}\sin\theta$:

$r\cos\alpha = 1$   $r\sin\alpha = \sqrt{2}$

$r = \sqrt{1 + 2} = \sqrt{3}$

$\tan\alpha = \dfrac{\sqrt{2}}{1} = \sqrt{2}$, so $\alpha = \arctan(\sqrt{2}) \approx 54.74^\circ$

Both $r\cos\alpha > 0$ and $r\sin\alpha > 0$, so $\alpha$ is in the first quadrant.

$\boxed{\cos\theta + \sqrt{2}\sin\theta = \sqrt{3}\cos(\theta - 54.74^\circ)}$

Part (ii): Maximum and minimum values

Maximum $= \sqrt{3}$, occurring when $\cos(\theta - 54.74^\circ) = 1$:

$\theta - 54.74^\circ = 0^\circ \Rightarrow \theta = 54.74^\circ$

Minimum $= -\sqrt{3}$, occurring when $\cos(\theta - 54.74^\circ) = -1$:

$\theta - 54.74^\circ = 180^\circ \Rightarrow \theta = 234.74^\circ$

$\boxed{\text{Max} = \sqrt{3} \text{ at } \theta \approx 54.7^\circ; \quad \text{Min} = -\sqrt{3} \text{ at } \theta \approx 234.7^\circ}$

Part (iii): Sketch the graph

The graph of $y = \sqrt{3}\cos(\theta - 54.74^\circ)$ is a cosine wave with:

• Amplitude $= \sqrt{3} \approx 1.73$

• Period $= 360^\circ$

• Maximum at $\theta = 54.74^\circ$, minimum at $\theta = 234.74^\circ$

• Crosses the $\theta$-axis at $\theta - 54.74^\circ = 90^\circ, 270^\circ$, i.e. $\theta = 144.74^\circ, 324.74^\circ$

• At $\theta = 0^\circ$: $y = 1$. At $\theta = 360^\circ$: $y = 1$.

Part (iv): Max/min of $3 + \cos\theta + \sqrt{2}\sin\theta$

The $+3$ shifts the graph vertically. The max/min occur at the same $\theta$ values as in part (ii).

Maximum $= 3 + \sqrt{3}$ at $\theta \approx 54.7^\circ$

Minimum $= 3 - \sqrt{3}$ at $\theta \approx 234.7^\circ$

$\boxed{\text{Max} = 3 + \sqrt{3} \text{ at } \theta \approx 54.7^\circ; \quad \text{Min} = 3 - \sqrt{3} \text{ at } \theta \approx 234.7^\circ}$

Exercise 3D Q8 Challenging

(i) Express $2\sin 2\theta + 3\cos 2\theta$ in the form $r\sin(2\theta + \alpha)$, where $r > 0$ and $0^\circ < \alpha < 90^\circ$.

(ii) State the maximum and minimum values of $2\sin 2\theta + 3\cos 2\theta$ and the smallest positive values of $\theta$ for which they occur.

(iii) Sketch the graph of $y = 2\sin 2\theta + 3\cos 2\theta$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.

(iv) Solve the equation $2\sin 2\theta + 3\cos 2\theta = 1$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.

💡 Hint

(i) Let $u = 2\theta$. Express $2\sin u + 3\cos u$ in the form $r\sin(u + \alpha)$. Match: $r\cos\alpha = 2$, $r\sin\alpha = 3$.

(ii) Max $= r$ when $\sin(2\theta + \alpha) = 1$; solve for $\theta$. The period is $180^\circ$ (due to $2\theta$), so be careful with the range.

(iv) Substitute: $r\sin(2\theta + \alpha) = 1$. Solve $\sin(2\theta + \alpha) = 1/r$. Since $\theta \in [0^\circ, 360^\circ]$, we have $2\theta \in [0^\circ, 720^\circ]$, so $2\theta + \alpha$ ranges over $720^\circ$ — expect four solutions.

👁 Reveal Solution
Part (i): Express in the form $r\sin(2\theta + \alpha)$

Expand $r\sin(2\theta + \alpha) = r\sin 2\theta\cos\alpha + r\cos 2\theta\sin\alpha$.

Match with $2\sin 2\theta + 3\cos 2\theta$:

$r\cos\alpha = 2$   $r\sin\alpha = 3$

$r = \sqrt{4 + 9} = \sqrt{13}$

$\tan\alpha = \dfrac{3}{2}$, so $\alpha = \arctan(1.5) \approx 56.31^\circ$

$\boxed{2\sin 2\theta + 3\cos 2\theta = \sqrt{13}\sin(2\theta + 56.31^\circ)}$

Part (ii): Maximum and minimum values

Maximum $= \sqrt{13}$, occurring when $\sin(2\theta + 56.31^\circ) = 1$:

$2\theta + 56.31^\circ = 90^\circ \Rightarrow 2\theta = 33.69^\circ \Rightarrow \theta = 16.85^\circ$

Minimum $= -\sqrt{13}$, occurring when $\sin(2\theta + 56.31^\circ) = -1$:

$2\theta + 56.31^\circ = 270^\circ \Rightarrow 2\theta = 213.69^\circ \Rightarrow \theta = 106.85^\circ$

$\boxed{\text{Max} = \sqrt{13} \text{ at } \theta \approx 16.8^\circ; \quad \text{Min} = -\sqrt{13} \text{ at } \theta \approx 106.8^\circ}$

Part (iii): Sketch the graph

The graph of $y = \sqrt{13}\sin(2\theta + 56.31^\circ)$ is a sine wave with:

• Amplitude $= \sqrt{13} \approx 3.61$

• Period $= \dfrac{360^\circ}{2} = 180^\circ$ (due to the $2\theta$)

• Two complete cycles in $[0^\circ, 360^\circ]$

• Maxima at $\theta = 16.85^\circ$ and $\theta = 196.85^\circ$

• Minima at $\theta = 106.85^\circ$ and $\theta = 286.85^\circ$

• At $\theta = 0^\circ$: $y = \sqrt{13}\sin(56.31^\circ) = 3$. At $\theta = 360^\circ$: $y = 3$.

Part (iv): Solve $2\sin 2\theta + 3\cos 2\theta = 1$

Substitute: $\sqrt{13}\sin(2\theta + 56.31^\circ) = 1$

$\sin(2\theta + 56.31^\circ) = \dfrac{1}{\sqrt{13}} \approx 0.2774$

Let $\beta = 2\theta + 56.31^\circ$. Since $\theta \in [0^\circ, 360^\circ]$, we have $\beta \in [56.31^\circ, 776.31^\circ]$.

Base solutions: $\beta_1 = \arcsin(0.2774) \approx 16.10^\circ$ (not in range), $\beta_2 = 180^\circ - 16.10^\circ = 163.90^\circ$ (in range).

Adding $360^\circ$ to find all solutions in range:

• $\beta = 163.90^\circ$: $2\theta = 107.59^\circ$, $\theta = 53.80^\circ$

• $\beta = 376.10^\circ$: $2\theta = 319.79^\circ$, $\theta = 159.90^\circ$

• $\beta = 523.90^\circ$: $2\theta = 467.59^\circ$, $\theta = 233.80^\circ$

• $\beta = 736.10^\circ$: $2\theta = 679.79^\circ$, $\theta = 339.90^\circ$

$\boxed{\theta \approx 53.8^\circ, \quad 159.9^\circ, \quad 233.8^\circ, \quad 339.9^\circ}$ (to 1 d.p.)

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