Figure 3.2: Harmonic Combination: $a\sin x + b\cos x = R\sin(x + \alpha)$
1. Compound Angle Addition and Subtraction Formulas
The addition and subtraction formulas express trigonometric ratios of compound arguments $(A \pm B)$ in terms of individual ratios of $A$ and $B$:
$$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$$ $$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$$ $$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$$Critical Sign Reversal: Note carefully that the cosine expansion reverses signs: $\cos(A + B)$ has a minus between the terms, whereas $\cos(A - B)$ has a plus.
2. Double Angle Formulas
Setting $A = B = \theta$ produces the standard double angle formulas:
$$\sin(2\theta) = 2\sin\theta\cos\theta$$ $$\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$ $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$Rearranging the three forms of $\cos(2\theta)$ yields the power-reduction identities, indispensable for Paper 1 calculus integration:
$$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}, \quad \sin^2\theta = \frac{1 - \cos(2\theta)}{2}$$3. Harmonic Linear Combination: $a\sin x + b\cos x$
Any linear sum of sine and cosine waves sharing identical period can be condensed into a single phase-shifted sinusoid. By expanding $R\sin(x + \alpha) = R\cos\alpha\sin x + R\sin\alpha\cos x$ and equating coefficients:
$$a\sin x + b\cos x = R\sin(x + \alpha) \quad \text{where} \quad R = \sqrt{a^2 + b^2}, \quad \tan\alpha = \frac{b}{a}$$ $$a\cos x + b\sin x = R\cos(x - \beta) \quad \text{where} \quad R = \sqrt{a^2 + b^2}, \quad \tan\beta = \frac{b}{a}$$Analytical Power of Harmonic Form
• Extrema & Range: Since $-1 \le \sin(x + \alpha) \le 1$, the expression $a\sin x + b\cos x$ has an absolute maximum of $+R$ and absolute minimum of $-R$. The range is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.
• Solving Linear Trig Equations: Equations of the form $a\sin x + b\cos x = c$ simplify directly to $\sin(x + \alpha) = \frac{c}{R}$. Real solutions exist if and only if $|c| \le R = \sqrt{a^2 + b^2}$.
Worked Examples
Worked Example 1: Exact Value via Addition Formula (Paper 1 Style · 5 marks)
Find the exact value of $\tan\left(\frac{5\pi}{12}\right)$ in the form $a + \sqrt{b}$ where $a, b \in \mathbb{Z}^+$.
Official Markscheme Solution:
Decompose the fraction: $\frac{5\pi}{12} = \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4}$ [M1].
Apply the tangent addition formula: [M1]
$\tan\left(\frac{\pi}{6} + \frac{\pi}{4}\right) = \frac{\tan(\pi/6) + \tan(\pi/4)}{1 - \tan(\pi/6)\tan(\pi/4)}$.
Substitute exact standard values $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ and $\tan\left(\frac{\pi}{4}\right) = 1$: [A1]
$= \frac{\frac{1}{\sqrt{3}} + 1}{1 - \left(\frac{1}{\sqrt{3}}\right)(1)} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1}$.
Rationalize the denominator by multiplying by $(\sqrt{3} + 1)$: [M1]
$= \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$ [A1].
Here $a = 2, b = 3 \implies 2 + \sqrt{3}$ [AG].
Worked Example 2: Solving Equations by Factoring Double Angles (Paper 1 Style · 6 marks)
Solve the equation $\sin(2x) = \sqrt{2}\cos x$ for $0 \le x \le 2\pi$.
Official Markscheme Solution:
Substitute double angle formula $\sin(2x) = 2\sin x\cos x$: [M1]
$2\sin x\cos x = \sqrt{2}\cos x \implies 2\sin x\cos x - \sqrt{2}\cos x = 0$ [A1].
Factor out $\cos x$: [M1]
$\cos x (2\sin x - \sqrt{2}) = 0$.
• Branch 1: $\cos x = 0$
In $[0, 2\pi]$: $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ [A1].
• Branch 2: $2\sin x - \sqrt{2} = 0 \implies \sin x = \frac{\sqrt{2}}{2}$
In $[0, 2\pi]$, sine is positive in Quadrants 1 and 2: [M1]
$x = \frac{\pi}{4}$ and $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ [A1].
All solutions: $x \in \left\{\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{3\pi}{2}\right\}$.
Examiner Warning: Dividing across by $\cos x$ immediately loses the two solutions $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, costing 3 out of 6 marks!
Worked Example 3: Harmonic Linear Combination Equation (Paper 2 Style · 7 marks)
Consider the function $f(x) = 3\sin(2x) - 4\cos(2x)$ for $0 \le x \le \pi$.
(a) Express $f(x)$ in the form $R\sin(2x - \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. [3 marks]
(b) Hence find the coordinates of the maximum point on the graph of $f(x)$ on $[0, \pi]$. [2 marks]
(c) Solve the equation $f(x) = 2.5$ for $0 \le x \le \pi$. [2 marks]
Official Markscheme Solution:
(a) Harmonic Form: [3 marks]
Expand $R\sin(2x - \alpha) = R\cos\alpha\sin(2x) - R\sin\alpha\cos(2x)$ [M1].
Equating coefficients: $R\cos\alpha = 3$ and $R\sin\alpha = 4$.
$R = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$ [A1].
$\tan\alpha = \frac{4}{3} \implies \alpha = \arctan(4/3) \approx 0.9273$ rad [A1].
$f(x) = 5\sin(2x - 0.927)$ rad.
(b) Maximum Point: [2 marks]
Maximum occurs when $\sin(2x - 0.9273) = 1$:
$2x - 0.9273 = \frac{\pi}{2} \approx 1.5708 \implies 2x = 2.4981 \implies x \approx 1.25$ rad [A1].
Maximum height is $R = 5$. Max point: $(1.25, 5)$ [A1].
(c) Equation Solving: [2 marks]
$5\sin(2x - 0.9273) = 2.5 \implies \sin(2x - 0.9273) = 0.5$ [M1].
For $x \in [0, \pi]$, the argument $u = 2x - 0.9273 \in [-0.9273, 2\pi - 0.9273] \approx [-0.927, 5.356]$:
$u_1 = \frac{\pi}{6} \approx 0.5236 \implies 2x = 0.5236 + 0.9273 = 1.4509 \implies x_1 \approx 0.725$ rad [A1].
$u_2 = \frac{5\pi}{6} \approx 2.6180 \implies 2x = 2.6180 + 0.9273 = 3.5453 \implies x_2 \approx 1.77$ rad [A1].
Solutions: $x \approx 0.725, 1.77$ rad.
Worked Example 4: Triple Angle Proof (Paper 1 Style · 6 marks)
Prove that $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$ for all $\theta \in \mathbb{R}$.
Official Markscheme Solution:
Start with LHS: write $3\theta = 2\theta + \theta$ [M1].
$\cos(3\theta) = \cos(2\theta + \theta) = \cos(2\theta)\cos\theta - \sin(2\theta)\sin\theta$ [M1].
Substitute the double angle identities: [M1]
$\cos(2\theta) = 2\cos^2\theta - 1$
$\sin(2\theta) = 2\sin\theta\cos\theta$
$= (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta$ [A1]
$= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta$.
Substitute $\sin^2\theta = 1 - \cos^2\theta$ to express in powers of $\cos\theta$ only: [M1]
$= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta$
$= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta$ [A1]
$= 4\cos^3\theta - 3\cos\theta = \text{RHS}$ [AG].
Worked Example 5: Definite Integration Using Power Reduction (Paper 1 Style · 7 marks)
Evaluate the exact value of $\int_0^{\pi/4} \cos^4 x\,dx$.
Official Markscheme Solution:
Write $\cos^4 x = (\cos^2 x)^2$ and apply power reduction: [M1]
$\cos^2 x = \frac{1 + \cos(2x)}{2} \implies \cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2 = \frac{1 + 2\cos(2x) + \cos^2(2x)}{4}$ [M1].
Apply power reduction to $\cos^2(2x)$: [M1]
$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$ [A1].
Substitute back into the expression:
$\cos^4 x = \frac{1}{4} \left[ 1 + 2\cos(2x) + \frac{1 + \cos(4x)}{2} \right] = \frac{1}{4} \left[ \frac{3}{2} + 2\cos(2x) + \frac{1}{2}\cos(4x) \right]$
$= \frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)$ [A1].
Integrate term-by-term: [M1]
$\int_0^{\pi/4} \left( \frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x) \right)\,dx = \left[ \frac{3}{8}x + \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) \right]_0^{\pi/4}$ [M1].
Evaluate at upper limit $x = \frac{\pi}{4}$:
$= \frac{3}{8}\left(\frac{\pi}{4}\right) + \frac{1}{4}\sin\left(\frac{\pi}{2}\right) + \frac{1}{32}\sin(\pi) = \frac{3\pi}{32} + \frac{1}{4}(1) + 0 = \frac{3\pi}{32} + \frac{1}{4}$ [A1].
At lower limit $x = 0$, all terms vanish.
Exact value: $\frac{3\pi + 8}{32}$ [A1].
Practice Exercises (10 Tiered Questions)
Work through these 10 examination-calibrated exercises ranging from core foundational skills to advanced Paper 1 and Paper 2 problems. Expand each exercise to view the full IB markscheme and examiner advice.
Given that $\sin\theta = \frac{3}{5}$ and $\theta$ is an acute angle, find the exact value of $\sin(2\theta)$ and $\cos(2\theta)$.
Show Full Markscheme
Since $\theta$ is acute, $\cos\theta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$ [M1].
• $\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right) = \frac{24}{25}$ [A1].
• $\cos(2\theta) = \cos^2\theta - \sin^2\theta = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16 - 9}{25} = \frac{7}{25}$ [A1].
Use the compound angle formula for $\tan(A - B)$ to find the exact value of $\tan(15^\circ)$ in simplest surd form.
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$\tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ\tan 30^\circ}$ [M1].
Substitute $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$:
$= \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ [A1].
Rationalize: $\frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$ [A1].
Express $\cos^2(3x)$ in terms of the first power of a cosine function.
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Apply identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ with $\theta = 3x$ [M1]:
$\cos^2(3x) = \frac{1 + \cos(6x)}{2} = \frac{1}{2} + \frac{1}{2}\cos(6x)$ [A1].
Solve $\cos(2x) + \cos x = 0$ for $0 \le x \le 2\pi$.
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Substitute $\cos(2x) = 2\cos^2 x - 1$: [M1]
$2\cos^2 x + \cos x - 1 = 0$ [A1].
Factor the quadratic: $(2\cos x - 1)(\cos x + 1) = 0$ [M1].
• $\cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}$ [A1].
• $\cos x = -1 \implies x = \pi$ [A1].
Solutions: $x \in \left\{\frac{\pi}{3}, \pi, \frac{5\pi}{3}\right\}$.
Let $\alpha$ and $\beta$ be acute angles such that $\tan\alpha = \frac{1}{2}$ and $\tan\beta = \frac{1}{3}$. Show that $\alpha + \beta = \frac{\pi}{4}$.
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Apply tangent addition formula: [M1]
$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{1/2 + 1/3}{1 - (1/2)(1/3)}$ [M1]
$= \frac{5/6}{1 - 1/6} = \frac{5/6}{5/6} = 1$ [A1].
Since $\alpha, \beta$ are acute ($0 < \alpha, \beta < \frac{\pi}{2}$) and $\tan\alpha, \tan\beta < 1$, we have $0 < \alpha, \beta < \frac{\pi}{4}$, so $0 < \alpha + \beta < \frac{\pi}{2}$ [R1].
Therefore, $\alpha + \beta = \arctan(1) = \frac{\pi}{4}$ [AG].
Express $\sin x - \sqrt{3}\cos x$ in the form $R\sin(x - \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$.
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Expand $R\sin(x - \alpha) = R\cos\alpha\sin x - R\sin\alpha\cos x$ [M1].
Equating coefficients: $R\cos\alpha = 1$ and $R\sin\alpha = \sqrt{3}$.
$R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$ [A1].
$\tan\alpha = \frac{\sqrt{3}}{1} = \sqrt{3} \implies \alpha = \frac{\pi}{3}$ [A1].
Result: $2\sin\left(x - \frac{\pi}{3}\right)$.
Solve $\tan(2x) = 4\tan x$ for $0 < x < \pi$.
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Substitute $\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}$: [M1]
$\frac{2\tan x}{1 - \tan^2 x} = 4\tan x \implies 2\tan x = 4\tan x(1 - \tan^2 x)$
$2\tan x [ 1 - 2(1 - \tan^2 x) ] = 0 \implies 2\tan x (2\tan^2 x - 1) = 0$ [M1].
• $\tan x = 0$: In $(0, \pi)$, no solutions (endpoints excluded).
• $2\tan^2 x - 1 = 0 \implies \tan^2 x = \frac{1}{2} \implies \tan x = \pm\frac{1}{\sqrt{2}}$ [A1].
$\tan x = \frac{1}{\sqrt{2}} \implies x = \arctan(1/\sqrt{2}) \approx 0.615$ rad [A1].
$\tan x = -\frac{1}{\sqrt{2}} \implies x = \pi - 0.615 \approx 2.53$ rad [A1].
Solutions: $x \approx 0.615, 2.53$ rad.
Show that $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$.
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$\sin(3\theta) = \sin(2\theta + \theta) = \sin(2\theta)\cos\theta + \cos(2\theta)\sin\theta$ [M1].
Substitute $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\cos(2\theta) = 1 - 2\sin^2\theta$: [M1]
$= (2\sin\theta\cos\theta)\cos\theta + (1 - 2\sin^2\theta)\sin\theta$
$= 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta$ [A1].
Replace $\cos^2\theta = 1 - \sin^2\theta$: [M1]
$= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta = 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta$
$= 3\sin\theta - 4\sin^3\theta$ [A1][AG].
Find the maximum value of $g(x) = \frac{12}{5\sin x + 12\cos x + 15}$ and state the value of $x \in [0, 2\pi]$ at which it occurs.
Show Full Markscheme
Convert denominator into harmonic form: $5\sin x + 12\cos x = R\sin(x + \alpha)$ [M1].
$R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13$ [A1].
$\tan\alpha = \frac{12}{5} \implies \alpha = \arctan(2.4) \approx 1.176$ rad.
Denominator is $13\sin(x + 1.176) + 15$.
The fraction $g(x)$ is maximized when its denominator is minimized [R1].
Minimum of denominator occurs when $\sin(x + 1.176) = -1$ [M1]:
$\text{Denominator}_{\min} = 13(-1) + 15 = 2$.
Maximum value of $g(x) = \frac{12}{2} = 6$ [A1].
Occurs at $x + 1.176 = \frac{3\pi}{2} \approx 4.7124 \implies x \approx 4.7124 - 1.176 = 3.54$ rad [A1].
Let $t = \tan\left(\frac{\theta}{2}\right)$. Prove that $\sin\theta = \frac{2t}{1 + t^2}$ and $\cos\theta = \frac{1 - t^2}{1 + t^2}$.
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Consider $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ [M1].
Divide and multiply by $\cos^2(\theta/2) + \sin^2(\theta/2) = 1$:
$\sin\theta = \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos^2(\theta/2) + \sin^2(\theta/2)}$ [M1].
Divide numerator and denominator by $\cos^2(\theta/2)$: [M1]
$= \frac{2\tan(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{2t}{1 + t^2}$ [A1][AG].
Similarly, $\cos\theta = \cos^2(\theta/2) - \sin^2(\theta/2) = \frac{\cos^2(\theta/2) - \sin^2(\theta/2)}{\cos^2(\theta/2) + \sin^2(\theta/2)}$ [M1].
Divide by $\cos^2(\theta/2)$:
$= \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - t^2}{1 + t^2}$ [A1][AG].
IB Formula Booklet References (Topic 3)
Syllabus 3.9: Compound & Double Angle Formulas
The official IB DP Mathematics AA Formula Booklet provides:
- $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
- $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
- $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$
- $\sin(2\theta) = 2\sin\theta\cos\theta$
- $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
- $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$
Key Formulas & Identities
1. Power-Reduction Identities (Crucial for Integration)
- $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$
- $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$
- $\sin^2\theta\cos^2\theta = \frac{1}{4}\sin^2(2\theta) = \frac{1 - \cos(4\theta)}{8}$
2. Harmonic Synthesis Form
- $a\sin x + b\cos x = R\sin(x + \alpha)$ where $R = \sqrt{a^2 + b^2}, \quad \cos\alpha = \frac{a}{R}, \quad \sin\alpha = \frac{b}{R}$
- $a\cos x + b\sin x = R\cos(x - \beta)$ where $R = \sqrt{a^2 + b^2}, \quad \cos\beta = \frac{a}{R}, \quad \sin\beta = \frac{b}{R}$
- Range: $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$; Maximum $= +R$, Minimum $= -R$.
3. Triple Angle and Half-Angle $t$-Formulas
- $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$
- $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$
- $t = \tan(\theta/2) \implies \sin\theta = \frac{2t}{1 + t^2}, \quad \cos\theta = \frac{1 - t^2}{1 + t^2}, \quad \tan\theta = \frac{2t}{1 - t^2}$
Common Exam Pitfalls
Pitfall 1: Sign Reversal in Cosine Addition
Remember that $\cos(A + B) = \cos A\cos B - \sin A\sin B$. The plus sign in the argument becomes a MINUS in the expansion!
Pitfall 2: Dividing by Variables and Losing Solutions
Never divide an equation like $\sin(2x) = \cos x$ by $\cos x$! This discards all solutions where $\cos x = 0$ (such as $x = \frac{\pi}{2}, \frac{3\pi}{2}$). Always factor: $\cos x (2\sin x - 1) = 0$.
Pitfall 3: Not Shifting the Domain for Compound Arguments
When solving $\sin(2x - \alpha) = c$ for $x \in [0, 2\pi]$, remember that the argument $u = 2x - \alpha$ spans $[-\alpha, 4\pi - \alpha]$. If you search only within $[0, 2\pi]$, you will miss valid solutions!
Topic Diagnostic Test
Complete this timed diagnostic test to verify your conceptual command of compound angles, double angles, and harmonic forms.
Section A: Diagnostic Multiple Choice Questions
If $\sin\theta = \frac{1}{3}$, what is the value of $\cos(2\theta)$?
What is the maximum value of $f(x) = 5\sin x - 12\cos x$?
How many real solutions exist for $\sin(2x) = \cos x$ in the interval $[0, 2\pi]$?
Which expression is identical to $\cos^2(4x)$?
Section B: Examination-Style Free Response Questions
Let $f(x) = \cos(2x) - 3\cos x + 2$ for $0 \le x \le 2\pi$.
(a) Show that $f(x) = 2\cos^2 x - 3\cos x + 1$. [2 marks]
(b) Hence solve $f(x) = 0$ for $x \in [0, 2\pi]$. [4 marks]
Show Full Markscheme
(a) Quadratic Form: [2 marks]
Substitute $\cos(2x) = 2\cos^2 x - 1$: [M1]
$f(x) = (2\cos^2 x - 1) - 3\cos x + 2 = 2\cos^2 x - 3\cos x + 1$ [A1][AG].
(b) Solving Equation: [4 marks]
Factorize the quadratic in $\cos x$: [M1]
$(2\cos x - 1)(\cos x - 1) = 0$.
• $\cos x = \frac{1}{2}$: In $[0, 2\pi]$, reference angle is $\frac{\pi}{3}$.
Quadrants 1 and 4: $x = \frac{\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ [A1][A1].
• $\cos x = 1$: In $[0, 2\pi]$, $x = 0$ and $x = 2\pi$ [A1].
Solutions: $x \in \left\{0, \frac{\pi}{3}, \frac{5\pi}{3}, 2\pi\right\}$ [A1].
The depth of water, $h$ meters, at an estuary entrance is modeled by $h(t) = 4\sin(0.5t) + 3\cos(0.5t) + 8$ where $t$ is the time in hours after midnight.
(a) Express $4\sin(0.5t) + 3\cos(0.5t)$ in the form $R\sin(0.5t + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. [3 marks]
(b) Find the maximum water depth and the earliest time after midnight at which it occurs. [3 marks]
(c) A ferry requires a minimum depth of 10.5 meters to enter. Determine the length of time in the first 12-hour period that the ferry can enter. [2 marks]
Show Full Markscheme
(a) Harmonic Form: [3 marks]
$R = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ [A1].
$R\sin(0.5t + \alpha) = R\cos\alpha\sin(0.5t) + R\sin\alpha\cos(0.5t)$
$R\cos\alpha = 4, R\sin\alpha = 3 \implies \tan\alpha = \frac{3}{4} \implies \alpha = \arctan(0.75) \approx 0.6435$ rad [A1].
$h(t) = 5\sin(0.5t + 0.6435) + 8$ [A1].
(b) Maximum Depth and Time: [3 marks]
Maximum depth: $h_{\max} = 5(1) + 8 = 13$ meters [A1].
Occurs when $\sin(0.5t + 0.6435) = 1 \implies 0.5t + 0.6435 = \frac{\pi}{2} \approx 1.5708$ [M1].
$0.5t = 1.5708 - 0.6435 = 0.9273 \implies t \approx 1.85$ hours after midnight (approx 01:51 AM) [A1].
(c) Duration Ferry Can Enter: [2 marks]
Set $h(t) \ge 10.5 \implies 5\sin(0.5t + 0.6435) + 8 \ge 10.5$
$\sin(0.5t + 0.6435) \ge \frac{2.5}{5} = 0.5$ [M1].
The sine function is $\ge 0.5$ over the interval $\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$, which has angular width $\frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$ rad.
Set $0.5 \Delta t = \frac{2\pi}{3} \implies \Delta t = \frac{4\pi}{3} \approx 4.19$ hours [A1].
The ferry can enter for approximately 4.19 hours (4 hours 11 minutes).