Trigonometry

1. Overview

Trigonometry in IGCSE Additional Mathematics (0606) extends your knowledge of sine, cosine, and tangent to encompass all angles, measured in both degrees and radians. You'll learn about the reciprocal trigonometric functions (secant, cosecant, and cotangent), trigonometric identities, and how to solve trigonometric equations. A key focus is understanding and manipulating trigonometric functions algebraically, often without a calculator (especially in Paper 1). This topic is crucial not only for its own sake, but also as a foundation for calculus involving trigonometric functions. Expect to see questions involving proving identities, solving equations, and analysing trigonometric graphs.

Key Definitions

• Cosecant ($\text{cosec } \theta$): The reciprocal of sine; $\text{cosec } \theta = \frac{1}{\sin \theta}$.
• Secant ($\sec \theta$): The reciprocal of cosine; $\sec \theta = \frac{1}{\cos \theta}$.
• Cotangent ($\cot \theta$): The reciprocal of tangent; $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$.
• Amplitude ($a$): The maximum displacement from the equilibrium (mid-line) of a sine or cosine graph.
• Period: The distance (in degrees or radians) taken for the graph to complete one full cycle.
• Principal Value: The specific solution to a trigonometric equation returned by a calculator (within a restricted range).

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Core Content

3.1 The Six Trigonometric Functions

You must be comfortable working with all six functions across all four quadrants (CAST diagram).

• Note: $\sec \theta$ is undefined where $\cos \theta = 0$. $\text{cosec } \theta$ is undefined where $\sin \theta = 0$.

3.2 Trigonometric Graphs

For the functions $y = a \sin(bx) + c$, $y = a \cos(bx) + c$, and $y = a \tan(bx) + c$:

• Amplitude ($a$): Only applies to sin/cos. If $a$ is negative, the graph is reflected in the x-axis.
• Period ($P$):
  • For sin/cos: $P = \frac{360^\circ}{b}$ or $\frac{2\pi}{b}$
  • For tan: $P = \frac{180^\circ}{b}$ or $\frac{\pi}{b}$
• Vertical Shift ($c$): Moves the entire graph up or down.
• Asymptotes: For $y = a \tan(bx) + c$, the asymptotes occur where $bx = 90^\circ, 270^\circ, \dots$

Worked example 1 — Graph Properties

State the amplitude and period (in radians) of $y = 4 \cos(\frac{1}{2}x) - 3$.

1. Amplitude: The coefficient $a = 4$.
   Amplitude = $4$
   The amplitude is the absolute value of the coefficient of the cosine function.

2. Period: The coefficient $b = \frac{1}{2}$.
   $Period = \frac{2\pi}{b}$
   The period formula for cosine functions.

3. Substitute $b = \frac{1}{2}$ into the period formula:
   $Period = \frac{2\pi}{1/2} = 4\pi$

4. Final Answer: Amplitude = $4$, Period = $\boxed{4\pi}$.

3.3 Trigonometric Identities

These are essential for simplifying expressions and proving "Show that" questions.

Key Trigonometric Identities:

• $\sin^2 A + \cos^2 A = 1$ (Given on formula sheet)
• $\sec^2 A = 1 + \tan^2 A$ (Given on formula sheet)
• $\text{cosec}^2 A = 1 + \cot^2 A$ (Given on formula sheet)

Important Trigonometric Ratios (Memorise):

• $\tan A = \frac{\sin A}{\cos A}$
• $\cot A = \frac{\cos A}{\sin A}$
• $\sec A = \frac{1}{\cos A}$
• $\text{cosec } A = \frac{1}{\sin A}$

Worked Example 2 — Proving an Identity

Prove that $\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta} = 2 \text{cosec}^2 \theta$.

1. Start with the left-hand side of the equation:
   $\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}$

2. Find a common denominator and add the fractions:
   $\frac{(1 + \cos \theta) + (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$

3. Simplify the numerator:
   $\frac{2}{(1 - \cos \theta)(1 + \cos \theta)}$

4. Expand the denominator:
   $\frac{2}{1 - \cos^2 \theta}$

5. Use the identity $\sin^2 \theta + \cos^2 \theta = 1$, so $1 - \cos^2 \theta = \sin^2 \theta$:
   $\frac{2}{\sin^2 \theta}$

6. Rewrite the expression:
   $2 \left(\frac{1}{\sin^2 \theta}\right)$

7. Use the identity $\text{cosec } \theta = \frac{1}{\sin \theta}$:
   $2 \text{cosec}^2 \theta$

8. Conclusion: LHS = RHS. $\boxed{\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta} = 2 \text{cosec}^2 \theta}$

Worked Example 3 — Proving an Identity (Advanced)

Prove the identity: $\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x$

1. Start with the left-hand side:
   $\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x}$

2. Combine the fractions using a common denominator:
   $\frac{\sin^2 x + (1 + \cos x)^2}{\sin x (1 + \cos x)}$

3. Expand the numerator:
   $\frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x (1 + \cos x)}$

4. Rearrange the terms in the numerator:
   $\frac{(\sin^2 x + \cos^2 x) + 1 + 2\cos x}{\sin x (1 + \cos x)}$

5. Apply the identity $\sin^2 x + \cos^2 x = 1$:
   $\frac{1 + 1 + 2\cos x}{\sin x (1 + \cos x)}$

6. Simplify the numerator:
   $\frac{2 + 2\cos x}{\sin x (1 + \cos x)}$

7. $\frac{2(1 + \cos x)}{\sin x (1 + \cos x)}$
   Factor out a 2 from the numerator.

8. Cancel the common factor of $(1 + \cos x)$:
   $\frac{2}{\sin x}$

9. Use the identity $\csc x = \frac{1}{\sin x}$:
   $2 \csc x$

10. Conclusion: LHS = RHS. $\boxed{\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x}$

3.4 Solving Trigonometric Equations

Always check the required domain ($0^\circ \le x \le 360^\circ$ or $0 \le x \le 2\pi$).

Worked Example 4 — Solving Equations
Solve $2 \cos^2 x + 3 \sin x = 0$ for $0^\circ \le x \le 360^\circ$.

1. Substitute Identity: Use $\cos^2 x = 1 - \sin^2 x$ to get the equation in terms of $\sin x$.
   $2(1 - \sin^2 x) + 3 \sin x = 0$

2. Expand and Rearrange: $2 - 2 \sin^2 x + 3 \sin x = 0 \implies 2 \sin^2 x - 3 \sin x - 2 = 0$.

3. Factorise: Let $u = \sin x$. $2u^2 - 3u - 2 = 0 \implies (2u + 1)(u - 2) = 0$.
   Substituting $u = \sin x$ simplifies the factorisation.

4. Solve for $\sin x$:

   • $\sin x = -\frac{1}{2}$
   • $\sin x = 2$ (No solution, as $-1 \le \sin x \le 1$)

5. Find Angles:

   • Reference angle $\alpha = \sin^{-1}(\frac{1}{2}) = 30^\circ$.
   • Sine is negative in Quadrants III and IV.
   • $x = 180^\circ + 30^\circ = 210^\circ$
   • $x = 360^\circ - 30^\circ = 330^\circ$

6. Final Answer: $\boxed{x = 210^\circ, 330^\circ}$

Worked Example 5 — Solving Equations (Radians)

Solve $3 \tan(2x) = 1$ for $0 \le x \le \pi$.

1. Isolate $\tan(2x)$: $\tan(2x) = \frac{1}{3}$
   Divide both sides by 3.

2. Find the principal value: $2x = \tan^{-1}(\frac{1}{3})$

3. Calculate the principal value (in radians): $2x \approx 0.32175$ radians
   Ensure your calculator is in radian mode.

4. Find all solutions for $2x$ in the interval $[0, 2\pi]$:
   Since $0 \le x \le \pi$, we need all angles $2x$ in the interval $[0, 2\pi]$. The tangent function repeats every $\pi$, so add $\pi$ to the principal value.

   • $2x_1 = 0.32175$
   • $2x_2 = \pi + 0.32175 \approx 3.46334$

5. Solve for $x$:

   • $x_1 = \frac{0.32175}{2} \approx 0.16088$
   • $x_2 = \frac{3.46334}{2} \approx 1.73167$

6. Final Answer: $x \approx 0.161, 1.732$ (to 3 decimal places). $\boxed{x = 0.161, 1.732}$

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In More Depth

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

Equation	Use Case	Notes
$\tan A = \frac{\sin A}{\cos A}$	Converting tan to sin/cos	Memorise
$\sin^2 A + \cos^2 A = 1$	Main identity for quadratic trig equations	Given on formula sheet
$\sec^2 A = 1 + \tan^2 A$	Linking secant and tangent	Given on formula sheet
$\text{cosec}^2 A = 1 + \cot^2 A$	Linking cosecant and cotangent	Given on formula sheet
$Period = \frac{360^\circ}{b}$ or $Period = \frac{2\pi}{b}$	Finding the width of a sin/cos cycle	Memorise
$Period = \frac{180^\circ}{b}$ or $Period = \frac{\pi}{b}$	Finding the width of a tan cycle	Memorise

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Common Mistakes to Avoid

• ❌ Wrong: Dividing both sides of an equation by $\sin x$ (e.g., $2 \sin x \cos x = \sin x \implies 2 \cos x = 1$).
• ✅ Right: Factorise instead: $\sin x (2 \cos x - 1) = 0$. Dividing by $\sin x$ loses valid solutions where $\sin x = 0$.
• ❌ Wrong: Forgetting to apply the chain rule when differentiating trig (e.g., $\frac{d}{dx}(\tan 3x) = \sec^2 3x$).
• ✅ Right: $\frac{d}{dx}(\tan 3x) = 3 \sec^2 3x$.
• ❌ Wrong: Giving answers in degrees when the domain is given in radians (e.g., $0 \le x \le 2\pi$).
• ✅ Right: Always check the unit required by the domain. Set your calculator to RAD mode for radian questions.
• ❌ Wrong: Providing a decimal answer for $\sin 60^\circ$ in Paper 1.
• ✅ Right: Use exact values: $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
• ❌ Wrong: Incorrectly applying the period formula, e.g., using $Period = \frac{2\pi}{b}$ for $y = a \tan(bx)$.
• ✅ Right: Remember that the period for tangent functions is $Period = \frac{\pi}{b}$.
• ❌ Wrong: Forgetting the $\pm$ when taking the square root in an identity problem.
• ✅ Right: When solving $\sin^2 x = \frac{1}{4}$, then $\sin x = \pm \frac{1}{2}$.
• ❌ Wrong: Not showing all steps when rationalising the denominator of a surd.
• ✅ Right: Show every step when multiplying by the conjugate.

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Exam Tips

• "Show That" Questions: Never work with both sides of the identity simultaneously. Start with the most complex side and manipulate it until it matches the other side. State the identity used at each step.
• The "b" Value: When finding the period, remember that $b$ is the coefficient of $x$. If you see $\cos(\frac{x}{3})$, then $b = \frac{1}{3}$, making the period $360 \div \frac{1}{3} = 1080^\circ$.
• Eliminating Trigonometry: In questions asking for a relationship between $x$ and $y$ (eliminating the parameter), you must use identities so that no $\sin, \cos, \tan$ remain in the final equation.
• Command Words: "Find all values of $x$ in the range..." means you must use the CAST diagram or graph to find every solution in the domain, not just the principal value.
• Formula Sheet: Familiarise yourself with the provided list. The Pythagorean identities are provided, but the definitions of $\sec, \text{cosec},$ and $\cot$ are not. You must memorize them.
• Exact Values (Paper 1): Be prepared to work with exact trigonometric values (e.g., $\sin 45^\circ = \frac{\sqrt{2}}{2}$) without a calculator.
• Calculator Mode (Paper 2): Double-check that your calculator is in the correct mode (degrees or radians) before starting any calculation. A wrong mode will lead to incorrect answers.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [9 marks]

Question:

(a) Solve the equation $2 \sin^2 x + 5 \cos x = 4$ for $0^\circ \le x \le 360^\circ$. [5]

(b) Show that $\frac{\cos \theta}{1 - \sin \theta} - \frac{\cos \theta}{1 + \sin \theta} = 2 \tan \theta$. [4]

Worked Solution:

(a)

1. Replace $\sin^2 x$ with $1 - \cos^2 x$:
   $2(1 - \cos^2 x) + 5 \cos x = 4$
   Using the identity $\sin^2 x + \cos^2 x = 1$

2. Expand and rearrange into a quadratic equation in $\cos x$:
   $2 - 2\cos^2 x + 5 \cos x - 4 = 0$
   $2\cos^2 x - 5 \cos x + 2 = 0$

3. Factorise the quadratic equation:
   $(2\cos x - 1)(\cos x - 2) = 0$

4. Solve for $\cos x$:
   $\cos x = \frac{1}{2}$ or $\cos x = 2$
   Note that $\cos x = 2$ has no solutions since $-1 \le \cos x \le 1$

5. Find the solutions for $x$ in the given domain:
   $x = 60^\circ, 300^\circ$
   Considering the quadrants where cosine is positive

Final Answer: $\boxed{x = 60^\circ, 300^\circ}$

How to earn full marks: Remember to use the trigonometric identity correctly and find all solutions within the specified range, considering the quadrants.

(b)

1. Combine the fractions on the left-hand side:
   $\frac{\cos \theta (1 + \sin \theta) - \cos \theta (1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}$
   Finding a common denominator

2. Simplify the numerator:
   $\frac{\cos \theta + \cos \theta \sin \theta - \cos \theta + \cos \theta \sin \theta}{1 - \sin^2 \theta}$
   $\frac{2 \cos \theta \sin \theta}{1 - \sin^2 \theta}$

3. Use the identity $\cos^2 \theta = 1 - \sin^2 \theta$:
   $\frac{2 \cos \theta \sin \theta}{\cos^2 \theta}$

4. Simplify the fraction:
   $\frac{2 \sin \theta}{\cos \theta} = 2 \tan \theta$
   Using the definition $\tan \theta = \frac{\sin \theta}{\cos \theta}$

Final Answer: $\boxed{\frac{\cos \theta}{1 - \sin \theta} - \frac{\cos \theta}{1 + \sin \theta} = 2 \tan \theta}$

How to earn full marks: Show each step clearly, especially when simplifying the fractions and applying trigonometric identities.

Common Pitfall: In part (a), remember to check if all solutions for $\cos x$ are valid within the range of the cosine function, which is -1 to 1. In part (b), make sure you correctly combine the fractions, paying close attention to the signs.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

(a) Given that $\tan x = \frac{8}{15}$ and $180^\circ < x < 270^\circ$, find the exact value of $\sin x$ and $\cos x$. [4]

(b) Solve the equation $2 \cot^2 y - 7 \operatorname{cosec} y = -5$ for $0^\circ < y < 360^\circ$. [4]

Worked Solution:

(a)

1. Since $180^\circ < x < 270^\circ$, $x$ is in the third quadrant, where both $\sin x$ and $\cos x$ are negative.

2. Use the identity $1 + \tan^2 x = \sec^2 x$ to find $\sec x$:
   $1 + \left(\frac{8}{15}\right)^2 = \sec^2 x$
   $1 + \frac{64}{225} = \sec^2 x$
   $\sec^2 x = \frac{289}{225}$
   $\sec x = \pm \frac{17}{15}$
   Since $x$ is in the third quadrant, $\cos x$ is negative, hence $\sec x$ is negative:
   $\sec x = -\frac{17}{15}$

3. Find $\cos x$:
   $\cos x = \frac{1}{\sec x} = -\frac{15}{17}$

4. Use $\tan x = \frac{\sin x}{\cos x}$ to find $\sin x$:
   $\sin x = \tan x \cdot \cos x = \frac{8}{15} \cdot \left(-\frac{15}{17}\right) = -\frac{8}{17}$

Final Answer: $\boxed{\sin x = -\frac{8}{17}, \cos x = -\frac{15}{17}}$

How to earn full marks: Clearly state the quadrant and ensure the signs of $\sin x$ and $\cos x$ are correct based on the quadrant.

(b)

1. Use the identity $\operatorname{cosec}^2 y = 1 + \cot^2 y$ to express the equation in terms of $\operatorname{cosec} y$:
   $2 (\operatorname{cosec}^2 y - 1) - 7 \operatorname{cosec} y = -5$
   $2 \operatorname{cosec}^2 y - 2 - 7 \operatorname{cosec} y = -5$
   $2 \operatorname{cosec}^2 y - 7 \operatorname{cosec} y + 3 = 0$

2. Factorise the quadratic equation:
   $(2 \operatorname{cosec} y - 1)(\operatorname{cosec} y - 3) = 0$

3. Solve for $\operatorname{cosec} y$:
   $\operatorname{cosec} y = \frac{1}{2}$ or $\operatorname{cosec} y = 3$

4. Solve for $y$:
   $\sin y = 2$ or $\sin y = \frac{1}{3}$
   Since $\sin y = 2$ has no solutions, we only consider $\sin y = \frac{1}{3}$.
   $y = \arcsin\left(\frac{1}{3}\right)$ and $y = 180^\circ - \arcsin\left(\frac{1}{3}\right)$
   $y \approx 19.47^\circ$ and $y \approx 160.53^\circ$

Final Answer: $\boxed{y = 19.47^\circ, 160.53^\circ}$

How to earn full marks: Remember to use the correct trigonometric identity and find all possible solutions within the given range.

Common Pitfall: In part (a), remember that the quadrant determines the signs of $\sin x$ and $\cos x$. In part (b), always check if the solutions for $\operatorname{cosec} y$ (or $\sin y$) are valid. The range of $\sin y$ is -1 to 1.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [10 marks]

Question:

(a) The function $f(x) = 2 \sin(3x) + 1$ is defined for $0^\circ \le x \le 180^\circ$.

(i) State the amplitude and period of $f(x)$. [2]
(ii) Sketch the graph of $y = f(x)$ on the axes below, showing the coordinates of the maximum and minimum points, and the points where the graph intersects the $x$-axis. [4]

(b) Solve the equation $3 \cos(x - 45^\circ) = 1$ for $0^\circ \le x \le 360^\circ$. [4]

Worked Solution:

(a)
(i)

1. Identify amplitude:
   Amplitude $= |2| = 2$
   The amplitude is the absolute value of the coefficient of the sine function.

2. Identify period:
   Period $= \frac{360^\circ}{3} = 120^\circ$
   The period is $360^\circ$ divided by the coefficient of $x$.

Final Answer: $\boxed{\text{Amplitude} = 2, \text{Period} = 120^\circ}$

How to earn full marks: State the amplitude and period clearly, showing the calculation for the period.

(ii)

1. Find maximum points: The maximum value of $2\sin(3x)$ is 2, so the maximum value of $f(x)$ is $2+1=3$. This occurs when $3x = 90^\circ, 450^\circ$, so $x = 30^\circ, 150^\circ$. Coordinates are $(30,3)$ and $(150,3)$

2. Find minimum points: The minimum value of $2\sin(3x)$ is -2, so the minimum value of $f(x)$ is $-2+1=-1$. This occurs when $3x = 270^\circ$, so $x = 90^\circ$. Coordinate is $(90,-1)$

3. Find x-intercepts: $2\sin(3x) + 1 = 0$, so $\sin(3x) = -\frac{1}{2}$. $3x = 210^\circ, 330^\circ$ (the next value, $570^\circ$, is outside $0^\circ \le 3x \le 540^\circ$). Hence $x=70^\circ, 110^\circ$. Coordinates are $(70,0)$ and $(110,0)$.

4. Sketch the graph with key points identified.

Final Answer: $\boxed{\text{See graph for sketch and key points}}$

How to earn full marks: Label all key points on the graph accurately, including maximum and minimum points, and x-intercepts.

(b)

1. Isolate the cosine function:
   $\cos(x - 45^\circ) = \frac{1}{3}$

2. Find the principal value:
   $x - 45^\circ = \arccos(\frac{1}{3}) = 70.53^\circ$ (to 2 dp)

3. Since $\cos$ has a period of $360^\circ$, find the other solution in the range $-45^\circ \le x - 45^\circ \le 315^\circ$:
   $x - 45^\circ = 360^\circ - 70.53^\circ = 289.47^\circ$

4. Solve for $x$:
   $x = 70.53^\circ + 45^\circ = 115.53^\circ$
   $x = 289.47^\circ + 45^\circ = 334.47^\circ$

Final Answer: $\boxed{x = 115.53^\circ, 334.47^\circ}$

How to earn full marks: Show all steps in solving the equation and find all solutions within the specified range.

Common Pitfall: When finding the period, make sure you divide $360^\circ$ (or $2\pi$ radians) by the coefficient of $x$ inside the trigonometric function. When solving trigonometric equations, remember to find all solutions within the specified range, using the symmetry and periodicity of the functions.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [11 marks]

Question:

(a) Show that $\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = \frac{2 \sin x}{\cos x}$. [3]

(b) Hence, solve the equation $\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 3 \cos x$ for $0^\circ < x < 360^\circ$. Give your answers to 1 decimal place. [4]

(c) A triangle $ABC$ has $AB = 7$ cm, $BC = 9$ cm, and angle $BAC = 35^\circ$. Show that only one triangle is possible and find its area. [4]

Worked Solution:

(a)

1. Combine the fractions on the left-hand side:
   $\frac{\cos x (1 + \sin x) - \cos x (1 - \sin x)}{(1 - \sin x)(1 + \sin x)}$
   Finding a common denominator

2. Expand and simplify the numerator:
   $\frac{\cos x + \cos x \sin x - \cos x + \cos x \sin x}{1 - \sin^2 x}$
   $\frac{2 \cos x \sin x}{1 - \sin^2 x}$

3. Use the identity $\cos^2 x = 1 - \sin^2 x$:
   $\frac{2 \cos x \sin x}{\cos^2 x}$

4. Simplify:
   $\frac{2 \sin x}{\cos x}$

Final Answer: $\boxed{\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = \frac{2 \sin x}{\cos x}}$

How to earn full marks: Show each step of the simplification process clearly, including the use of trigonometric identities.

(b)

1. Substitute the result from part (a) into the equation:
   $\frac{2 \sin x}{\cos x} = 3 \cos x$
   $2 \tan x = 3 \cos x$

2. Multiply both sides by $\cos x$ and use $\cos^2 x = 1 - \sin^2 x$ to form a quadratic in $\sin x$:
   $2 \sin x = 3 \cos^2 x$
   $2 \sin x = 3 (1 - \sin^2 x)$
   $3 \sin^2 x + 2 \sin x - 3 = 0$

3. Solve for $\sin x$ using the quadratic formula:
   $\sin x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-3)}}{2(3)} = \frac{-2 \pm \sqrt{40}}{6} = \frac{-1 \pm \sqrt{10}}{3}$
   $\sin x = 0.72076$ or $\sin x = -1.3874$
   Since $-1 \le \sin x \le 1$, we only consider $\sin x = 0.72076$

4. Find the solutions for $x$ in the given domain:
   $x = \arcsin(0.72076) = 46.1^\circ$
   $x = 180^\circ - 46.1^\circ = 133.9^\circ$

Final Answer: $\boxed{x = 46.1^\circ, 133.9^\circ}$ (to 1 dp)

How to earn full marks: Remember to use the quadratic formula correctly and check for extraneous solutions, giving your answers to the specified decimal place.

(c)

1. Use the sine rule to find the possible values of angle $ACB$:
   $\frac{\sin C}{7} = \frac{\sin 35^\circ}{9}$
   $\sin C = \frac{7 \sin 35^\circ}{9} = \frac{7(0.5736)}{9} = 0.4461$

2. Find the two possible values for angle $C$:
   $C_1 = \arcsin(0.4461) = 26.49^\circ$
   $C_2 = 180^\circ - 26.49^\circ = 153.51^\circ$

3. Find the corresponding values for angle $B$:
   $B_1 = 180^\circ - (35^\circ + 26.49^\circ) = 118.51^\circ$
   $B_2 = 180^\circ - (35^\circ + 153.51^\circ) = -8.51^\circ$ (invalid)

4. Because $B_2$ is invalid, there is only one possible triangle.
   $B = 180 - (35 + 26.49) = 118.51$
   Area $= \frac{1}{2} a c \sin B = \frac{1}{2} (7)(9) \sin(118.51) = \frac{1}{2} (7)(9) (0.8787) = 27.68$

Final Answer: $\boxed{27.7 \text{ cm}^2}$

How to earn full marks: Use the sine rule correctly, find both possible angles, and check for valid triangles before calculating the area.

Common Pitfall: In part (a), remember to use trigonometric identities to simplify the expression. In part (b), be careful when solving the quadratic equation for $\sin x$ and check for extraneous solutions. In part (c), remember to check if both possible values for the angle lead to valid triangles (angles must be positive and sum to less than 180 degrees).