Trigonometry : Sine & Cosine.

Grade 12 Mathematics: Section 1.5 - Trigonometry (Identities, Equations, Graphs, Sine/Cosine Rules)

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Lesson Objectives

  • Recall and apply fundamental trigonometric identities
  • Prove trigonometric identities using algebraic manipulation
  • Solve trigonometric equations for given intervals
  • Sketch and interpret graphs of sine, cosine, and tangent functions
  • Apply the sine rule to find unknown sides and angles in triangles
  • Apply the cosine rule to find unknown sides and angles in triangles
  • Solve real-world problems involving triangles using trigonometry

Introduction to Trigonometry

Trigonometry is the study of relationships between angles and sides in triangles. It has applications in navigation, physics, engineering, architecture, and many other fields. This section covers trigonometric identities, solving equations, graphing trigonometric functions, and applying the sine and cosine rules to solve triangles.

Fundamental Trigonometric Identities
$\sin^2 \theta + \cos^2 \theta = 1$
$\tan \theta = \frac{\sin \theta}{\cos \theta}$
$\sec^2 \theta = 1 + \tan^2 \theta$
$\csc^2 \theta = 1 + \cot^2 \theta$
$\sin(90^\circ - \theta) = \cos \theta$, $\cos(90^\circ - \theta) = \sin \theta$
Key Definitions:
• Trigonometric Identity: An equation involving trigonometric functions that is true for all values of the variable.
• General Solution: All angles that satisfy a trigonometric equation, expressed using $n \in \mathbb{Z}$.
• Principal Values: Solutions within a specific interval (usually $0^\circ \leq \theta < 360^\circ$ or $0 \leq \theta < 2\pi$).
• Period: The length of one complete cycle of a trigonometric function (sine/cosine: $360^\circ$ or $2\pi$).
• Amplitude: The maximum displacement from the mean value of a periodic function.

Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the variable. They are used to simplify expressions, prove other identities, and solve equations. The most important identities include the Pythagorean identity, quotient identities, and compound angle formulas.

Key Trigonometric Identities to Memorise:
1. $\sin^2 \theta + \cos^2 \theta = 1$
2. $1 + \tan^2 \theta = \sec^2 \theta$
3. $1 + \cot^2 \theta = \csc^2 \theta$
4. $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
5. $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
6. $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$
7. $\sin 2A = 2 \sin A \cos A$
8. $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
9. $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$
Example 1: Proving an Identity
Problem: Prove that $\tan^2 \theta + 1 = \sec^2 \theta$.

Solution:
$\tan^2 \theta + 1 = \frac{\sin^2 \theta}{\cos^2 \theta} + 1 = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta$ ✓
Example 2: Simplifying an Expression
Problem: Simplify $\frac{\sin^2 \theta}{1 - \cos \theta}$.

Solution:
$\frac{\sin^2 \theta}{1 - \cos \theta} = \frac{1 - \cos^2 \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta} = 1 + \cos \theta$
Answer: $1 + \cos \theta$
Example 3: Double Angle Identity
Problem: Simplify $\sin 2\theta \cos \theta + \cos 2\theta \sin \theta$.

Solution:
Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$ with $A = 2\theta$, $B = \theta$:
$\sin 2\theta \cos \theta + \cos 2\theta \sin \theta = \sin(2\theta + \theta) = \sin 3\theta$
Answer: $\sin 3\theta$

Practice for Identities

  1. Prove that $\frac{\sin \theta}{\cos \theta} = \tan \theta$.
  2. Simplify $\cos^2 \theta - \sin^2 \theta$.
  3. Simplify $\frac{1 - \sin^2 \theta}{\cos \theta}$.
  4. Prove that $\csc^2 \theta - \cot^2 \theta = 1$.
  5. Simplify $\sin(90^\circ - \theta) \cdot \cos \theta + \cos(90^\circ - \theta) \cdot \sin \theta$.

Solving Trigonometric Equations

Solving trigonometric equations involves finding all angles that satisfy the equation. Solutions can be expressed as general solutions (including all periodic repetitions) or principal solutions within a specific interval.

Step-by-Step Method for Solving Trigonometric Equations:
1. Use identities to simplify the equation if necessary.
2. Isolate a single trigonometric function (e.g., $\sin \theta = k$).
3. Find the reference angle: $\alpha = \sin^{-1}|k|$, $\cos^{-1}|k|$, or $\tan^{-1}|k|$.
4. Determine the quadrants where the function has the required sign.
5. Write solutions using the general form or within the given interval.
General Solutions for Trigonometric Equations
$\sin \theta = k$: $\theta = n \cdot 180^\circ + (-1)^n \alpha$ or $\theta = n\pi + (-1)^n \alpha$
$\cos \theta = k$: $\theta = n \cdot 360^\circ \pm \alpha$ or $\theta = 2n\pi \pm \alpha$
$\tan \theta = k$: $\theta = n \cdot 180^\circ + \alpha$ or $\theta = n\pi + \alpha$
where $\alpha$ is the reference angle in degrees/radians, $n \in \mathbb{Z}$.
Example 1: Basic Sine Equation
Problem: Solve $\sin \theta = \frac{1}{2}$ for $0^\circ \leq \theta < 360^\circ$.

Solution:
Reference angle: $\alpha = \sin^{-1}(\frac{1}{2}) = 30^\circ$
Sine is positive in Quadrants I and II.
Solutions: $\theta = 30^\circ$ and $\theta = 180^\circ - 30^\circ = 150^\circ$
Answer: $\theta = 30^\circ, 150^\circ$
Example 2: Cosine Equation
Problem: Solve $\cos \theta = -\frac{\sqrt{3}}{2}$ for $0^\circ \leq \theta < 360^\circ$.

Solution:
Reference angle: $\alpha = \cos^{-1}(\frac{\sqrt{3}}{2}) = 30^\circ$
Cosine is negative in Quadrants II and III.
Quadrant II: $\theta = 180^\circ - 30^\circ = 150^\circ$
Quadrant III: $\theta = 180^\circ + 30^\circ = 210^\circ$
Answer: $\theta = 150^\circ, 210^\circ$
Example 3: Tangent Equation
Problem: Solve $\tan \theta = \sqrt{3}$ for $0^\circ \leq \theta < 360^\circ$.

Solution:
Reference angle: $\alpha = \tan^{-1}(\sqrt{3}) = 60^\circ$
Tangent is positive in Quadrants I and III.
Quadrant I: $\theta = 60^\circ$
Quadrant III: $\theta = 180^\circ + 60^\circ = 240^\circ$
Answer: $\theta = 60^\circ, 240^\circ$
Example 4: Equation Requiring Identity
Problem: Solve $2\sin^2 \theta - \sin \theta - 1 = 0$ for $0^\circ \leq \theta < 360^\circ$.

Solution:
Let $x = \sin \theta$: $2x^2 - x - 1 = 0$ → $(2x + 1)(x - 1) = 0$
$\sin \theta = -\frac{1}{2}$ or $\sin \theta = 1$
For $\sin \theta = 1$: $\theta = 90^\circ$
For $\sin \theta = -\frac{1}{2}$: $\alpha = 30^\circ$, sine negative in Quadrants III and IV → $\theta = 210^\circ, 330^\circ$
Answer: $\theta = 90^\circ, 210^\circ, 330^\circ$
Example 5: General Solution
Problem: Find the general solution for $\sin \theta = \frac{\sqrt{3}}{2}$.

Solution:
$\alpha = \sin^{-1}(\frac{\sqrt{3}}{2}) = 60^\circ$
General solution: $\theta = n \cdot 180^\circ + (-1)^n \cdot 60^\circ$, $n \in \mathbb{Z}$
Answer: $\theta = n \cdot 180^\circ + (-1)^n \cdot 60^\circ$
Watch Out!
When solving trigonometric equations, be careful not to lose solutions when dividing by a trigonometric function. Always consider the case where the function equals zero before dividing.

Practice for Solving Equations

  1. Solve $\cos \theta = \frac{\sqrt{2}}{2}$ for $0^\circ \leq \theta < 360^\circ$.
  2. Solve $\tan \theta = -1$ for $0^\circ \leq \theta < 360^\circ$.
  3. Solve $2\cos^2 \theta - \cos \theta - 1 = 0$ for $0^\circ \leq \theta < 360^\circ$.
  4. Solve $\sin 2\theta = \frac{1}{2}$ for $0^\circ \leq \theta < 360^\circ$.
  5. Find the general solution for $\cos \theta = -\frac{1}{2}$.

Trigonometric Graphs

The graphs of sine, cosine, and tangent functions are periodic and have characteristic shapes. Understanding amplitude, period, phase shift, and vertical shift allows us to sketch transformed trigonometric functions.

Key Features of Trigonometric Graphs:
For $y = a \sin(bx + c) + d$ or $y = a \cos(bx + c) + d$:
• Amplitude: $|a|$ (half the distance between max and min)
• Period: $\frac{360^\circ}{|b|}$ or $\frac{2\pi}{|b|}$ (length of one cycle)
• Phase Shift: $-\frac{c}{b}$ (horizontal shift)
• Vertical Shift: $d$ (shifts the midline)
For $y = a \tan(bx + c) + d$, the period is $\frac{180^\circ}{|b|}$ or $\frac{\pi}{|b|}$.
Key Values for Sine and Cosine Graphs
$\theta$$0^\circ$$90^\circ$$180^\circ$$270^\circ$$360^\circ$
$\sin \theta$$0$$1$$0$$-1$$0$
$\cos \theta$$1$$0$$-1$$0$$1$

For $\tan \theta$: $\tan 0^\circ = 0$, $\tan 45^\circ = 1$, $\tan 90^\circ$ undefined, $\tan 135^\circ = -1$, $\tan 180^\circ = 0$

Example 1: Sketching a Sine Graph
Problem: Sketch $y = 2\sin x$ for $0^\circ \leq x \leq 360^\circ$.

Solution:
Amplitude = 2, Period = $360^\circ$
Key points: $(0,0)$, $(90^\circ,2)$, $(180^\circ,0)$, $(270^\circ,-2)$, $(360^\circ,0)$
Graph is a sine wave with maximum 2, minimum -2.
Example 2: Cosine with Phase Shift
Problem: Describe the graph of $y = \cos(x - 60^\circ)$.

Solution:
Amplitude = 1, Period = $360^\circ$
Phase shift: $60^\circ$ to the right (since $x - 60^\circ = 0$ when $x = 60^\circ$)
The graph of $\cos x$ is shifted $60^\circ$ right.
Example 3: Tangent Graph Features
Problem: Determine the period of $y = \tan(2x)$.

Solution:
Period of $\tan x$ is $180^\circ$
For $y = \tan(2x)$, period = $\frac{180^\circ}{2} = 90^\circ$
Answer: $90^\circ$

Practice for Graphs

  1. What is the amplitude of $y = 3\sin x$?
  2. What is the period of $y = \cos(3x)$?
  3. What is the phase shift of $y = \sin(x + 30^\circ)$?
  4. Describe the transformation from $y = \cos x$ to $y = 2\cos(x - 45^\circ) + 1$.
  5. State the period of $y = \tan(\frac{x}{2})$.

The Sine Rule

The sine rule relates the sides of a triangle to the sines of its opposite angles. It is used to solve triangles when given two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA - ambiguous case).

The Sine Rule
For any triangle $ABC$ with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Using the Sine Rule:
1. Identify the given sides and angles.
2. Set up the proportion: $\frac{a}{\sin A} = \frac{b}{\sin B}$ or equivalent.
3. Cross-multiply and solve for the unknown.
4. For the ambiguous case (SSA), check for two possible triangles.
Example 1: AAS (Two Angles and a Side)
Problem: In triangle ABC, $A = 40^\circ$, $B = 60^\circ$, $a = 10$ cm. Find side $b$.

Solution:
$\frac{a}{\sin A} = \frac{b}{\sin B}$ → $\frac{10}{\sin 40^\circ} = \frac{b}{\sin 60^\circ}$
$b = \frac{10 \times \sin 60^\circ}{\sin 40^\circ} \approx \frac{10 \times 0.8660}{0.6428} \approx 13.47$ cm
Answer: $b \approx 13.5$ cm
Example 2: SSA (Two Sides and a Non-included Angle)
Problem: In triangle ABC, $a = 8$ cm, $b = 10$ cm, $A = 30^\circ$. Find angle $B$.

Solution:
$\frac{a}{\sin A} = \frac{b}{\sin B}$ → $\frac{8}{\sin 30^\circ} = \frac{10}{\sin B}$
$\frac{8}{0.5} = \frac{10}{\sin B}$ → $16 = \frac{10}{\sin B}$ → $\sin B = \frac{10}{16} = 0.625$
$B = \sin^{-1}(0.625) \approx 38.68^\circ$ or $B = 180^\circ - 38.68^\circ = 141.32^\circ$
Both are possible. Check if $A + B < 180^\circ$: $30^\circ + 38.68^\circ = 68.68^\circ$ ✓; $30^\circ + 141.32^\circ = 171.32^\circ$ ✓
Answer: $B \approx 38.7^\circ$ or $141.3^\circ$ (ambiguous case)
Watch Out!
The sine rule ambiguous case (SSA) can produce 0, 1, or 2 triangles. Always check if $\sin B \leq 1$ and if the sum of angles is less than $180^\circ$.

Practice for Sine Rule

  1. In triangle ABC, $A = 50^\circ$, $B = 70^\circ$, $a = 12$ cm. Find side $b$.
  2. In triangle ABC, $a = 15$ cm, $b = 20$ cm, $A = 40^\circ$. Find angle $B$.
  3. In triangle ABC, $B = 45^\circ$, $C = 75^\circ$, $b = 8$ cm. Find side $c$.
  4. Explain why SSA is called the ambiguous case.
  5. In triangle ABC, $a = 5$ cm, $b = 7$ cm, $A = 30^\circ$. Determine if 0, 1, or 2 triangles exist.

The Cosine Rule

The cosine rule relates the sides of a triangle to the cosine of one of its angles. It is used to solve triangles when given three sides (SSS) or two sides and the included angle (SAS).

The Cosine Rule
For any triangle $ABC$ with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively:
$a^2 = b^2 + c^2 - 2bc\cos A$
$b^2 = a^2 + c^2 - 2ac\cos B$
$c^2 = a^2 + b^2 - 2ab\cos C$
Alternatively: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Using the Cosine Rule:
1. For SAS (two sides and included angle): Use $a^2 = b^2 + c^2 - 2bc\cos A$ to find the third side.
2. For SSS (three sides): Use $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ to find an angle.
3. Always ensure the correct sides correspond to the angle being used.
Example 1: SAS (Two Sides and Included Angle)
Problem: In triangle ABC, $b = 8$ cm, $c = 10$ cm, $A = 60^\circ$. Find side $a$.

Solution:
$a^2 = b^2 + c^2 - 2bc\cos A = 8^2 + 10^2 - 2(8)(10)\cos 60^\circ$
$a^2 = 64 + 100 - 160 \times 0.5 = 164 - 80 = 84$
$a = \sqrt{84} = 2\sqrt{21} \approx 9.17$ cm
Answer: $a \approx 9.17$ cm
Example 2: SSS (Three Sides)
Problem: In triangle ABC, $a = 7$ cm, $b = 8$ cm, $c = 9$ cm. Find angle $A$.

Solution:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{8^2 + 9^2 - 7^2}{2 \times 8 \times 9} = \frac{64 + 81 - 49}{144} = \frac{96}{144} = \frac{2}{3} \approx 0.6667$
$A = \cos^{-1}(\frac{2}{3}) \approx 48.19^\circ$
Answer: $A \approx 48.2^\circ$
Example 3: Finding an Angle Given Three Sides
Problem: In triangle ABC, $a = 6$ cm, $b = 7$ cm, $c = 8$ cm. Find the largest angle.

Solution:
Largest angle is opposite largest side (c = 8 cm, opposite angle C).
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{36 + 49 - 64}{2 \times 6 \times 7} = \frac{21}{84} = 0.25$
$C = \cos^{-1}(0.25) \approx 75.52^\circ$
Answer: $C \approx 75.5^\circ$
Watch Out!
When using the cosine rule, ensure you are using the correct angle. The formula $a^2 = b^2 + c^2 - 2bc\cos A$ uses angle $A$ opposite side $a$.

Practice for Cosine Rule

  1. In triangle ABC, $b = 12$ cm, $c = 15$ cm, $A = 45^\circ$. Find side $a$.
  2. In triangle ABC, $a = 10$ cm, $b = 12$ cm, $c = 14$ cm. Find angle $B$.
  3. In triangle ABC, $a = 8$ cm, $b = 9$ cm, $C = 120^\circ$. Find side $c$.
  4. Find the smallest angle in a triangle with sides 5 cm, 6 cm, 7 cm.
  5. When would you use the cosine rule instead of the sine rule?

Methods & Techniques

Mastering trigonometry requires systematic approaches and verification strategies. Use these techniques to ensure accuracy.

Verification / Checking Strategy:
1. For identities: Test with specific angle values (e.g., $30^\circ$, $45^\circ$, $60^\circ$).
2. For equations: Substitute solutions back into the original equation.
3. For sine/cosine rules: Check that the sum of angles is $180^\circ$ and that sides satisfy the triangle inequality.
4. For graphs: Verify key points (max, min, intercepts, asymptotes).
Example: Checking a Trigonometric Equation
Original: $\sin \theta = \frac{1}{2}$, solution $\theta = 30^\circ$.

Check:
$\sin 30^\circ = 0.5 = \frac{1}{2}$ ✓
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Forgetting the ambiguous case in sine rule (SSA) → Solution: Always check for two possible angles.
• Pitfall 2: Losing solutions when dividing by a trigonometric function → Solution: Factor instead of dividing.
• Pitfall 3: Using degrees and radians inconsistently → Solution: Always check calculator mode.
• Pitfall 4: Confusing sine and cosine rules → Solution: Sine rule for AAS/ASA/SSA; Cosine rule for SAS/SSS.
• Pitfall 5: Sign errors in compound angle formulas → Solution: Memorise the patterns carefully.

Technique Practice

  1. Verify that $\sin^2 30^\circ + \cos^2 30^\circ = 1$.
  2. Check the solution $\theta = 150^\circ$ for $\sin \theta = \frac{1}{2}$.
  3. For triangle with sides $a=3$, $b=4$, $c=5$, verify the cosine rule for angle $C$.
  4. Identify the error: A student wrote $\cos(A+B) = \cos A + \cos B$. Correct this.

Real-World Applications

Trigonometry is essential in navigation, surveying, engineering, physics, and many other fields.

Application 1: Navigation
Scenario: A ship sails 20 km on a bearing of $060^\circ$, then 30 km on a bearing of $150^\circ$. Find the distance from the start.
Problem: Use cosine rule to find the displacement.

Solution:
The angle between the two paths is $150^\circ - 60^\circ = 90^\circ$
By Pythagoras: distance = $\sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} \approx 36.06$ km
Application 2: Surveying (Sine Rule)
Scenario: A surveyor measures two angles of a triangular plot as $35^\circ$ and $55^\circ$, and the side between them is 100 m. Find the other sides.
Problem: Use sine rule.

Solution:
Third angle = $180^\circ - 35^\circ - 55^\circ = 90^\circ$
$\frac{a}{\sin 35^\circ} = \frac{100}{\sin 90^\circ}$ → $a = 100 \sin 35^\circ \approx 57.4$ m
$\frac{b}{\sin 55^\circ} = \frac{100}{\sin 90^\circ}$ → $b = 100 \sin 55^\circ \approx 81.9$ m
Application 3: Height of a Building
Scenario: From a point 50 m from the base of a building, the angle of elevation to the top is $30^\circ$. Find the height.
Problem: Use tangent ratio.

Solution:
$\tan 30^\circ = \frac{h}{50}$ → $h = 50 \tan 30^\circ = 50 \times \frac{\sqrt{3}}{3} \approx 28.9$ m

Cross-Curricular Connections

  • Physics: Projectile motion, waves, oscillations, vectors
  • Engineering: Structural analysis, surveying, signal processing
  • Navigation: GPS, bearings, triangulation
  • Astronomy: Parallax, celestial navigation
  • Music: Sound waves and harmonics

Cumulative Practice Exercises

Try these problems on your own. Show all working steps. Use the verification strategies to check your answers.

  1. Prove that $\frac{\sin^2 \theta}{1 - \cos \theta} = 1 + \cos \theta$.
  2. Solve $\cos \theta = \frac{\sqrt{3}}{2}$ for $0^\circ \leq \theta < 360^\circ$.
  3. Solve $2\sin^2 \theta - \sin \theta = 0$ for $0^\circ \leq \theta < 360^\circ$.
  4. What is the period of $y = \sin(4x)$?
  5. What is the amplitude of $y = 5\cos x + 2$?
  6. In triangle ABC, $A = 35^\circ$, $B = 65^\circ$, $a = 15$ cm. Find side $b$.
  7. In triangle ABC, $a = 10$ cm, $b = 14$ cm, $c = 18$ cm. Find angle $C$.
  8. In triangle ABC, $b = 12$ cm, $c = 15$ cm, $A = 50^\circ$. Find side $a$.
  9. Find the general solution for $\tan \theta = -1$.
  10. Sketch $y = 2\cos(x - 60^\circ)$ for $0^\circ \leq x \leq 360^\circ$ (describe key points).
  11. A ladder 5 m long leans against a wall. The foot is 2 m from the wall. Find the angle the ladder makes with the ground.
  12. Two ships leave a port. One sails 40 km on bearing $030^\circ$, the other sails 60 km on bearing $120^\circ$. Find the distance between them.
  13. Error analysis: A student solved $\sin 2\theta = \frac{1}{2}$ and got $\theta = 15^\circ$ only. What mistake was made? Find all solutions for $0^\circ \leq \theta < 360^\circ$.
  14. Prove that $\tan^2 \theta + 1 = \sec^2 \theta$ using $\sin^2 \theta + \cos^2 \theta = 1$.
  15. A triangle has sides 7 cm, 8 cm, 9 cm. Find the largest angle using the cosine rule.
Show/Hide Answers

Answers to Cumulative Exercises

  1. Problem: Prove $\frac{\sin^2 \theta}{1 - \cos \theta} = 1 + \cos \theta$.
    Answer: $\frac{1-\cos^2 \theta}{1-\cos \theta} = \frac{(1-\cos \theta)(1+\cos \theta)}{1-\cos \theta} = 1 + \cos \theta$ ✓
  2. Problem: $\cos \theta = \frac{\sqrt{3}}{2}$ for $0^\circ \leq \theta < 360^\circ$.
    Answer: $\theta = 30^\circ, 330^\circ$
  3. Problem: $2\sin^2 \theta - \sin \theta = 0$.
    Answer: $\sin \theta(2\sin \theta - 1) = 0$ → $\sin \theta = 0$ or $\sin \theta = \frac{1}{2}$ → $\theta = 0^\circ, 180^\circ, 360^\circ, 30^\circ, 150^\circ$
  4. Problem: Period of $y = \sin(4x)$.
    Answer: $\frac{360^\circ}{4} = 90^\circ$
  5. Problem: Amplitude of $y = 5\cos x + 2$.
    Answer: $|5| = 5$
  6. Problem: $A = 35^\circ$, $B = 65^\circ$, $a = 15$ cm. Find $b$.
    Answer: $\frac{15}{\sin 35^\circ} = \frac{b}{\sin 65^\circ}$ → $b = \frac{15 \times 0.9063}{0.5736} \approx 23.7$ cm
  7. Problem: $a = 10$, $b = 14$, $c = 18$. Find $C$.
    Answer: $\cos C = \frac{10^2+14^2-18^2}{2\times10\times14} = \frac{100+196-324}{280} = \frac{-28}{280} = -0.1$ → $C = \cos^{-1}(-0.1) \approx 95.74^\circ$
  8. Problem: $b = 12$, $c = 15$, $A = 50^\circ$. Find $a$.
    Answer: $a^2 = 144 + 225 - 2\times12\times15\times\cos 50^\circ = 369 - 360\times0.6428 = 369 - 231.4 = 137.6$ → $a \approx 11.73$ cm
  9. Problem: General solution for $\tan \theta = -1$.
    Answer: $\theta = n \cdot 180^\circ - 45^\circ$, $n \in \mathbb{Z}$
  10. Problem: Sketch $y = 2\cos(x - 60^\circ)$.
    Answer: Amplitude 2, period $360^\circ$, phase shift $60^\circ$ right. Key points: at $x=60^\circ$, $y=2$; $x=150^\circ$, $y=0$; $x=240^\circ$, $y=-2$; $x=330^\circ$, $y=0$; $x=420^\circ$, $y=2$.
  11. Problem: Ladder 5 m, foot 2 m from wall. Angle with ground?
    Answer: $\cos \theta = \frac{2}{5} = 0.4$ → $\theta = \cos^{-1}(0.4) \approx 66.4^\circ$
  12. Problem: Ships: 40 km at $030^\circ$, 60 km at $120^\circ$. Angle between = $90^\circ$. Distance = $\sqrt{40^2+60^2} = \sqrt{1600+3600} = \sqrt{5200} \approx 72.1$ km
  13. Problem: Error: $\sin 2\theta = \frac{1}{2}$ → $2\theta = 30^\circ$ or $150^\circ$ → $\theta = 15^\circ$ or $75^\circ$. Also add period: $2\theta = 30^\circ + n\cdot360^\circ$ and $2\theta = 150^\circ + n\cdot360^\circ$ → $\theta = 15^\circ + n\cdot180^\circ$ or $75^\circ + n\cdot180^\circ$. For $0^\circ \leq \theta < 360^\circ$: $\theta = 15^\circ, 75^\circ, 195^\circ, 255^\circ$
  14. Problem: Prove $\tan^2 \theta + 1 = \sec^2 \theta$.
    Answer: $\frac{\sin^2 \theta}{\cos^2 \theta} + 1 = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta$
  15. Problem: Triangle sides 7, 8, 9 cm. Largest angle opposite side 9 cm.
    Answer: $\cos C = \frac{7^2+8^2-9^2}{2\times7\times8} = \frac{49+64-81}{112} = \frac{32}{112} = 0.2857$ → $C = \cos^{-1}(0.2857) \approx 73.4^\circ$

Conclusion & Summary

Trigonometry is a powerful tool for solving problems involving angles and triangles. The fundamental identities form the basis for simplification and proof. Solving trigonometric equations requires understanding of reference angles and quadrants. Trigonometric graphs have characteristic features of amplitude, period, and phase shift. The sine and cosine rules enable us to solve any triangle, given sufficient information.

Key Takeaways:
1. Identities: $\sin^2 \theta + \cos^2 \theta = 1$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
2. Solving equations: Use reference angles and quadrants; general solutions include periodic repetitions.
3. Graphs: $y = a\sin(bx+c)+d$ has amplitude $|a|$, period $\frac{360^\circ}{|b|}$, phase shift $-\frac{c}{b}$.
4. Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ — use for AAS, ASA, SSA (ambiguous).
5. Cosine Rule: $a^2 = b^2 + c^2 - 2bc\cos A$ — use for SAS, SSS.
6. Applications: Navigation, surveying, physics, engineering.

Keep practising with different types of triangles and equations. Trigonometry is essential for calculus, physics, and engineering!

Video Resource

Watch this video for more examples of solving trigonometric equations and applying the sine/cosine rules.

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