Lesson 3.2: Advanced Trigonometric Functions

Master Advanced Trigonometric Functions & Applications

Dive deep into trigonometric functions! Learn fundamental definitions, core identities, special angle values, and systematic approaches to solving right triangles. Master the art of trigonometric calculations and real-world applications including angle of elevation and depression problems.

Learning Objectives

Define and apply trigonometric functions

Use fundamental trigonometric identities

Calculate special angle values

Solve right triangles systematically

Apply trigonometry to real-world problems

Use complementary angle relationships

Core Concepts & Theoretical Foundation

Trigonometric Function Definitions

In right triangle ABC with $\angle C = 90°$ and $\angle A$ as the reference angle:

• Sine: $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB}$

• Cosine: $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AC}{AB}$

• Tangent: $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC}$

• Key insight: These ratios depend only on the angle, not the triangle size

• Range: Sine and cosine values are between -1 and 1; tangent can be any real number

Fundamental Trigonometric Identities

Essential relationships between trigonometric functions:

• Pythagorean Identity: $\sin^2 A + \cos^2 A = 1$

- Derived from Pythagorean theorem: $\text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2$

- Divide by hypotenuse² to get the identity

• Quotient Identity: $\tan A = \frac{\sin A}{\cos A}$

- Direct consequence of the definitions

• Reciprocal Identities:

- $\csc A = \frac{1}{\sin A}$ (cosecant)

- $\sec A = \frac{1}{\cos A}$ (secant)

- $\cot A = \frac{1}{\tan A}$ (cotangent)

Complementary Angle Relationships

Special relationships for complementary angles:

• Co-function Identities:

- $\sin A = \cos(90° - A)$

- $\cos A = \sin(90° - A)$

- $\tan A = \cot(90° - A)$

• Examples:

- $\sin 30° = \cos 60° = \frac{1}{2}$

- $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$

• Application: Reduces the number of values to memorize

Special Angle Values

Memorize these essential values:

Angle	$\sin \theta$	$\cos \theta$	$\tan \theta$
30°	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
45°	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	1
60°	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

Memory aids: 30°-60°-90° triangle has sides in ratio 1:√3:2; 45°-45°-90° triangle has sides in ratio 1:1:√2

Detailed Worked Examples

Example 1: Trigonometric Identity Application

Given that $\tan A = \frac{3}{4}$ for acute angle A, find $\sin A$ and $\cos A$ , and verify the Pythagorean identity.

Step 1: Set up the triangle

Since $\tan A = \frac{3}{4} = \frac{\text{opposite}}{\text{adjacent}}$ , we can set:

Opposite side = 3k, Adjacent side = 4k (where k > 0)

Step 2: Find the hypotenuse

Using Pythagorean theorem:

$\text{hypotenuse} = \sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$

Step 3: Calculate trigonometric values

$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3k}{5k} = \frac{3}{5}$

$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4k}{5k} = \frac{4}{5}$

Step 4: Verify Pythagorean identity

$\sin^2 A + \cos^2 A = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2$

$= \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$ ✓

Step 5: Verify quotient identity

$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{5} \times \frac{5}{4} = \frac{3}{4}$ ✓

Key Insight: When given one trigonometric ratio, we can construct a right triangle and find all other ratios. The identities provide powerful verification tools and alternative calculation methods.

Example 2: Solving Right Triangles

In right triangle ABC with $\angle C = 90°$ , $\angle A = 30°$ , and AB = 12 cm. Find the lengths of the other sides and the measure of $\angle B$ .

Step 1: Find the missing angle

Since $\angle A + \angle B + \angle C = 180°$ :

$30° + \angle B + 90° = 180°$

$\angle B = 180° - 120° = 60°$

Step 2: Use trigonometric ratios to find sides

Given: AB = 12 cm (hypotenuse), $\angle A = 30°$

Step 3: Find BC (opposite to angle A)

$\sin 30° = \frac{BC}{AB}$

$\frac{1}{2} = \frac{BC}{12}$

$BC = 12 \times \frac{1}{2} = 6$ cm

Step 4: Find AC (adjacent to angle A)

$\cos 30° = \frac{AC}{AB}$

$\frac{\sqrt{3}}{2} = \frac{AC}{12}$

$AC = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$ cm

Step 5: Verification using Pythagorean theorem

$AC^2 + BC^2 = (6\sqrt{3})^2 + 6^2 = 108 + 36 = 144 = 12^2 = AB^2$ ✓

Systematic Approach: When solving right triangles, always identify what you know and what you need to find. Use the appropriate trigonometric ratio based on the given information.

Example 3: Angle of Elevation Problem

An observer at point A is 20 meters horizontally from the base of a tower. The angle of elevation from A to the top of the tower is 60°. Find the height of the tower (ignoring the observer's height).

Step 1: Set up the right triangle

Let B be the base of the tower, C be the top, and A be the observer

Right triangle ABC with $\angle B = 90°$ , AB = 20 m, $\angle BAC = 60°$

Step 2: Identify what to find

We need to find BC (height of the tower)

Step 3: Choose the appropriate trigonometric ratio

We have the adjacent side (AB = 20 m) and need the opposite side (BC)

Use tangent: $\tan A = \frac{\text{opposite}}{\text{adjacent}}$

Step 4: Apply the tangent ratio

$\tan 60° = \frac{BC}{AB}$

$\sqrt{3} = \frac{BC}{20}$

$BC = 20 \times \sqrt{3} \approx 20 \times 1.732 \approx 34.64$ meters

Step 5: Interpretation

The tower is approximately 34.64 meters tall

Real-World Application: Angle of elevation problems are common in surveying, architecture, and navigation. The key is to identify the right triangle formed by the horizontal line, vertical line, and line of sight.

Example 4: Complementary Angle Application

In right triangle PQR with $\angle R = 90°$ , $\sin P = \frac{4}{5}$ . Find $\cos Q$ and $\tan P$ .

Step 1: Use complementary angle relationship

Since $\angle P + \angle Q = 90°$ (complementary angles in right triangle):

$\cos Q = \sin P = \frac{4}{5}$

Step 2: Find cos P using Pythagorean identity

$\sin^2 P + \cos^2 P = 1$

$\left(\frac{4}{5}\right)^2 + \cos^2 P = 1$

$\frac{16}{25} + \cos^2 P = 1$

$\cos^2 P = 1 - \frac{16}{25} = \frac{9}{25}$

$\cos P = \frac{3}{5}$ (positive since P is acute)

Step 3: Find tan P using quotient identity

$\tan P = \frac{\sin P}{\cos P} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{5} \times \frac{5}{3} = \frac{4}{3}$

Step 4: Verification

We can also verify: $\sin Q = \cos P = \frac{3}{5}$

And: $\sin^2 Q + \cos^2 Q = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1$ ✓

Complementary Power: The co-function identities are incredibly useful for reducing calculations. When you know one trigonometric value, you automatically know its complement's co-function value.

Advanced Techniques & Problem-Solving Strategies

Right Triangle Solving Strategies

Systematic approaches based on given information:

• Given: One angle and one side

- Use trigonometric ratios to find other sides

- Use angle sum to find remaining angle

• Given: Two sides

- Use Pythagorean theorem for third side

- Use inverse trigonometric functions for angles

• Given: Special angles (30°, 45°, 60°)

- Use exact values instead of decimal approximations

Angle of Elevation and Depression

Key concepts for real-world applications:

• Angle of elevation: Angle from horizontal up to line of sight

• Angle of depression: Angle from horizontal down to line of sight

• Horizontal line: Always parallel to ground level

• Line of sight: Direct line from observer to object

• Common applications: Height of buildings, distance across rivers, navigation

Exact vs. Approximate Values

When to use exact values and when to approximate:

• Use exact values when:

- Working with special angles (30°, 45°, 60°)

- Final answer should be exact

- Further calculations require precision

• Use approximations when:

- Real-world measurements are involved

- Practical applications require decimal answers

- Calculator values are more appropriate

Common Pitfalls & Error Prevention

Pitfall 1: Confusing Opposite and Adjacent Sides

Error: Misidentifying which side is opposite or adjacent to the reference angle.

Solution: Always identify the reference angle first, then determine opposite (across from) and adjacent (next to) sides.

Pitfall 2: Using Wrong Trigonometric Ratio

Error: Choosing sine when you need cosine, or tangent when you need sine.

Solution: Identify what you have and what you need, then choose the ratio that connects them.

Pitfall 3: Calculator Mode Errors

Error: Using calculator in wrong mode (degrees vs. radians).

Solution: Always check that your calculator is in degree mode for angle measurements in degrees.

Pitfall 4: Rounding Too Early

Error: Rounding intermediate calculations, leading to inaccurate final answers.

Solution: Keep exact values in intermediate steps, round only the final answer.

Pitfall 5: Misinterpreting Angle of Elevation/Depression

Error: Confusing angle of elevation with angle of depression.

Solution: Remember that both are measured from the horizontal line, elevation goes up, depression goes down.

Comprehensive Practice Problems

Problem 1: Trigonometric Values

In right triangle ABC with $\angle C = 90°$ , if $\sin A = \frac{5}{13}$ , find $\cos A$ and $\tan A$ .

Show Solution

Using Pythagorean identity: $\cos A = \frac{12}{13}$

Using quotient identity: $\tan A = \frac{5}{12}$

Problem 2: Solving Right Triangle

In right triangle DEF with $\angle F = 90°$ , $\angle D = 45°$ , and EF = 7 cm. Find DE and DF.

Show Solution

Since 45°-45°-90° triangle: DF = EF = 7 cm

Hypotenuse: $DE = 7\sqrt{2}$ cm

Problem 3: Angle of Elevation

A ladder leans against a wall at an angle of 60° to the ground. If the ladder is 10 meters long, how high up the wall does it reach?

Show Solution

Using sine: $\sin 60° = \frac{h}{10}$

Height: $h = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ meters

Problem 4: Complementary Angles

In right triangle GHI with $\angle I = 90°$ , if $\cos G = \frac{3}{5}$ , find $\sin H$ and $\tan G$ .

Show Solution

Co-function identity: $\sin H = \cos G = \frac{3}{5}$

Find sin G: $\sin G = \frac{4}{5}$

Quotient identity: $\tan G = \frac{4}{3}$

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