Trigonometry Fundamentals: Angles and Functions

Trigonometry is a pivotal branch of mathematics that studies the relationships between angles and sides of triangles, extending to periodic phenomena like waves and oscillations. It introduces functions such as sine, cosine, and tangent, which are essential for solving geometric problems, modeling real-world scenarios, and analyzing periodic behavior. Trigonometry is indispensable in fields like navigation for determining directions, engineering for structural design, physics for wave analysis, astronomy for celestial measurements, and even music theory for understanding sound frequencies. In this guide, we’ll explore trigonometric functions, their applications in various contexts, graphical representations, and practical examples, providing a comprehensive foundation for understanding this versatile subject.

Basic Trigonometric Functions

In a right triangle, trigonometric functions define the relationships between an angle \( \theta \) and the ratios of the triangle’s sides. The three primary functions are:

Sine (\( \sin(\theta) \)): \( \frac{\text{opposite}}{\text{hypotenuse}} \), representing the vertical component of the angle relative to the hypotenuse.
Cosine (\( \cos(\theta) \)): \( \frac{\text{adjacent}}{\text{hypotenuse}} \), representing the horizontal component.
Tangent (\( \tan(\theta) \)): \( \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin(\theta)}{\cos(\theta)} \), the ratio of the opposite to the adjacent side.

Additionally, the reciprocal functions are:

Cosecant (\( \csc(\theta) \)): \( \frac{1}{\sin(\theta)} = \frac{\text{hypotenuse}}{\text{opposite}} \)
Secant (\( \sec(\theta) \)): \( \frac{1}{\cos(\theta)} = \frac{\text{hypotenuse}}{\text{adjacent}} \)
Cotangent (\( \cot(\theta) \)): \( \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\text{adjacent}}{\text{opposite}} \)

These functions are often memorized using the mnemonic SOH-CAH-TOA (Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent). Trigonometric functions also extend beyond right triangles to the unit circle, where for an angle \( \theta \), \( \sin(\theta) \) and \( \cos(\theta) \) are the \( y \)- and \( x \)-coordinates of a point on a circle of radius 1, making them periodic with a period of \( 2\pi \) radians.

Examples: Right Triangles, Angle of Elevation, Non-Right Triangle, Finding Angles, Using Identities, Periodic Motion

Let’s explore trigonometry through various scenarios with detailed calculations.

Example 1: Right Triangles

Case 1: The 3-4-5 Right Triangle

Consider a right triangle with legs 3 and 4, and hypotenuse 5 (a Pythagorean triple). For the angle \( \theta \) opposite the side of length 3:

\[ \text{Opposite} = 3, \text{ Adjacent} = 4, \text{ Hypotenuse} = 5 \] \[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \] \[ = \frac{3}{5} \] \[ = 0.6 \] \[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \] \[ = \frac{4}{5} \] \[ = 0.8 \] \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \] \[ = \frac{3}{4} \] \[ = 0.75 \]

The angle \( \theta \) is:

\[ \theta = \sin^{-1}(0.6) \] \[ \approx 36.87^\circ \]

Case 2: A 5-12-13 Right Triangle

Now consider a triangle with legs 5 and 12, and hypotenuse 13. For the angle \( \phi \) opposite the side of length 5:

\[ \sin(\phi) = \frac{5}{13} \] \[ \approx 0.3846 \] \[ \cos(\phi) = \frac{12}{13} \] \[ \approx 0.9231 \] \[ \tan(\phi) = \frac{5}{12} \] \[ \approx 0.4167 \]

The angle \( \phi \):

\[ \phi = \sin^{-1}(0.3846) \] \[ \approx 22.62^\circ \]

Example 2: Angle of Elevation

A person stands 50 meters from the base of a tree and measures the angle of elevation to the top as \( 30^\circ \). Find the height of the tree.

\[ \tan(\theta) = \frac{\text{height}}{\text{distance}} \] \[ \theta = 30^\circ, \text{ distance} = 50 \] \[ \tan(30^\circ) = \frac{\text{height}}{50} \] \[ \tan(30^\circ) = \frac{\sqrt{3}}{3} \approx 0.577 \] \[ \text{height} = 50 \times \tan(30^\circ) \] \[ \approx 50 \times 0.577 \] \[ \approx 28.85 \, \text{meters} \]

The tree is approximately 28.85 meters tall, demonstrating how trigonometry aids in indirect measurements.

Example 3: Non-Right Triangle (Law of Sines)

In a triangle with angles \( A = 40^\circ \), \( B = 60^\circ \), and side \( a = 10 \) (opposite angle \( A \)), find side \( b \) (opposite angle \( B \)). First, find angle \( C \):

\[ A + B + C = 180^\circ \] \[ 40^\circ + 60^\circ + C = 180^\circ \] \[ C = 180^\circ - 100^\circ \] \[ C = 80^\circ \]

Use the Law of Sines: \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \)

\[ \frac{10}{\sin(40^\circ)} = \frac{b}{\sin(60^\circ)} \] \[ \sin(40^\circ) \approx 0.6428, \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \] \[ b = 10 \times \frac{\sin(60^\circ)}{\sin(40^\circ)} \] \[ = 10 \times \frac{0.866}{0.6428} \] \[ \approx 10 \times 1.347 \] \[ \approx 13.47 \]

Side \( b \) is approximately 13.47 units, showing how trigonometry extends to non-right triangles.

Example 4: Finding Missing Angles

In a right triangle, the hypotenuse is 10, and one leg is 6. Find the acute angles.

\[ \text{Opposite leg} = 6, \text{ Hypotenuse} = 10 \] \[ \text{Adjacent leg} = \sqrt{10^2 - 6^2} \] \[ = \sqrt{100 - 36} \] \[ = \sqrt{64} \] \[ = 8 \]

For the angle \( \alpha \) opposite the leg of length 6:

\[ \sin(\alpha) = \frac{6}{10} \] \[ = 0.6 \] \[ \alpha = \sin^{-1}(0.6) \] \[ \approx 36.87^\circ \]

The other acute angle \( \beta \):

\[ \beta = 90^\circ - \alpha \] \[ = 90^\circ - 36.87^\circ \] \[ \approx 53.13^\circ \]

Alternatively, using cosine for \( \beta \):

\[ \cos(\beta) = \frac{6}{10} \] \[ \beta = \cos^{-1}(0.6) \] \[ \approx 53.13^\circ \]

Example 5: Using Trigonometric Identities

Verify the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) for \( \theta = 45^\circ \), and use it to find \( \cos(\theta) \) if \( \sin(\theta) = \frac{3}{5} \).

Case 1: Verify for \( \theta = 45^\circ \)

\[ \sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071 \] \[ \sin^2(45^\circ) + \cos^2(45^\circ) = \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 \] \[ = 0.5 + 0.5 \] \[ = 1 \]

Case 2: Find \( \cos(\theta) \)

\[ \sin(\theta) = \frac{3}{5} \] \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] \[ \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 \] \[ \frac{9}{25} + \cos^2(\theta) = 1 \] \[ \cos^2(\theta) = 1 - \frac{9}{25} \] \[ = \frac{16}{25} \] \[ \cos(\theta) = \pm \sqrt{\frac{16}{25}} \] \[ = \pm \frac{4}{5} \]

Assuming \( \theta \) is acute, \( \cos(\theta) = \frac{4}{5} \).

Example 6: Periodic Motion

A pendulum’s position follows \( y = 5 \sin(2t) \), where \( y \) is displacement (cm), \( t \) is time (seconds). Find the position at \( t = \frac{\pi}{4} \), amplitude, and period.

\[ y = 5 \sin(2t) \] \[ t = \frac{\pi}{4} \] \[ 2t = 2 \times \frac{\pi}{4} \] \[ = \frac{\pi}{2} \] \[ \sin\left(\frac{\pi}{2}\right) = 1 \] \[ y = 5 \times 1 \] \[ = 5 \, \text{cm} \] \[ \text{Amplitude} = 5 \, \text{(coefficient of sine)} \] \[ \text{Period of } \sin(kt) = \frac{2\pi}{k} \] \[ k = 2 \] \[ \text{Period} = \frac{2\pi}{2} \] \[ = \pi \, \text{seconds} \]

The pendulum is at 5 cm at \( t = \frac{\pi}{4} \), with a maximum displacement of 5 cm and a period of \( \pi \) seconds.

Graphical Representation

Trigonometric functions are periodic, making their graphs essential for understanding their behavior. Let’s plot sine, cosine, and tangent.

Sine Function: \( y = \sin(x) \)

Graph of \( y = \sin(x) \) from \( x = -5 \) to \( 5 \) radians:

Cosine Function: \( y = \cos(x) \)

Graph of \( y = \cos(x) \):

Tangent Function: \( y = \tan(x) \)

Graph of \( y = \tan(x) \), noting vertical asymptotes at \( x = \pm \frac{\pi}{2} \):

Applications of Trigonometry

Trigonometry is integral to many fields. Here are detailed examples with calculations:

Navigation - Ship Triangulation: A ship is 100 km from port A at a bearing of \( 30^\circ \), and 150 km from port B at a bearing of \( 120^\circ \). Approximate the distance between ports:
\[ \text{Angle at ship} = 120^\circ - 30^\circ = 90^\circ \] \[ \text{Distance} = \sqrt{100^2 + 150^2} \] \[ = \sqrt{10000 + 22500} \] \[ = \sqrt{32500} \] \[ \approx 180.28 \, \text{km} \]
Trigonometry confirms the angle makes it a right triangle, simplifying the calculation.
Engineering - Bridge Design: A bridge support forms a right triangle with a base of 20 m and an angle of \( 60^\circ \). Find the height:
\[ \tan(60^\circ) = \frac{\text{height}}{20} \] \[ \tan(60^\circ) = \sqrt{3} \approx 1.732 \] \[ \text{height} = 20 \times 1.732 \] \[ \approx 34.64 \, \text{m} \]
This height ensures structural stability.
Physics - Wave Modeling: A sound wave is modeled as \( y = 3 \sin(4\pi t) \). Find the frequency:
\[ y = A \sin(2\pi f t) \] \[ 2\pi f = 4\pi \] \[ f = \frac{4\pi}{2\pi} \] \[ = 2 \, \text{Hz} \]
The frequency is 2 Hz, typical for sound wave analysis.
Astronomy - Star Altitude: A star is at an altitude angle of \( 45^\circ \), observed from a point 1000 m above sea level. Observer’s height above ground is 2 m. Distance to star’s ground projection:
\[ \tan(45^\circ) = \frac{1000}{d} \] \[ \tan(45^\circ) = 1 \] \[ d = 1000 \, \text{m} \]
The star’s ground projection is 1000 m away.
Architecture - Roof Pitch: A roof has a pitch angle of \( 35^\circ \), base width 10 m. Find the roof height:
\[ \text{Half base} = 5 \, \text{m} \] \[ \tan(35^\circ) = \frac{\text{height}}{5} \] \[ \tan(35^\circ) \approx 0.7002 \] \[ \text{height} = 5 \times 0.7002 \] \[ \approx 3.501 \, \text{m} \]
The roof peak is 3.5 m high.
Music Theory - Sound Frequency: A note’s frequency is 440 Hz (A4). The next note (A#4) has frequency \( 440 \times 2^{1/12} \). Model its wave:
\[ 2^{1/12} \approx 1.0595 \] \[ f = 440 \times 1.0595 \] \[ \approx 466.18 \, \text{Hz} \] \[ y = \sin(2\pi \times 466.18 t) \]
This models the sound wave for A#4.