Chapter 3: Trigonometric Functions

3.1 Angles

An angle is considered as the figure obtained by rotating a given ray about its endpoint. The original ray is called the initial side and the final position of the ray after rotation is called the terminal side of the angle.

• Positive angle: Anti-clockwise rotation.
• Negative angle: Clockwise rotation.

Measure of an Angle

1. Degree Measure: If a rotation from the initial side to terminal side is \( \frac{1}{360} \) of a revolution, the angle is said to have a measure of one degree (1°).
   • 1 degree = 60 minutes (1° = 60’)
   • 1 minute = 60 seconds (1’ = 60“)
2. Radian Measure: Angle subtended at the center by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian.
   • If an arc of length \( l \) subtends an angle \( \theta \) radians at the center of a circle of radius \( r \), then \( \theta = \frac{l}{r} \).
3. Relation between Degree and Radian: \[ \pi \text{ radians} = 180^\circ \]

3.2 Trigonometric Functions

Let a unit circle with center at origin of the coordinate axes. The angle \( x \) subtends from point \( P(a,b) \) on the unit circle. We define:

• \( \cos x = a \)
• \( \sin x = b \)

Since \( P(a,b) \) lies on the unit circle \( x^2 + y^2 = 1 \), we get: \[ \sin^2 x + \cos^2 x = 1 \]

Domain and Range of Trigonometric Functions

• Sine & Cosine: Domain = \( \mathbb{R} \), Range = \( [-1, 1] \).
• Tangent: Domain = \( \mathbb{R} - { x : x = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} } \), Range = \( \mathbb{R} \).

Graphs of Trigonometric Functions

The periodic nature of trigonometric functions can be visualized easily:

3.3 Trigonometric Identities

Sum and difference of two angles:

1. \( \sin(x \pm y) = \sin x \cos y \pm \cos x \sin y \)
2. \( \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y \)
3. \( \tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} \)

Double and triple angles:

• \( \sin 2x = 2 \sin x \cos x = \frac{2 \tan x}{1 + \tan^2 x} \)
• \( \cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \)
• \( \sin 3x = 3 \sin x - 4 \sin^3 x \)

Sums into products:

• \( \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \)
• \( \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \)

Competency-Based Questions

Question 1:

A Ferris wheel has a radius of 20 meters and its center is 25 meters above the ground. The wheel turns at a constant speed, making one complete rotation every 4 minutes.

If a passenger starts at the lowest point of the wheel, express their height \( h \) above the ground as a function of time \( t \) (in minutes). Construct the equation of height.

Answer 1:

The motion is periodic and circular, so we will use a trigonometric function.

The wheel makes a full revolution in 4 minutes, so its period \( T = 4 \).

The angular speed \( \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \) radians per minute.

The minimum height of the passenger = Center height - Radius = 25 - 20 = 5m.

The maximum height = Center height + Radius = 25 + 20 = 45m.

The amplitude of the function is the radius, \( A = 20 \).

The vertical shift is the center position, \( D = 25 \).

Since the passenger starts at the lowest point at \( t = 0 \), a negative cosine function is the best fit, as \( \cos(0) = 1 \), and we want the minimum value. Structure:

\[ h(t) = -A \cos(\omega t) + D \]

\[ h(t) = -20 \cos\left(\frac{\pi}{2} t\right) + 25 \]

Conclusion: The equation representing the passenger’s height is \( h(t) = 25 - 20 \cos\left(\frac{\pi}{2} t\right) \).

Question 2:

Prove the identity using the properties of trigonometric functions:

\[ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x \]

Answer 2:

We use the sum-to-product formulas for sine and cosine:

\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

Apply these formulas to the numerator and denominator:

Numerator:

\[ \sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin 4x \cos x \]

Denominator:

\[ \cos 5x + \cos 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \cos 4x \cos x \]

Substitute back into the expression:

\[ \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x} \]

Assuming \( \cos x \neq 0 \), we can cancel \( 2 \cos x \) from both numerator and denominator:

\[ \frac{\sin 4x}{\cos 4x} = \tan 4x \]

Conclusion: Hence proved.