Trigonometry, derived from the Greek words trigon (triangle) and metron (measure), originally focused on measuring the sides and angles of a triangle. Modern trigonometry has expanded into a foundational branch of mathematics with applications in seismology, electrical circuit design, atomic state description, oceanography, and acoustics.
An angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side, and the final position of the ray after rotation is called the terminal side. The point of rotation is the vertex.
Angles are primarily measured in two systems: degree measure and radian measure.
Radian measures and real numbers can be considered as one and the same. By placing a real number line tangent to a unit circle at point $A(1,0)$, wrapping the positive line anticlockwise and the negative line clockwise associates every real number with a unique radian angle.
Since a complete revolution of a circle subtends $2\pi$ radians at the centre, we establish the fundamental equivalence:
$$2\pi \text{ radians} = 360^\circ \implies \pi \text{ radians} = 180^\circ$$This yields the following conversion formulas:
The table below lists the radian equivalents of common degree measures:
| Degree Measure | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ | $180^\circ$ | $270^\circ$ | $360^\circ$ |
|---|---|---|---|---|---|---|---|
| Radian Measure | $\frac{\pi}{6}$ | $\frac{\pi}{4}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\pi$ | $\frac{3\pi}{2}$ | $2\pi$ |
Trigonometric ratios are generalized to trigonometric functions for any real angle using a unit circle coordinate system.
Let $P(a, b)$ be any point on a unit circle centered at the origin $O(0,0)$ such that the angle $AOP = x$ radians. The arc length $AP = x$. We define:
$$\cos x = a \quad \text{and} \quad \sin x = b$$Since $P(a,b)$ lies on the unit circle $a^2 + b^2 = 1$, we obtain the fundamental identity:
$$\cos^2 x + \sin^2 x = 1$$Angles that are integral multiples of $\frac{\pi}{2}$ are called quadrantal angles. Their coordinates yield:
Since a complete revolution returns to the same point, trigonometric functions are periodic with period $2\pi$:
$$\sin(2n\pi + x) = \sin x, \quad \cos(2n\pi + x) = \cos x \quad (n \in \mathbb{Z})$$The domain and range of the six trigonometric functions are defined as follows:
| Function | Domain | Range |
|---|---|---|
| $\sin x$ | $\mathbb{R}$ | $[-1, 1]$ |
| $\cos x$ | $\mathbb{R}$ | $[-1, 1]$ |
| $\tan x$ | $\mathbb{R} \setminus \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ | $\mathbb{R}$ |
| $\cot x$ | $\mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$ | $\mathbb{R}$ |
| $\sec x$ | $\mathbb{R} \setminus \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ | $(-\infty, -1] \cup [1, \infty)$ |
| $\csc x$ | $\mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$ | $(-\infty, -1] \cup [1, \infty)$ |
The signs of the trigonometric functions depend on the quadrant of the terminal side:
| Quadrant | I ($0 < x < \frac{\pi}{2}$) | II ($\frac{\pi}{2} < x < \pi$) | III ($\pi < x < \frac{3\pi}{2}$) | IV ($\frac{3\pi}{2} < x < 2\pi$) |
|---|---|---|---|---|
| $\sin x$, $\csc x$ | $+$ | $+$ | $-$ | $-$ |
| $\cos x$, $\sec x$ | $+$ | $-$ | $-$ | $+$ |
| $\tan x$, $\cot x$ | $+$ | $-$ | $+$ | $-$ |
Trigonometric identities establish relationships between functions for sums, differences, and multiples of angles.
Double Angle Formulas:
Triple Angle Formulas:
Sum-to-Product Formulas:
Product-to-Sum Formulas:
Mathematical Problem Statement: Convert $40^\circ 20'$ into radian measure.
Step-by-Step Algorithmic Solution Trace:
- Convert minutes to degrees: $20' = \frac{20}{60}^\circ = \frac{1}{3}^\circ$.
- Express total angle in degrees: $40^\circ 20' = 40\frac{1}{3}^\circ = \frac{121}{3}^\circ$.
- Apply conversion factor (multiply by $\frac{\pi}{180}$): $$\text{Radian measure} = \frac{121}{3} \times \frac{\pi}{180} = \frac{121\pi}{540}$$
Final Evaluation Output: $\frac{121\pi}{540}\text{ radians}$
Mathematical Problem Statement: Find the distance moved by the tip of a $1.5\text{ cm}$ long minute hand of a watch in $40\text{ minutes}$ (Use $\pi = 3.14$).
Step-by-Step Algorithmic Solution Trace:
- Determine the fraction of a full revolution in 40 minutes: $\frac{40}{60} = \frac{2}{3}$ of a revolution.
- Convert the revolution to radians: $\theta = \frac{2}{3} \times 2\pi = \frac{4\pi}{3}$ radians.
- Use the arc length formula $l = r\theta$ with $r = 1.5\text{ cm}$: $$l = 1.5 \times \frac{4\pi}{3} = 2\pi \text{ cm}$$
- Substitute $\pi \approx 3.14$: $l = 2 \times 3.14 = 6.28\text{ cm}$.
Final Evaluation Output: $6.28\text{ cm}$
Mathematical Problem Statement: If $\cos x = -\frac{3}{5}$ and $x$ lies in the third quadrant, find the values of the other five trigonometric functions.
Step-by-Step Algorithmic Solution Trace:
- Find $\sec x$ as the reciprocal of $\cos x$: $\sec x = -\frac{5}{3}$.
- Use the identity $\sin^2 x = 1 - \cos^2 x$ to find $\sin x$: $$\sin^2 x = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \implies \sin x = \pm \frac{4}{5}$$
- Since $x$ lies in the third quadrant, sine is negative: $\sin x = -\frac{4}{5}$.
- Find $\csc x$ as the reciprocal of $\sin x$: $\csc x = -\frac{5}{4}$.
- Calculate $\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \frac{4}{3}$.
- Calculate $\cot x = \frac{1}{\tan x} = \frac{3}{4}$.
Final Evaluation Output: $\sin x = -\frac{4}{5}, \csc x = -\frac{5}{4}, \tan x = \frac{4}{3}, \cot x = \frac{3}{4}, \sec x = -\frac{5}{3}$
Mathematical Problem Statement: Prove that $\frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$.
Step-by-Step Algorithmic Solution Trace:
- Rearrange the numerator: $(\sin 5x + \sin x) - 2\sin 3x$.
- Apply the sum-to-product formula to the numerator group: $$\sin 5x + \sin x = 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2} = 2\sin 3x \cos 2x$$
- Substitute back into the numerator: $2\sin 3x \cos 2x - 2\sin 3x = 2\sin 3x(\cos 2x - 1)$.
- Apply the sum-to-product formula to the denominator: $$\cos 5x - \cos x = -2\sin\frac{5x+x}{2}\sin\frac{5x-x}{2} = -2\sin 3x \sin 2x$$
- Form the ratio: $$\frac{2\sin 3x(\cos 2x - 1)}{-2\sin 3x \sin 2x} = \frac{\cos 2x - 1}{-\sin 2x} = \frac{1 - \cos 2x}{\sin 2x}$$
- Use double angle identities ($1 - \cos 2x = 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$): $$\frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x$$
Final Evaluation Output: $\text{L.H.S.} = \text{R.H.S.} = \tan x$ (Proved)
Historical Chronicle: Indian Origins of Trigonometry
Historical Chronicle: Greek and Western Developments