TRIGONOMETRIC FUNCTIONS-Notes
Maths - Notes
Angles
An angle is formed when a ray revolves about its endpoint in a plane. The fixed endpoint is called the vertex, and the initial position of the ray is known as the initial side, while its position after rotation is called the terminal side. The amount of rotation from the initial side to the terminal side determines the measure of the angle.
In trigonometry, angles are studied not only as geometric figures but also as quantities representing rotation. If the ray rotates anticlockwise, the angle is considered positive, whereas a clockwise rotation represents a negative angle. This convention allows angles of any magnitude, including those greater than one complete revolution.
Angles can be measured using two principal systems: the degree measure and the radian measure. In the degree system, a full rotation is divided into 360 equal parts, each part being one degree. In the radian system, the measure of an angle is defined by the ratio of the length of the arc subtended by the angle at the centre of a circle to the radius of the circle. One complete revolution corresponds to \(2\pi\) radians.
Relation between radian and real numbers
In trigonometry, the radian measure of an angle establishes a natural and precise connection between geometry and real numbers. Unlike the degree measure, which depends on an arbitrary division of a circle, the radian measure arises directly from the geometric properties of a circle, as emphasized in the NCERT approach.
Consider a circle of radius \( r \) with centre at \( O \). Let an angle \( \theta \) at the centre subtend an arc of length \( l \) on the circle. The radian measure of the angle \( \theta \) is defined as the ratio of the length of the arc to the radius of the circle. Thus,
\[ \begin{aligned} \theta &= \frac{\text{length of arc}}{\text{radius}} \\ &= \frac{l}{r} \end{aligned} \]
Since both \( l \) and \( r \) are measured in the same unit, their ratio \( \frac{l}{r} \) is a pure number. Hence, the radian measure of an angle is a real number. This shows that every angle measured in radians corresponds to a real number, and conversely, each real number represents an angle of a certain magnitude in radian measure.
If the length of the arc is equal to the radius of the circle, that is \( l = r \), then
\[ \begin{aligned} \theta &= \frac{r}{r} \\ &= 1 \end{aligned} \]
Therefore, an angle that subtends an arc equal in length to the radius of the circle measures one radian.
For one complete revolution of a circle, the length of the circumference is \( 2\pi r \). Hence, the radian measure of a complete angle is
\[ \begin{aligned} \text{Complete angle} &= \frac{2\pi r}{r} \\ &= 2\pi \end{aligned} \]
This gives the fundamental relation \[ 360^\circ = 2\pi \text{ radians} \]
Identifying angles with real numbers through radian measure is fundamental in NCERT, as it allows trigonometric functions to be defined for all real values and studied using algebraic and graphical methods in higher mathematics.
The relation between degree measures and radian measure of some common angles are given in the following table:
| Degree | \(30^\circ\) | \(45^\circ\) | \(60^\circ\) | \(90^\circ\) | \(180^\circ\) | \(270^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|---|---|
| Radian | \(\dfrac{\pi}{6}\) | \(\dfrac{\pi}{4}\) | \(\dfrac{\pi}{3}\) | \(\dfrac{\pi}{2}\) | \(\pi\) | \(\dfrac{3\pi}{2}\) | \(2\pi\) |
Notational Convention
In trigonometry, angles may be measured either in degrees or in radians. To avoid confusion between these two systems, NCERT adopts a clear and consistent notational convention while representing angles. This convention plays an important role in interpreting trigonometric expressions correctly.
When an angle is written with the symbol \( ^\circ \), such as \( \theta^\circ \), it explicitly indicates that the angle is measured in degrees. Thus, \( \theta^\circ \) represents an angle whose degree measure is \( \theta \). On the other hand, when an angle is written simply as \( \beta \), without any degree symbol, it is understood that the angle is measured in radians. Therefore, the symbol \( \beta \) denotes an angle whose radian measure is \( \beta \).
It is important to note that when angles are expressed in radians, the word “radian” is usually omitted. For example, the angle corresponding to \( 180^\circ \) is written as \( \pi \), and the angle \( 45^\circ \) is written as \( \frac{\pi}{4} \), with the implicit understanding that \( \pi \) and \( \frac{\pi}{4} \) are radian measures. This convention is widely used in trigonometry and higher mathematics.
The relationship between degree measure and radian measure follows directly from the geometry of a circle. Since one complete revolution corresponds to \( 360^\circ \) or \( 2\pi \) radians, we obtain the following fundamental relations:
\[ \begin{aligned} 180^\circ &= \pi \text{ radians}, \\ 1^\circ &= \frac{\pi}{180} \text{ radians}. \end{aligned} \]
Hence, if an angle has a degree measure of \( \theta^\circ \), its radian measure is given by
\[ \begin{aligned} \text{Radian measure} &= \frac{\pi}{180} \times \text{Degree measure}. \end{aligned} \]
Conversely, if the radian measure of an angle is known, its degree measure can be obtained as
\[ \begin{aligned} \text{Degree measure} &= \frac{180}{\pi} \times \text{Radian measure}. \end{aligned} \]
These conventions ensure uniformity in notation and interpretation throughout the study of trigonometric functions. By clearly distinguishing between degree and radian measures and by treating radian measures as real numbers, NCERT lays a strong conceptual foundation for defining trigonometric functions for all real values.
Trigonometric Functions
Consider a circle of radius \( r \) with centre at the origin \( O \) in a Cartesian plane. Let an angle \( \theta \) be measured from the positive direction of the \( x \)-axis, and let the terminal side of the angle intersect the circle at a point \( P(x,y) \). The trigonometric functions of the angle \( \theta \) are then defined in terms of the coordinates of the point \( P \).
The basic trigonometric functions are defined as
\[ \begin{aligned} \sin \theta &= \frac{y}{r}, \\\\ \cos \theta &= \frac{x}{r}, \\\\ \tan \theta &= \frac{y}{x}, \quad (x \neq 0) \end{aligned} \]
From these definitions, the remaining trigonometric functions are obtained as reciprocals:
\[ \begin{aligned} \text{cosec}\ \theta &= \frac{1}{\sin \theta} = \frac{r}{y}, \quad (y \neq 0), \\\\ \sec \theta &= \frac{1}{\cos \theta} = \frac{r}{x}, \quad (x \neq 0), \\\\ \cot \theta &= \frac{1}{\tan \theta} = \frac{x}{y}, \quad (y \neq 0) \end{aligned} \]
These definitions are valid for all angles for which the corresponding ratios are defined. The signs of the trigonometric functions depend on the signs of \( x \) and \( y \), which in turn depend on the quadrant in which the terminal side of the angle lies. This provides a consistent method for determining the values of trigonometric functions for angles greater than \( 90^\circ \), negative angles, and angles exceeding one complete revolution.
An important consequence of these definitions follows from the equation of the circle, \[ x^2 + y^2 = r^2 \] Dividing both sides by \( r^2 \), we obtain
\[ \begin{aligned} \frac{x^2}{r^2} + \frac{y^2}{r^2} &= 1 \\\\ \cos^2 \theta + \sin^2 \theta &= 1 \end{aligned} \]
This identity forms the basis for many other trigonometric identities studied later. Thus, trigonometric functions, as presented in the NCERT textbook, provide a powerful framework for relating angles to real numbers and for analyzing geometric and algebraic problems in a unified manner.
Sign of trigonometric functions
In trigonometry, once angles are extended beyond acute angles, it becomes necessary to adopt a clear sign convention for trigonometric functions. This convention, as followed in the NCERT textbook, is based on the position of the terminal side of an angle in the Cartesian plane and ensures that trigonometric functions are well-defined for all real values of the angle.
Consider a circle of radius \( r \) with centre at the origin \( O \). Let an angle \( \theta \) be measured from the positive direction of the \( x \)-axis, and let its terminal side intersect the circle at a point \( P(x,y) \). The trigonometric functions are defined as
\[ \begin{aligned} \sin \theta &= \frac{y}{r} \\ \cos \theta &= \frac{x}{r} \\ \tan \theta &= \frac{y}{x} \quad (x \neq 0) \end{aligned} \]
Since the radius \( r \) is always positive, the sign of each trigonometric function depends entirely on the signs of the coordinates \( x \) and \( y \). These signs vary according to the quadrant in which the point \( P \) lies.
If the terminal side of the angle lies in the first quadrant, both \( x \) and \( y \) are positive. Therefore,
\[ \begin{aligned} \sin \theta &> 0 \\ \cos \theta &> 0 \\ \tan \theta &> 0 \end{aligned} \]
In the second quadrant, the \( x \)-coordinate is negative while the \( y \)-coordinate is positive. Hence,
\[ \begin{aligned} \sin \theta &> 0 \\ \cos \theta &< 0 \\ \tan \theta &< 0 \end{aligned} \]
In the third quadrant, both \( x \) and \( y \) are negative. Consequently,
\[ \begin{aligned} \sin \theta &< 0 \\ \cos \theta &< 0 \\ \tan \theta &> 0 \end{aligned} \]
Finally, in the fourth quadrant, the \( x \)-coordinate is positive and the \( y \)-coordinate is negative. Thus,
\[ \begin{aligned} \sin \theta &< 0 \\ \cos \theta &> 0 \\ \tan \theta &< 0 \end{aligned} \]
The signs of the remaining trigonometric functions follow directly from their definitions as reciprocals:
\[ \begin{aligned} \text{cosec}\ \theta &= \frac{1}{\sin \theta} \\ \sec \theta &= \frac{1}{\cos \theta} \\ \cot \theta &= \frac{1}{\tan \theta} \end{aligned} \]
Hence, each reciprocal function has the same sign as the corresponding basic trigonometric function wherever it is defined. This systematic sign convention enables us to determine the sign of any trigonometric function for any angle without evaluating its exact value.
Trigonometrical Ratios
Graphs of Trigonometrical Functions
Example-1
If \(\cos x = -\dfrac{3}{5},\; x\) lies in the third quadrant, find the values of other five trigonometric functions.
Solution
We are given that \(\cos x = -\dfrac{3}{5}\) and \(x\) lies in the third quadrant. Using the Pythagorean identity, \(\sin^2 x + \cos^2 x = 1\), we determine the remaining trigonometric functions.
\[ \begin{aligned} \sin^2 x &= 1 - \cos^2 x \\ &= 1 - \left(-\dfrac{3}{5}\right)^2 \\ &= 1 - \dfrac{9}{25} \\ &= \dfrac{16}{25} \end{aligned} \]
Taking square roots, we obtain \[ \begin{aligned} \sin x &= \pm \sqrt{\dfrac{16}{25}} \\ &= \pm \dfrac{4}{5} \end{aligned} \] Since \(x\) lies in the third quadrant, \(\sin x\) is negative, hence \(\sin x = -\dfrac{4}{5}\).
Now, \[ \begin{aligned} \tan x &= \dfrac{\sin x}{\cos x} \\ &= \dfrac{-\frac{4}{5}}{-\frac{3}{5}} \\ &= \dfrac{4}{3} \end{aligned} \]
Therefore, \[ \begin{aligned} \cot x &= \dfrac{1}{\tan x} = \dfrac{3}{4} \\ \sec x &= \dfrac{1}{\cos x} = -\dfrac{5}{3} \\ \csc x &= \dfrac{1}{\sin x} = -\dfrac{5}{4} \end{aligned} \]
Hence, the values of the remaining five trigonometric functions are \(\sin x = -\dfrac{4}{5}\), \(\tan x = \dfrac{4}{3}\), \(\cot x = \dfrac{3}{4}\), \(\sec x = -\dfrac{5}{3}\), and \(\csc x = -\dfrac{5}{4}\).
Example-2
Find the value of \(\sin \dfrac{31π}{3}\).
Solution
We are required to find the value of \(\sin \dfrac{31\pi}{3}\). To simplify the given angle, we use the periodicity property of the sine function, namely \(\sin(\theta + 2n\pi) = \sin \theta\), where \(n\) is an integer.
\[ \begin{aligned} \sin \dfrac{31\pi}{3} \\ = \sin \left( \dfrac{30\pi}{3} + \dfrac{\pi}{3} \right) \\ = \sin \left( 10\pi + \dfrac{\pi}{3} \right) \\ = \sin \left( 5(2\pi) + \dfrac{\pi}{3} \right) \\ = \sin \dfrac{\pi}{3} \\ = \dfrac{\sqrt{3}}{2} \end{aligned} \]
Hence, the value of \(\sin \dfrac{31\pi}{3}\) is \(\dfrac{\sqrt{3}}{2}\).
Example-3
Prove that \(3\sin \dfrac{\pi}{6}\sec\dfrac{\pi}{3}-4\sin\dfrac{5\pi}{6}\cot \dfrac{\pi}{4}=1\)
Solution
We are required to prove that \(3\sin \dfrac{\pi}{6}\sec \dfrac{\pi}{3}-4\sin \dfrac{5\pi}{6}\cot \dfrac{\pi}{4}=1\). We begin with the left-hand side of the given identity.
\[ \begin{aligned} \text{LHS} &= 3\sin \dfrac{\pi}{6}\sec \dfrac{\pi}{3}-4\sin \dfrac{5\pi}{6}\cot \dfrac{\pi}{4} \\ &= 3 \cdot \dfrac{1}{2} \cdot \dfrac{2}{1} - 4 \cdot \sin 150^\circ \cdot 1 \\ &= 3 - 4 \cdot \dfrac{1}{2} \\ &= 3 - 2 \\ &= 1 \end{aligned} \]
Since the left-hand side evaluates to \(1\), which is equal to the right-hand side, the given identity is proved. Hence, \( \text{LHS} = \text{RHS} \).
Example-4
Find the value of \(\sin 15^\circ\).
Solution
We are required to find the value of \(\sin 15^\circ\). Using the sine subtraction formula, \(\sin(A - B) = \sin A \cos B - \cos A \sin B\), we proceed as follows.
\[ \begin{aligned} \sin 15^\circ &= \sin \left(45^\circ - 30^\circ\right) \\ &= \sin 45^\circ \cdot \cos 30^\circ - \cos 45^\circ \cdot \sin 30^\circ \\ &= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} \\ &= \dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}} \\ &= \dfrac{\sqrt{3} - 1}{2\sqrt{2}} \end{aligned} \]
Hence, the value of \(\sin 15^\circ\) is \(\dfrac{\sqrt{3} - 1}{2\sqrt{2}}\).
Example-5
Find the value of \(\tan \dfrac{13\pi}{12}\)
Solution
We are required to find the value of \(\tan \dfrac{13\pi}{12}\). Using the periodicity property of the tangent function, \(\tan(\theta + \pi) = \tan \theta\), we simplify the given expression.
\[ \begin{aligned} \tan \dfrac{13\pi}{12} \\ = \tan \left( \pi + \dfrac{\pi}{12} \right) \\ = \tan \dfrac{\pi}{12} \\ = \tan \dfrac{180^\circ}{12} \\ = \tan 15^\circ \end{aligned} \]
To evaluate \(\tan 15^\circ\), we write \(15^\circ = 45^\circ - 30^\circ\) and apply the tangent subtraction formula \(\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}\).
\[ \begin{aligned} \tan 15^\circ \\ = \dfrac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \\ = \dfrac{1 - \dfrac{1}{\sqrt{3}}}{1 + 1 \cdot \dfrac{1}{\sqrt{3}}} \\ = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} \\ = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ = \dfrac{(\sqrt{3} - 1)^2}{3 - 1} \\ = \dfrac{3 + 1 - 2\sqrt{3}}{2} \\ = 2 - \sqrt{3} \end{aligned} \]
Hence, the value of \(\tan \dfrac{13\pi}{12}\) is \(2 - \sqrt{3}\).
Example-6
Show that \(\tan 3 x \tan 2 x \tan x = \tan 3x – \tan 2 x – \tan x\)
Solution
We are required to show that \(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\). We begin with the left-hand side of the given identity.
\[ \begin{aligned} \tan 3x \cdot \tan 2x \cdot \tan x &= \tan 3x - \tan 2x - \tan x \end{aligned} \]
Using the tangent addition formula, \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A \tan B}\), we write
\[ \begin{aligned} \tan 3x &= \tan(2x+x) \\ &= \dfrac{\tan 2x + \tan x}{1 - \tan 2x \cdot \tan x} \end{aligned} \]
Multiplying both sides by \(1 - \tan 2x \cdot \tan x\), we get \[ \begin{aligned} \tan 3x - \tan 3x \cdot \tan 2x \cdot \tan x &= \tan 2x + \tan x \end{aligned} \]
Rearranging the terms, we obtain \[ \begin{aligned} \tan 3x - \tan 2x - \tan x &= \tan 3x \cdot \tan 2x \cdot \tan x \end{aligned} \]
Hence, \(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\), which proves the given identity.
Example-7
Prove that \(\dfrac{\cos 7x+ \cos 5x}{\sin 7x-\sin 5x}=\cot x\)
Solution
We are required to prove that \(\dfrac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x\). We begin with the left-hand side of the given identity.
\[ \begin{aligned} \text{LHS} &= \dfrac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} \\ &= \dfrac{2\cos \left(\dfrac{7x+5x}{2}\right)\cos \left(\dfrac{7x-5x}{2}\right)} {2\cos \left(\dfrac{7x+5x}{2}\right)\sin \left(\dfrac{7x-5x}{2}\right)} \end{aligned} \]
\[ \begin{aligned} &= \dfrac{\cos 6x \cdot \cos x}{\cos 6x \cdot \sin x} \\ &= \dfrac{\cos x}{\sin x} \\ &= \cot x \end{aligned} \]
Thus, the left-hand side is equal to the right-hand side and hence \(\dfrac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x\). Hence proved.
