Table of Contents
Last modified on January 29th, 2025
Trigonometric functions, like all mathematical functions, can be classified as even, odd, both even and odd, or neither. These functions are categorized based on their symmetry.
A trigonometric function f(x) is:
Sometimes, trigonometric functions can be even or odd at the same time or neither.
The following table lists the even and odd trigonometric functions:
Cosine and secant functions are symmetric about the y-axis. Mathematically,
Sine, tangent, and their reciprocals (cosecant and cotangent) are symmetric about the origin. Mathematically,
Since the graphs of even trigonometric functions are symmetric with respect to the y-axis, for any two opposite values of x, the function values remain identical.
The graphs of odd trigonometric functions are symmetric with respect to the origin. For any two opposite values of x, the function values have the same magnitude but opposite signs.
Let us now prove why the above-discussed trigonometric functions are even or odd. To prove this, we will use a unit circle.
In the unit circle, the sine of an angle (let θ) corresponds to the y-coordinate of the point where the end side of the angle intersects the circle.
For angle -θ, which is the reflection of θ across the x-axis, the y-coordinate becomes the negative of that for θ.
Thus, sin(-θ) = -sin θ
Since sin(-θ) = -sin θ, the sine function is odd.
In the unit circle, the cosine of an angle (let θ) corresponds to the x-coordinate of the point where the end side of the angle intersects the circle.
Reflecting θ across the x-axis to get -θ does not change the x-coordinate.
Thus, cos(-θ) = cos θ
Since cos(-θ) = cos θ, the cosine function is even.
By the definition of the tangent function,
${\tan \theta =\dfrac{\sin \theta }{\cos \theta }}$
From the properties of sine and cosine, we get
${\tan \left( -\theta \right) =\dfrac{\sin \left( -\theta \right) }{\cos \left( -\theta \right) }}$
= ${\dfrac{-\sin \theta }{\cos \theta }}$
= ${-\tan \theta}$
Since tan(-θ) = -tan θ, the cosine function is odd.
By the definition of the cotangent function,
${\cot \theta =\dfrac{\cos \theta }{\sin \theta }}$
From the properties of sine and cosine, we get
${\cot \left( -\theta \right) =\dfrac{\cos \left( -\theta \right) }{\sin \left( -\theta \right) }}$
= ${\dfrac{\cos \theta }{-\sin \theta }}$
= ${-\cot \theta}$
Since cot(-θ) = -cot θ, the cosine function is odd.
By the definition of the cosecant function,
${\text{cosec}\, \theta =\dfrac{1}{\sin \theta }}$
By using the property of sine, we get
${\text{cosec}\, \left( -\theta \right) =\dfrac{1}{\sin \left( -\theta \right) }}$
= ${\dfrac{1}{-\sin \theta }}$
= ${-\text{cosec}\, \theta }$
Since cosec(-θ) = -cosec θ, the cosine function is odd.
By the definition of the cosecant function,
${\sec \theta =\dfrac{1}{\cos \theta }}$
By using the property of cosine, we get
${\sec \left( -\theta \right) =\dfrac{1}{\cos \left( -\theta \right) }}$
= ${\dfrac{1}{\cos \theta }}$
= ${\sec \theta }$
Since sec(-θ) = sec θ, the cosine function is even.
The difference between two even functions gives an even function, whereas the difference between two odd functions results in an odd function.
Now, let us find whether the function f(x) = cosec x – cot x is even, odd, or neither.
Step 1: Substituting -x into f(x)
f(-x) = cosec(-x) – cot(-x)
Since cosec(-x) = -cosec x and cot(-x) = -cot x
Here, f(-x) = -cosec x – (-cot x) = -cosec x + cot x
Step 2: Comparing f(x) and f(-x)
f(x) = cosec x – cot x
f(-x) = -cosec x + cot x = -(cosec x – cot x)
Now, comparing f(x) and f(-x), we get
f(-x) = -f(x)
Thus, the function is odd.
Find whether the function g(x) = x3 sin2 x cos x is even, odd, or neither.
Given, g(x) = x3 sin2 x cos x …..(i)
Substituting -x into g(x),
g(-x) = (-x)3 sin2(-x) cos(-x)
Since the sine function is odd and the cosine function is even.
Here, (-x)3 = -x3, sin2(-x) = (sin(-x))2 = (-sin x)2 = sin2 x, and cos(-x) = cos x
Now, g(-x) = -x3 sin2 x cos x …..(ii)
Comparing equations (i) and (ii), we get
g(-x) = -g(x)
Thus, g(x) is an odd function.
Determine whether k(x) = tan3 x – cot x even, odd, or neither.
Given, k(x) = tan3 x – cot x …..(i)
Substituting -x into k(x),
k(-x) = tan3(-x) – cot(-x)
Since the tangent and cotangent functions are odd.
Here, tan3(-x) = (tan(-x))3 = -tan3 x and cot(-x) = -cot x
Now, k(-x) = -tan3 x – (-cot x) = -tan3 x + cot x = -(tan3 x – cot x) …..(ii)
Comparing equations (i) and (ii), we get
k(-x) = -k(x)
Thus, k(x) is an odd function.
Verify whether the function n(x) = sin x tan x is an even.
Given, n(x) = sin x tan x …..(i)
Substituting -x into n(x),
n(-x) = sin(-x) tan(-x)
Since the sine and tangent functions are odd.
Here, sin(-x) = -sin x and tan(-x) = -tan x
Now, n(-x) = (-sin x)(-tan x) = sin x tan x …..(ii)
Comparing equations (i) and (ii), we get
n(-x) = n(x)
Thus, n(x) is an even function.
Which of the following trigonometric functions is an odd function, and why?
a) sin x
b) cos x
c) cot2 x
Here, option a) sin x is an odd function as sin(-x) = -sin x
Last modified on January 29th, 2025