Trigonometric Ratios of Allied Angles

Trigonometric Ratios of Allied Angles

What Are Allied Angles in Trigonometry?

When the sum or difference of two angles is either zero or a multiple of $90^\circ$, the angles are called allied angles.
For example, $30^\circ$ and $60^\circ$ are allied angles because their sum is $90^\circ$.

For an angle $x$ (in degrees), the allied angles include:
$-x$, $90^\circ \pm x$, $180^\circ \pm x$, $270^\circ \pm x$, and $360^\circ \pm x$.

In general form, allied angles can be written as:
$n \times 90^\circ \pm x$ or $n \times \frac{\pi}{2} \pm x$, where $n \in \mathbb{Z}$.

Quadrant-Based Sign Rule for Allied Angles (ASTC Rule)

The Cartesian plane is divided into four quadrants, each spanning $90^\circ$.
The sign of trigonometric ratios depends on the quadrant in which the terminal side of the angle lies:

• First Quadrant ($0^\circ$ to $90^\circ$): All trigonometric ratios are positive

• Second Quadrant ($90^\circ$ to $180^\circ$): $\sin$ and $\csc$ are positive

• Third Quadrant ($180^\circ$ to $270^\circ$): $\tan$ and $\cot$ are positive

• Fourth Quadrant ($270^\circ$ to $360^\circ$): $\cos$ and $\sec$ are positive

This quadrant rule is essential for determining the sign of allied angle ratios.

Trigonometric Ratios of Allied Angles When $n$ Is Even

Allied Angles of the Form $180^\circ \pm \theta$

• $\sin(180^\circ-\theta)=\sin\theta$

• $\cos(180^\circ-\theta)=-\cos\theta$

• $\tan(180^\circ-\theta)=-\tan\theta$

• $\csc(180^\circ-\theta)=\csc\theta$

• $\sec(180^\circ-\theta)=-\sec\theta$

• $\cot(180^\circ-\theta)=-\cot\theta$

Allied Angles of the Form $180^\circ+\theta$

• $\sin(180^\circ+\theta)=-\sin\theta$

• $\cos(180^\circ+\theta)=-\cos\theta$

• $\tan(180^\circ+\theta)=\tan\theta$

• $\csc(180^\circ+\theta)=-\csc\theta$

• $\sec(180^\circ+\theta)=-\sec\theta$

• $\cot(180^\circ+\theta)=\cot\theta$

Trigonometric Ratios of Allied Angles When $n$ Is Odd

Allied Angles of the Form $90^\circ-\theta$

• $\sin(90^\circ-\theta)=\cos\theta$

• $\cos(90^\circ-\theta)=\sin\theta$

• $\tan(90^\circ-\theta)=\cot\theta$

• $\csc(90^\circ-\theta)=\sec\theta$

• $\sec(90^\circ-\theta)=\csc\theta$

• $\cot(90^\circ-\theta)=\tan\theta$

Allied Angles of the Form $90^\circ+\theta$

• $\sin(90^\circ+\theta)=\cos\theta$

• $\cos(90^\circ+\theta)=-\sin\theta$

• $\tan(90^\circ+\theta)=-\cot\theta$

• $\csc(90^\circ+\theta)=\sec\theta$

• $\sec(90^\circ+\theta)=-\csc\theta$

• $\cot(90^\circ+\theta)=-\tan\theta$

Key Rules to Remember Allied Angle Identities

Rule 1: Even Multiple of $\frac{\pi}{2}$

All trigonometric functions of angles of the form
$2n\left(\frac{\pi}{2}\right)\pm x$
are numerically equal to the same function of $x$, with the sign decided by the quadrant.

Example:
$\cos(\pi+x)=-\cos x$
because $(\pi+x)$ lies in the third quadrant, where cosine is negative.

Rule 2: Odd Multiple of $\frac{\pi}{2}$

All trigonometric functions of angles of the form
$(2n+1)\frac{\pi}{2}\pm x$
are numerically equal to the co-function of $x$, with the sign determined by the quadrant.

Co-function pairs:

• $\sin \leftrightarrow \cos$

• $\tan \leftrightarrow \cot$

• $\sec \leftrightarrow \csc$

Example:
$\sec\left(\frac{\pi}{2}+x\right)=-\csc x$
since $\frac{\pi}{2}+x$ lies in the second quadrant, where secant is negative.

Comprehensive Table of Trigonometric Ratios of Allied Angles

Algorithm to Find Trigonometric Ratios of Any Angle Using a Positive Acute Angle

To evaluate the trigonometric ratios of any angle (positive, negative, or greater than $90^\circ$), we follow a systematic algorithm that converts the given angle into a positive acute angle. This method is extremely useful in Class 11 trigonometry, board exams, and competitive exams like JEE.

Step I: Convert a Negative Angle into a Positive Angle

First, check whether the given angle $a$ is negative.

If the angle is negative, use the standard identities:

• $\sin(-x) = -\sin x$

• $\cos(-x) = \cos x$

• $\tan(-x) = -\tan x$

This step ensures that we work with a positive angle while keeping track of the correct sign.

Step II: Express the Angle in the Form $a = n \times 90^\circ \pm X$

Rewrite the positive angle obtained in Step I in the form:

$a = n \times 90^\circ \pm X$

where:

• $X$ is a positive acute angle ($0^\circ < X < 90^\circ$)

• $n$ is an integer

This representation helps identify whether the angle is an even or odd multiple of $90^\circ$, which is crucial for applying allied angle rules.

Step III: Identify the Quadrant of the Angle

Determine the quadrant in which the terminal side of the angle lies.

The quadrant decides the sign of the trigonometric ratio, according to the ASTC rule:

• First quadrant: all ratios positive

• Second quadrant: $\sin$, $\csc$ positive

• Third quadrant: $\tan$, $\cot$ positive

• Fourth quadrant: $\cos$, $\sec$ positive

Step IV: Apply Allied Angle Rules Based on the Value of $n$

Case 1: When $n$ Is an Odd Integer

If $n$ is odd, the trigonometric function converts into its co-function:

• $\sin a = \pm \cos X$

• $\cos a = \pm \sin X$

• $\tan a = \pm \cot X$

• $\sec a = \pm \csc X$

• $\csc a = \pm \sec X$

The sign is taken from the quadrant determined in Step III.

Case 2: When $n$ Is an Even Integer

If $n$ is even, the trigonometric function remains the same function:

• $\sin a = \pm \sin X$

• $\cos a = \pm \cos X$

• $\tan a = \pm \tan X$

• $\sec a = \pm \sec X$

• $\csc a = \pm \csc X$

Again, the sign depends on the quadrant identified in Step III.

Solved Examples Based on Trigonometry Ratios of Allied Angles

Example 1: The value of $\cos^3\left(\frac{\pi}{8}\right)\cdot\cos\left(\frac{3\pi}{8}\right)+\sin^3\left(\frac{\pi}{8}\right)\cdot\sin\left(\frac{3\pi}{8}\right)$ is: [JEE MAINS 2020]

Solution:

Using

$\cos\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{2}-\frac{3\pi}{8}\right)$

$=\sin\left(\frac{\pi}{8}\right)$

Now,

$\cos^3\frac{\pi}{8}\cos\frac{3\pi}{8}+\sin^3\frac{\pi}{8}\sin\frac{3\pi}{8}$

$=\cos^3\frac{\pi}{8}\sin\frac{\pi}{8}+\sin^3\frac{\pi}{8}\cos\frac{\pi}{8}$

$=\sin\frac{\pi}{8}\cos\frac{\pi}{8}\left(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}\right)$

$=\sin\frac{\pi}{8}\cos\frac{\pi}{8}$

$=\frac{1}{2}\left(2\sin\frac{\pi}{8}\cos\frac{\pi}{8}\right)$

$=\frac{1}{2}\sin\frac{2\pi}{8}$

$=\frac{1}{2\sqrt{2}}$

Example 2: Let $f$ be an odd function defined on the set of real numbers such that for $x\ge0$, $f(x)=3\sin x+4\cos x$ Then $f(x)$ at $x=-\frac{11\pi}{6}$ is equal to: [JEE MAINS 2014]

Solution:

$f(x)=3\sin x+4\cos x$

$f\left(-\frac{11\pi}{6}\right)=3\sin\left(-\frac{11\pi}{6}\right)+4\cos\left(-\frac{11\pi}{6}\right)$

$f\left(-\frac{11\pi}{6}\right)=-3\sin\left(\frac{11\pi}{6}\right)+4\cos\left(\frac{11\pi}{6}\right)$

$f\left(-\frac{11\pi}{6}\right)=-3\sin\left(2\pi-\frac{\pi}{6}\right)+4\cos\left(2\pi-\frac{\pi}{6}\right)$

$f\left(-\frac{11\pi}{6}\right)=3\sin\left(\frac{\pi}{6}\right)+4\cos\left(\frac{\pi}{6}\right)$

$f\left(-\frac{11\pi}{6}\right)=\frac{3}{2}+2\sqrt{3}$

Hence, the answer is $\frac{3}{2}+2\sqrt{3}$.

Example 3: The value of $\sin^2(1^\circ)+\sin^2(2^\circ)+\cdots+\sin^2(90^\circ)$ is

Solution:

Using allied angles:

$\sin(90^\circ-\theta)=\cos\theta$

$\cos(90^\circ-\theta)=\sin\theta$

Now,

$\sin^2(1^\circ)+\sin^2(2^\circ)+\cdots+\sin^2(90^\circ)$

$=\sin^2(1^\circ)+\cdots+\sin^2(44^\circ)+\sin^2(45^\circ)$

$+\cos^2(44^\circ)+\cdots+\cos^2(1^\circ)+\cos^2(0^\circ)$

Rearranging,

$=\left[\sin^2(1^\circ)+\cos^2(1^\circ)\right]+\cdots+\left[\sin^2(44^\circ)+\cos^2(44^\circ)\right]$

$+\sin^2(45^\circ)+\cos^2(0^\circ)$

$=44+\frac{1}{2}+1$

$=45.5$

Hence, the answer is 45.5.

Example 4: The value of $\frac{\sin37^\circ}{\cos53^\circ}+\frac{\cos53^\circ}{\sin37^\circ}$ is

Solution:

Using allied angles:

$\sin(90^\circ-\theta)=\cos\theta$

$\cos(90^\circ-\theta)=\sin\theta$

Now,

$\frac{\sin37^\circ}{\cos53^\circ}+\frac{\cos53^\circ}{\sin37^\circ}$

$=\frac{\sin(90^\circ-53^\circ)}{\cos53^\circ}+\frac{\cos(90^\circ-37^\circ)}{\sin37^\circ}$

$=\frac{\cos53^\circ}{\cos53^\circ}+\frac{\sin37^\circ}{\sin37^\circ}$

$=2$

Hence, the answer is 2.

Example 5: The value of $\sin(-\theta)+\cos(-\theta)+\cot(-\theta)\times\tan(-\theta)$ is

Solution:

Using signs of trigonometric functions:

$\sin(-\theta)=-\sin\theta$

$\cos(-\theta)=\cos\theta$

$\cot(-\theta)=-\cot\theta$

$\tan(-\theta)=-\tan\theta$

So, the expression becomes

$-\sin\theta+\cos\theta+\cot\theta\cdot\tan\theta$

$=-\sin\theta+\cos\theta+1$

Hence, the answer is

$-\sin\theta+\cos\theta+1$

List of Topics Related to the Trigonometric Ratios of Allied Angles

This section provides a quick overview of key topics closely connected to trigonometric ratios of allied angles, helping you strengthen concepts and improve problem-solving speed.

NCERT Resources

Given below is the list of NCERT resources, including NCERT Notes, solutions and exemplar solutions for class 11, chapter 3 Trigonometric Functions:

NCERT Class 11 Chapter 3 Trigonometric Functions Notes

NCERT Class 11 Chapter 3 Trigonometric Functions Solutions

NCERT Exemplar Class 11 Chapter 3 Trigonometric Functions

Practice Questions on Trigonometric Ratios of Allied Angles

This section includes exam-oriented practice questions on trigonometric ratios of allied angles to help you apply identities and quadrant rules with confidence. These problems are ideal for concept strengthening, revision, and improving accuracy in board and competitive exams.

Trigonometric Ratios Of Allied Angles - Practice Question

We have provided the list of practice questions based on the following topics: