Trigonometric Ratios of Allied Angles

Trigonometric Ratios of Allied Angles

Edited By Komal Miglani | Updated on Oct 12, 2024 12:49 PM IST

Allied Angles are a pair of angles whose sum or difference is a multiple of 90 degrees or π/2 radians. 40° and 140° are allied angles because their sum is 180°. So, Allied angles are a pair of angles. In real life, we use the Trigonometric Ratios of Allied Angles to measure the height of a building or mountain.

Trigonometric Ratios of Allied Angles
Trigonometric Ratios of Allied Angles

In this article, we will cover the concept of Trigonometric Ratios of Allied Angles. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Questions based on this topic have been asked frequently in JEE Mains.

Trigonometric Ratios of Allied Angles

When the sum or difference of two angles is either zero or a multiple of 90, then they are called allied angles. e.g., 30 and 60 are allied angles because their sum is 90. Angles x,90±x,180±x, and 360±x, are angles allied to the angle x if measured in degrees.

The Cartesian Plane is divided into four quadrants each comprising 90. Different trigonometric ratios have distinct properties in these quadrants such as all T -ratios being positive in the 1 st quadrant which ranges from 0 to 90 and only sine and cosecant being positive in the 2nd quadrant, tan and cot being positive in 3rd quadrant and cos and sec being positive in 4th quadrant. Thus, when one angle transitions from one quadrant to another, the trigonometric ratio changes its sign or its complementary.
In general, for angle x, allied angles are (n×90±x) or nπ/2±x,n I

Trigonometric Ratios of Allied Angles when n is even

- sin(180θ)=sin(θ)
- cos(180θ)=cos(θ)
- tan(180θ)=tan(θ)
- csc(180θ)=csc(θ)
- sec(180θ)=sec(θ)
- cot(180θ)=cot(θ)
- sin(180+θ)=sin(θ)
- cos(180+θ)=cos(θ)
- tan(180+θ)=tan(θ)
- csc(180+θ)=csc(θ)
- sec(180+θ)=sec(θ)
- cot(1800+θ)=cot(θ)

Trigonometric Ratios of Allied Angles when n is odd

- sin(90θ)=cos(θ)
- cos(90θ)=sin(θ)
- tan(90θ)=cot(θ)
- csc(900θ)=sec(θ)
- sec(90θ)=csc(θ)
- cot(90θ)=tan(θ)
- sin(90+θ)=cos(θ)
- cos(90+θ)=sin(θ)
- tan(90+θ)=cot(θ)
- csc(90+θ)=sec(θ)
- sec(900+θ)=csc(θ)
- cot(90+θ)=tan(θ)

AID TO REMEMBER

1. All the trigonometric functions of a real number of the form 2n(π/2)±x(nI)(i.e. an even multiple of π/2±x) are numerically equal to the same function of x, with sign depending on the quadrant in which terminal side of the angles lies.

For example cos(π+x)=cos(2(π/2)+x)=cos(x), the -ve sign was chosen because (π+x) lies in 3rd quadrant and 'cos' is -ve in third quadrant.
2. All the trigonometric functions of a real number of the form (2n+1)π/2±x(nI) (i.e. an even multiple of π/2±x) is numerically equal to the co-function of x, with sign depending on the quadrant in which terminal side of the angles lies.

Note that 'sin' and 'cos' are co-functions of each other, 'tan' and 'cot' are co-functions of each other, and 'sec' and 'cosec' are co-functions of each other.
For example: sec(π/2+x)=cosec(x), as (π/2+x) lies in the 2 nd quadrant and 'sec' is ve in 2nd quadrant.

Algorithm to Find Trigonometric Ratios of Any Angle in Terms of Those of Positive Acute Angle

Step I: See whether the given angle a is positive or negative if it is negative, make it positive as follows:

sin(X)=sinX,cos(X)=cosX,tan(X)=tanX


Step II: Express the positive angle obtained in step I in the form a=90degn±X where X is an acute angle.
Step III: Determine the quadrant in which the terminal side of the angle lies which determines the sign of the given trigonometric function.
Step IV: In step II, if n is an odd integer, then
sina=±cosX,cosa=±sinX,tana=±cotX,seca=cosec, etc. The sign on RHS will be the sign obtained in step III.
In step II, if n is even integer, then sina=±sinX,cosa=±cosX,tana=±tanX,seca=±secX, etc. The sign on RHS will be the sign obtained in step III.

Summary

The allied angles help us to find the trignometric value of larger angles by breaking them into a multiple of 90 and the remainder being the smaller angle. Allied angles help us in determining the behavior of T-ratios across the quadrants, wave motions, harmonic motions, signal processing, etc. The study of trigonometric allied angles enriches the understanding and application of trigonometry by providing tools to simplify expressions, solve equations, and explore relationships between angles and trigonometric functions.

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Solved Examples Based on Trigonometry Allied Angles

Example 1: The value of

cos3(π8)cos(3π8)+sin3(π8)sin(3π8) is:

[JEE MAINS 2020]
Solution
Using cos(3π8)=sin(π23π8)

=sin(π8)


Now,

cos3π8cos3π8+sin3π8sin3π8=cos3π8sinπ8+sin3π8cosπ8=sinπ28cosπ8(cos2π8+sin2π8)=12(2sinπ8cosπ8)=12sin2π8=122

Example 2: Let f be an odd function defined on the set of real numbers such that for

x0f(x)=3sinx+4cosx
Then f(x) atx =11π6 is equal to:
[JEE MAINS 2014]
Solution

f(x)=3sinx+4cosxf(11π6)=3sin(11π6)+4cos(11π6)f(11π6)=3sin(11π6)+4cos(11π6)f(11π6)=3sin(2ππ6)+4cos(2ππ6)f(11π6)=+3sin(π6)+4cos(π6)f(11π6)=32+432
Hence, the answer is 32+23

Example 3: The value of sin2(1)+sin2(2).+sin2(90) is
Solution

Allied Angles
- sin(90θ)=cos(θ)
- cos(90θ)=sin(θ)

Now,

sin2(1o)+sin2(2).+sin2(90)=sin2(1)+sin2(2o)sin2(44)+sin2(45)+cos2(44)++cos2(1)+cos2(0)
Rearranging

=[sin2(1)+cos2(1)]+[sin2(2)+cos2(2)]++[sin2(44)+cos2(44)]+sin2(45)+cos2(0)=44+12+1=4512
Hence, the answer is 45.5 .

Example 4:The value of sin37cos53+cos53sin37 is
Solution
Allied Angles
- sin(90θ)=cos(θ)
- cos(90θ)=sin(θ)

Now,

=sin37cos53+cos53sin37=sin(9053)cos53+cos(9037)sin37=cos53cos53+sin37sin37=2
Hence, the answer is 2 .

Example 5: The value of the expression

sin(θ)+cos(θ)+cot(θ)×tan(θ) is


Solution
We know, signs of Trigonometric Functions
In the fourth quadrant, the angle is measured as negative, x is positive and y is negative, so only cosθ and secθ are positive.
Since,

sin(θ)=sinθcos(θ)=cosθcot(θ)=cotθ


and tan(θ)=tanθ
So, the given expression becomes

sinθ+cosθ+cotθtanθ=sinθ+cosθ+1
Hence, the answer is sinθ+cosθ+1



Frequently Asked Questions (FAQs)

1. What are Allied angles?

When the sum or difference of two angles is either zero or a multiple of $90^{\circ}$, then they are called allied angles.

2. How do we represent allied angles in general?

In general, for angle $x$, allied angles are $(n \times 90 \pm x)$ or $n * \pi / 2 \pm x, n$ belongs to $I$.

3. What is the value of $\cos (\pi+x)$ ?

$\cos (\pi+x)=\cos (2(\pi / 2)+x)=-\cos (x)$, -ve sign chosen because $(\pi+x)$ lies in 3rd quadrant and 'cos' is -ve in third quadrant.

4. What is the value of $\sec (\pi / 2+x)$ ?

$\sec (\pi / 2+x)=-\operatorname{cosec}(x)$, as $(\pi / 2+x)$ lies in the 2 nd quadrant and 'sec' is -ve in $2 n$ quadrant.

5. $ \text { if } \sin \theta-\cos \theta=\frac{1}{\sqrt{2}} \text { then find the value of } \theta $

$
\begin{aligned}
& \quad \sin \theta-\cos \theta=\frac{1}{\sqrt{2}} \\
& \sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta\right)=\frac{1}{\sqrt{2}} \\
& \sqrt{2} \sin \left(\theta-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\
& \theta-\frac{\pi}{4}=\frac{\pi}{6} \\
& \theta=\frac{5 \pi}{12}
\end{aligned}
$

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