Trigonometric Functions Class 11th Notes - Free NCERT Class 11 Maths Chapter 3 notes - Download PDF

NCERT Notes for Class 11 Chapter 3 Trigonometric Functions: Free PDF Download

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Trigonometric Functions Class 11 Notes

These Trigonometric Functions Class 11th Notes, prepared by Careers360 experts, are designed to make your revision easy and efficient.

Trigonometry: Trigon means three sides, and it is a triangle.
And Metry means measurement.

Angle

Angle is the measure of rotation of a ray from its initial point, which is generally denoted by θ.

We have both positive and negative angles. They are represented as follows:

Degree

One degree has 60 minutes, and one minute is further divided into 60 seconds. If a ray rotates from one initial point to another end or terminal point, then it is said to cover $\frac1{360}^{\text{th}}$, then it is said to be 1° (one degree)

Radian

It is another unit to measure angles. The angle measured at the centre forming an arc of 1 unit length in a unit circle with a radius being 1 is called a measure of 1 radian.

Formula:
$\theta=\frac{\text { arc length }}{\text { radius }}=\frac lr$
$⇒ l= rθ$

l = length made by the arc

r = radius of the circle

θ = angle measure.

Relation Between Degree And Radian

To find the radian measure when the degree is given and vice versa.

1 radian = $\frac{180°}{π}$

$\begin{array}{c|c|c}
\text { degrees } & \text { radians } & \text { revolutions } \\
\hline 0 & 0 & 0 \\
30 & \pi / 6 & 1 / 12 \\
45 & \pi / 4 & 1 / 8 \\
60 & \pi / 3 & 1 / 6 \\
90 & \pi / 2 & 1 / 4 \\
120 & 2 \pi / 3 & 1 / 3 \\
135 & 3 \pi / 4 & 3 / 8 \\
180 & \pi & 1 / 2 \\
225 & 5 \pi / 4 & 5 / 8 \\
270 & 3 \pi / 2 & 3 / 4 \\
315 & 7 \pi / 4 & 7 / 8 \\
360 & 2 \pi & 1
\end{array}$

Conventional Measure

$
\text { Radian measure }=\left(\frac{\pi}{180^{\circ}}\right) \times \text { degree value }
$
$
\text { Degree measure }=\left(\frac{180^{\circ}}{\pi}\right) \times \text { radian value }
$

Trigonometric Functions

$\begin{aligned}
& \sin \theta=\frac{\text { opposite side }}{\text { hypotenuse }} \\
& \operatorname{cosec} \theta=\frac{\text { hypotenuse }}{\text { opposite } \operatorname{side}}=\frac{1}{\sin \theta} \\
& \cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }} \\
& \sec \theta=\frac{\text { hypotenuse }}{\text { adjacent } \operatorname{side}}=\frac{1}{\cos \theta} \\
& \tan \theta=\frac{\text { opposite side }}{\text { adjacent } \operatorname{side}}=\frac{\sin \theta}{\cos \theta} \\
& \cot \theta=\frac{\text { adjacent side }}{\text { opposite } \operatorname{side}}=\frac{\cos \theta}{\sin \theta}=\frac{1}{\tan \theta}
\end{aligned}$

By the hypotenuse theorem:
$(\text { opposite side })^2+(\text { adjacent side })^2=(\text { hypotenuse })^2$
From the above right-angle theorem:

$(a)^2+(b)^2=(h)^2$

We have a few identities:

$\begin{aligned} & \sin ^2 \theta+\cos ^2 \theta=1 \\ & 1+\tan ^2 \theta=\sec ^2 \theta \\ & 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta\end{aligned}$

Table to be remembered

$\begin{array}{|c|c|c|c|c|c|c|}
\hline \hline \mathrm{A} & \sin \mathrm{~A} & \cos \mathrm{~A} & \tan \mathrm{~A} & \cot \mathrm{~A} & \sec \mathrm{~A} & \operatorname{cosec} \mathrm{~A} \\
\hline \hline 0^{\circ} & 0 & 1 & 0 & \infty & 1 & \infty \\
\hline 30^{\circ} & 1 / 2 & \sqrt{3} / 2 & 1 / \sqrt{3} & \sqrt{3} & 2 / \sqrt{3} & 2 \\
\hline 45^{\circ} & \sqrt{2} / 2 & \sqrt{2} / 2 & 1 & 1 & \sqrt{2} & \sqrt{2} \\
\hline 60^{\circ} & \sqrt{3} / 2 & 1 / 2 & \sqrt{3} & 1 / \sqrt{3} & 2 & 2 / \sqrt{3} \\
\hline 90^{\circ} & 1 & 0 & \infty & 0 & \infty & 1 \\
\hline \hline
\end{array}$

Sign of Trigonometric Functions

From the above diagram, we can find the sign of the trigonometric functions.

Some points to remember

sin(-x) = -sin(x)

cos(-x) = cos(x)

tan(-x) = -tan(x)

cot(-x) = -cot(x)

sec(-x)= sec(x)

cosec(-x) = -cosec(x)

Domain and Range of Trigonometric Functions

Formula:

$\begin{aligned} & \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y \\ & \sin (x-y)=\sin x \cdot \cos y-\cos x \cdot \sin y \\ & \cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y \\ & \cos (x-y)=\cos x \cdot \cos y+\sin x \cdot \sin y \\ & \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \cdot \tan y} \\ & \tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \cdot \tan y} \\ & \cot (x+y)=\frac{\cot x \cdot \cot y-1}{\cot x+\cot y} \\ & \cot (x-y)=\frac{\cot x \cdot \cot y+1}{\cot y-\cot x}\end{aligned}$

Graphs of Trigonometric Functions

Transformation Formulas

$\begin{aligned} & \sin (x+y)+\sin (x-y)=2 \sin x \cdot \cos y \\ & \sin (x+y)-\sin (x-y)=2 \cos x \cdot \sin y \\ & \cos (x+y)+\cos (x-y)=2 \cos x \cdot \cos y \\ & \cos (x-y)-\cos (x+y)=2 \sin x \cdot \sin y \\ & \sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) \\ & \sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) \\ & \cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) \\ & \cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) \\ & \sin 2 x=2 \sin x \cos x \\ & \cos 2 x=\cos ^2 x-\sin ^2 x=1-2 \sin ^2 x=2 \cos ^2 x-1 \\ & \sin 3 x=3 \sin x-4 \sin ^3 x \\ & \cos 3 x=4 \cos ^3 x-3 \cos ^2 x \\ & \tan 3 x=\frac{\left(3 \tan x-\tan ^3 x\right)}{1-3 \tan ^2 x} \\ & \cot 3 x=\frac{3 \cot x-\cot ^3 x}{1-3 \cot ^2 x}\end{aligned}$

With this topic, we conclude the Trigonometric Functions Class 11th Notes.

Trigonometric Functions: Previous Year Question and Answer

Given below are selected previous year question answers for NCERT Class 11 Maths Chapter 3 Trigonometric Functions, collected from various examinations.

Question 1: If $\operatorname{cosec} \theta+\cot \theta=\mathbf{s}$, then what is the value of $\cos \theta$?

Solution:

Given, $\operatorname{cosec} \theta+\cot \theta=\mathbf{s}$
We know, $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin\theta}$
⇒ $\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=s$
⇒ $\frac{1+\cos\theta}{\sin\theta}=s$
Squaring both sides, we get
⇒ $\frac{(1+\cos\theta)^2}{\sin^2\theta}=s^2$
We know, $\cos^2\theta+\sin^2\theta=1$
⇒ $\frac{1+\cos^2\theta+2\cos\theta}{1-\cos^2\theta}=s^2$
Applying componendo and dividendo,
⇒ $\frac{1+\cos^2\theta+2\cos\theta+1-\cos^2\theta}{1+\cos^2\theta+2\cos\theta-(1-\cos^2\theta)}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{1+\cos^2\theta+2\cos\theta+1-\cos^2\theta}{1+\cos^2\theta+2\cos\theta-1+\cos^2\theta}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{2+2\cos\theta}{2\cos^2\theta+2\cos\theta}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{2(1+\cos\theta)}{2\cos\theta(\cos\theta+1)}=\frac{s^2+1}{s^2-1}$
⇒ $\frac{1}{\cos\theta}=\frac{s^2+1}{s^2-1}$
⇒ $\cos\theta=\frac{s^2-1}{s^2+1}$
Hence, the correct answer is $\frac{s^2-1}{s^2+1}$.

Question 2:
$\triangle $KLM is a right-angled triangle. $\angle$M = 90$^{\circ}$, KM = 12 cm and LM = 5 cm. What is the value of $\sec$ L?

Solution:
By Pythagoras' theorem,
KL² = KM² + ML²
⇒ KL = $\sqrt{12^2 + 5^2}$ = $\sqrt {144+69}$ = $\sqrt{169}$ = 13
$\therefore\sec$ L = $\frac{\text{KL}}{\text{ML}}$ = $\frac{13}{5}$
Hence, the correct answer is $\frac{13}{5}$.

Question 3:
If $\frac{\sin \theta}{\cot \theta+\operatorname{cosec} \theta}=1$, then what is the value of $\theta$?

Solution:
$\frac{\sin \theta}{\cot \theta+\operatorname{cosec} \theta}=1$
⇒ $\frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} +\frac{1}{\sin \theta}}=1$
⇒ $\frac{\sin \theta}{\frac{1+\cos \theta}{\sin \theta}}=1$
⇒ $\frac{\sin^2 \theta}{1+\cos \theta}=1$
⇒ $\sin^2 \theta = 1+\cos \theta$
⇒ $1-\sin^2 \theta+\cos \theta = 0$
⇒ $\cos^2 \theta + \cos \theta = 0$ ($\because$ $ \sin^2 \theta+\cos^2 \theta =1$ )
⇒ $\cos \theta(1+ \cos \theta) = 0$
⇒ $\cos \theta= 0$ or $\cos \theta = –1$
⇒ $\theta = 90^{\circ}$ or $180^{\circ}$
Hence, the correct answer is $90^{\circ}$.

Importance of NCERT Class 11 Maths Chapter 3 Notes

NCERT Class 11 Maths Chapter 3 Notes play a vital role in helping students grasp the core concepts of the chapter easily and effectively, so that they can remember these concepts for a long time. Some important points of these notes are:

• Effective Revision: These notes provide a detailed overview of all the important theorems and formulas, so that students can revise the chapter quickly and effectively.
• Clear Concepts: With these well-prepared notes, students can understand the basic concepts effectively. Also, these notes will help the students remember the key concepts by breaking down complex topics into simpler and easier-to-understand points.
• Time Saving: Students can look to save time by going through these notes instead of reading the whole lengthy chapter.
• Exam Ready Preparation: These notes also highlight the relevant content for various exams, so that students can get the last-minute guidance for exams.

NCERT Class 11 Maths Notes – Chapter-Wise Links

For students' preparation, Careers360 has gathered all Class 11 Maths NCERT Notes here for quick and convenient access.

NCERT Exemplar Solutions Subject-Wise

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.

• NCERT Exemplar Class 11 Solutions
• NCERT Exemplar Class 11 Maths
• NCERT Exemplar Class 11 Physics
• NCERT Exemplar Class 11 Chemistry
• NCERT Exemplar Class 11 Biology

Subject-Wise NCERT Solutions

Students can also check these well-structured, subject-wise solutions.

• NCERT Solutions for Class 11 Mathematics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 11 Biology

NCERT Books and NCERT Syllabus

Students should always analyse the latest syllabus before making a study routine. The following links will help them check the syllabus and some reference books.

• NCERT Books Class 11 Maths
• NCERT Syllabus Class 11 Maths
• NCERT Books Class 11
• NCERT Syllabus Class 11