trigonometria iit

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Note to the Students
The IIT-JEE is one of the hardest exams to crack for students, for a very simple 
reason – concepts cannot be learned by rote, they have to be absorbed, and IIT
believes in strong concepts. Each question in the IIT-JEE entrance exam is meant
to push the analytical ability of the student to its limit. That is why the questions 
are called brainteasers!
Students find Mathematics the most difficult part of IIT-JEE. We understand that 
it is difficult to get students to love mathematics, but one can get students to love 
succeeding at mathematics. In order to accomplish this goal, the book has been 
written in clear, concise, and inviting writing style. It can be used as a self-study
text as theory is well supplemented with examples and solved examples. Wher-
ever required, figures have been provided for clear understanding.
If you take full advantage of the unique features and elements of this textbook, 
we believe that your experience will be fulfilling and enjoyable. Let’s walk
through some of the special book features that will help you in your efforts to 
crack IIT-JEE.
To crack mathematics paper for IIT-JEE the five things to remember are:
1. Understanding the concepts 
2. Proper applications of concepts 
3. Practice
4. Speed 
5. Accuracy
About the Cover Picture
Medieval mathematician and businessman Fibonacci (Leonardo Pisano) posed
the following problem in his treatise Liber Abaci (pub. 1202): 
How many pairs of rabbits will be produced in a year, beginning with a single pair,
if in every month each pair bears a new pair which becomes productive from the 
second month on?
The solution to this problem leads to the recursive sequence, obeying the simple 
rule that to calculate the next term one simply sums the preceding two. The Fibonacci 
numbers are recursively defined by FnFF +1 = FnFF + FnFF –1, where F0FF = 0, F1FF = 1. The first
few are
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, …
The image on the cover is that of a Fibonacci spiral.
Special attention has been paid to present
an engaging, clear, precise narrative in the
layout that is easy to use and designed to
reduce math anxiety students may have.
CLEAR, CONCISE, AND INVITING WRITING
Every new topic or concept starts with de-
fining the concept for students. Related ex-
amples to aid the understanding follow the
definition.
DEFINITIONS
DEFINITION 7.1 Random Experiment A random experiment is an experiment in which
1. The experiment can be repeated any number of times under identical conditions.
2. All possible outcomes of the experiment are known in advance.
3. The actual outcome in a particular experiment is not known in advance.
(1) Rolling of an Unbiased Die
Rolling of an unbiased die is a random experiment in
which all the possible outcomes are 1, 2, 3, 4, 5 and 6, if 
we denote the six faces of the die with the numbers 1, 2,
3, 4, 5 and 6. The faces of the die may also contain dots 
in numbers 1, 2, 3, 4, 5 and 6. In any case, we identify
the faces of a die with the numbers 1, 2, 3, 4, 5 and 6. 
The actual outcome in a particular experiment (rolling
of the die) is the number that appears on the upper-
most face of the die and this is not known in advance. 
This experiment can be performed any number of times 
under identical conditions.
(2) Tossing of an Unbiased Coin
Tossing of an unbiased coin is a random experiment in
which there are only two possible outcomes, namely, Head
(H) and Tail (T). In a particular experiment (tossing of the 
coin), the outcome is not known in advance. This experiment
can also be performed any number of times under iden-
tical conditions and therefore this is a random experiment.
Examples
“Tossing of an unbiased coin till tail appears” is also a random experiment. However, experiments such as 
“measuring the acceleration due to gravity using a compound pendulum” is not a random experiment, since the
experiment cannot be repeated under identical conditions and the possible outcomes are not known in advance.
Throughout this chapter, by a coin or die we always mean an unbiased coin (or a fair coin) or unbiased die (or a
fair die) unless otherwise mentioned.
DEFINITION 7.2 Some of the important definitions are as follows:
1. A set of possible outcomes of an experiment is called an event.
2. Two or more events are said to be exhaustive if the performance of the experiment always
results in the occurance of atleast one of them.
3. Two or more events are said to be mutually exclusive if the occurance of one of the events
prevents the occurance of any of the remaining events.
4. Two or more events are said to be equally likely (or equiprobable) if there is no reason to
expect one of them in preference to others.
Each chapter starts with an opening vignette, defini-
tion of the topic, and contents of the chapter that give
you an overview of the chapter to help you see the 
big picture.
CHAPTER OPENER
Trigonometric Ratios: The
relationships between the 
angles and the sides of a 
right triangle are expressed 
in terms of trigonometric
ratios.
Contents
1.1 Angles and Their 
Measures
1.2 Trigonometric Ratios
1.3 Periodicity and
Variance
1.4 Trigonometric Ratios 
of Compound Angles
1.5 Trigonometric Ratios 
of Multiple and
Submultiple Angles
1.6 Sum and Product 
Transformations
Worked-Out Problems
Summary
Exercises
Answers
Trigonometric Ratios 
and Transformations 1
T
ri
g
o
n
o
m
e
tr
ic
 R
a
ti
o
s 
a
n
d
T
ra
n
sf
o
rm
a
ti
o
n
s
X
Y
y = sin x
–1
1
X
Y
p p 2p 3pO p/2p/2 3p/2 5p/2
Y
Y
O
1
1
p
XX
y = cos x
2pp/2p/2 3p/2 5p/2
O
X
Y y = tan x
X
pp/2p/2 3p/2
0
X
Y
y = cot x
X
p 2pp/2p/2 3p/2 5p/2
Y
2
1
0
1
2
X
p
y = sec x
X
p/2p/2 3p/2
Y
2
1
O
1
2
X
p
y = cosec x
X
2pp/2 3p/2
A. PEDAGOGY 
Examples pose a specific problem
using concepts already presented
and then work through the solution. 
These serve to enhance the students' 
understanding of the subject matter.
EXAMPLESExample 7.3 
Suppose that an integer is picked from among 1 to 20
(both inclusive). What is the probability of picking a 
prime?
Solution: There are 20 outcomes of the experiment
of picking an integer. The primes between 1 and 20 are
2, 3, 5, 7, 11, 13, 17 and 19 and these are 8 in number.
Therefore, 8 are favourable to the event of picking a 
prime and hence the probability of picking a prime is
8
20
2
5
=
Example 7.4 
none of the dice shows 3 (on the upper most face).
Solution: Any outcome of “throwing 8 dice” can be 
expressed as an 8-tuple of integers from 1 to 6 and hence 
the total number of possible outcomes is 68.
An outcome that none of the dice shows 3 can be
expressed as 8-tuple of integers from the 5-element set
{1, 2, 4, 5, 6} and there are 58 such outcomes. Thus, the 
probability that none of the dice shows 3 is
5
6
5
6
8
8
8
= æèç
ö
ø÷
Example 7.5 
Suppose that a bag contains 6 red, 5 black and 4 blue
balls. Find the probability that three balls drawn simul-
taneously are one blue, one black and one red.
Solution: The total number of balls is
6 + 5 + 4 = 15
Out of these 15 balls, 3 balls can be drawn in 15C3 ways.
Therefore, 3 balls can be drawn simultaneously in
15 14 13
1 2 3
455
× ×
× ×
= ways
Drawing one blue, one black and one red ball simulta-
neously can be expressed as a triple (a1, a2, a3), where 
1 £ a1 £ 4, 1 £ a2 £ 5 and 1 £ a3 £ 6. The number of such
tuples is
4 5 6 120´ ´ =
Thus, the probability that 3 balls drawn simultaneously
are one blue, one black and one red is
120
455
24
91
=
Relevant theorems are provided along 
with proofs to emphasize conceptual un-
derstanding rather than rote learning.
THEOREMS DEFINITION 1.8 Two angles are said to be complementary if their sum is a right 
angle. Therefore, for 0 £ q < p/2,pp q and q p/2 pp - q are complementary.q
THEOREM 1.6 Let q and q y be complementary angles. Theny
1. sin q = cos y and cosy q = sin y
2. tan q =cot y and coty q = tan y
3. sec q = cosec y and cosecy q = sec y
PROOF Draw a right-angled triangle OAP with OAP� = °90 and AOP� = q (Figure 1.12). Since the sum of 
the angles of any triangle is 180° and since q + y = 90°, it follows that OPA� = y .
O
q
y
P
90°
FIGURE 1.12 Theorem 1.6.
Some important formulae and con-
cepts that do not require exhaustive 
explanation, but their mention is im-
portant, are presented in this section.
These are marked with a magnifying
glass.
QUICK LOOKQUICK LOOK 4
Unlike the trigonometric functions sine and cosine, the
trigonometric functions secant, tangent, cosecant and
cotangent are not defined on the whole real line. The
functions sec x and tanx are not defined at x = (2n + 1)
p/2, pp n �, since cosx = 0 at these x. Similarly cosecx
and cotx are not defined at x = np,pp n �, since sin x = 0
at these x. The domains of all the trigonometric funct-
ions are collected in Table 1.1 for convenience and for 
quick reference.
Within each chapter the stu-
dents would find problems 
to reinforce and check their 
understanding. This would 
help build confidence as one 
progresses in the chapter.
These are marked with a 
pointed finger.
TRY IT OUT
At the end of every
chapter, a summary is
presented that organ-
izes the key formulae
and theorems in an
easy to use layout. The
related topics are indi-
cated so that one can
quickly summarize a
chapter.
SUMMARY
Example 7.8 
Let S = {1, 2, 3, 4, 5, 6} be the sample space of a random 
experiment. Define P E P s
s E
( ) ( )=
Î
å for any E £ S, where
P
P P
P
P P
( ) .
( ) . ( )
( ) .
( ) . ( )
1 0 1
2 0 2 3
4 0 4
5 0 05 6
=
= =
=
= =
P is a probability function.P
Solution: We have
(1) Positive axiom:
P E P s
s E
( ) ( )= >
Î
å 0
since P(s) > 0 for all s ÎS.
(2) Completeness axiom:
P S P s P P P P P P
s S
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
. . . .
= = + + + + +
= + + +
Î
å 1 2 3 4 5 6
0 1 0 2 0 2 0 44 0 05 0 05
1
+ +
=
. .
(3) Union axiom: For any E1 and E2 ÎÃ(S) with
E1 Ç E2 = f, we have
P E E P s P s P s P E P E
s E E s E s E
( ) ( ) ( ) ( ) ( ) ( )1 2 1 2
1 2 1 2
È = = + = +
Î È Î Î
å å å
since s ÎE1 È E2 if and only if s ÎE1 or s ÎE2 but not
both.
Thus P is a probability function.P
Try it out Let S = {H, T} be the sample space of a random experiment T x and definex
and 
P H P T
P
P S
( ) ( )
( )
( )
= =
=
=
1
2
0
1
f
Then verify that P is a probability function.P
 SUMMARY
7.1 Random experiment: An experiment is called
random experiment if the following conditions are 
satisfied.
(1) The experiment can be repeated any number of 
times under similar conditions.
(2) All possible outcomes of the experiment are 
known in advance
(3) The actual outcome in a particular experiment 
cannot be exactly predicted.
7.2 Sample space and event: The set of all possible 
outcomes of a random experiment is called sample
space of the experiment and any subset of the sample 
space is called an event.
7.3 Exhaustive events: Two or more events are called 
exhaustive events if the performance of the experi-
ment results in the occurrence of at least one of 
these events.
7.4 Mutually exclusive events: Two or more events are 
said to be mutually exclusive if the occurrence of one
of the events prevents the occurrence of any one of 
the other events.
7.5 Equally likely events: Two or more events are said 
to be equally likely (or equiprobable) if there is no
reason to expect one of them in preference to the 
others.
7.6 Probability (classical definition): Suppose in a 
random experiment there are n exhaustive, mutually
exclusive, equally likely outcomes. If m of them are 
favourable to an event E, then the probability P(E)
of E is defined asE
 7.8 Usual probability: The classical probability is also 
called usual probability.
 7.9 Sample points and sample space: Any possible 
outcome of a random experiment is called a sample 
point and the set of all sample points is called
the sample space of the random experiment. An
elementary event means a sample point. Generally 
sample space is denoted by S.
7.10 Finite sample space: A set A is called finite if either
A is an empty set or it is bijective with the set {1,
2, 3, …, n} for some positive integer n. If a sample 
space is finite, then it is called a finite sample space.
7.11 Countably infinite sample space: A set A is called 
countably infinite set if it is bijective with the set 
�+ of all positive integers. If the sample space of a 
random experiment is countably infinite set, then 
the sample space is called countably infinite. For
example, tossing a fair coin till head appears has a
countably infinite sample space.
7.12 Definition: Here afterwards events mean subsets 
of the sample space. If A and B are two events, then 
A È B means at least one of A or B and A Ç B
means both A and B. Impossible event is denoted
by empty set f and a certain event means the entire f
sample space.
7.13 Various events in set theoretical form:
(1) Events E1, E2, …, En are said to be mutually 
exclusive if Ei Ç EjE = f forf i ¹ j.
(2) E1, E2, …, En are called exhaustive events if E1 È 
E2 È � È EnE = S (sample space).
B. WORKED- OUT PROBLEMS AND ASSESSMENT – AS PER IIT-JTT EE PATTERNPP
In-depth solutions are provided to all worked-out problems for students to understand the logic behind and 
formula used.
WORKED-OUT PROBLEMS
Mere theory is not enough. It is also important to practice and test what has been
proved theoretically. The worked-out problems and exercise at the end of each 
chapter are in resonance with the IIT-JEE paper pattern. Keeping the IIT-JEE
pattern in mind, the worked-out problems and exercises have been divided into:
1. Single Correct Choice Type Questions
2. Multiple Correct Choice Type QuestionsTT
3. Matrix-Match Type Questions
4. Comprehension-Type Questions
5. Assertion–Reasoning Type QuestionsTT
6. Integer Answer Type Questions
 WORKED-OUT PROBLEMS
Single Correct Choice Type Questions
1. If (sin q + cosecqc )qq 2 + (cos q + sec qc )q 2 = tan= 2 q + cot2 q + k, 
then k equalsk
(A) 9 (B) 7 (C) 5 (D) 8
Solution:
LHS = sin2 q +q 2 + cosec2 q +q cos2 q +q 2 + sec2 q
= + + + +
= + +
5 1 1
7
2 2
2 2
( cot ) ( tan )
tan cot
q q
q q
Answer: (B) 
2. If sin q +q sin2 q = q 1, then cos12 q +q 3cos10 q +q 3cos8 q +q
cos6 q is equal toq
(A) 0 (B) 2 (C) 1 (D) 4
Solution: Given that sin q + q sin2 q = 1. This implies 
sin q = q cos2 q.
Now
1 2 2
4 2 2 4
2 2 2 2
= +
= +
= +
cos sin
( )
q q
m n m n
m n m n
Answer: (D) 
4. If a sec q = 1 - b tan q andq a2 sec2 q = q 5 + b2 tan2 q, then 
a2b2 + 4a2 = kb2 where value of k is
(A) 3 (B) 4 (C) 5 (D) 9
Solution: (1 - b tan q)qq 2 = a2 sec2 q =q 5 + b2 tan3 q
Therefore
1 2 5
2
2 2 2 2- + = +
=
-
b b b
b
tan tan tan
tan
q q q
q
Multiple correct choice type
questions have four choices
provided, but one or more of 
the choices provided may be 
correct.
MULTIPLE 
CORRECT CHOICE 
TYPE QUESTIONS
1. A = -7/25 and A lies between 450° and 540° then
(A) sin
A
2
4
5
= - (B) cos A
2
3
5
= -
(C) sin
A
2
3
5
=
-
 (D) cos
A
2
4
5
= -
Solution: By hypothesis 450° < A < 540°. This implies
225° < A/2 < 270°. So
sin
cos ( / )A A
2
1
2
1 7 25
2
4
5
= - = + =
Now A/2 lies in the third quadrant. This means
sin
A
2
4
5
=
-
Again
cos
cos ( / )A A
2
1
2
1 7 25
2
3
5
= + = - =
sin cos22
1
2
1 22
1
2
1
2 2
4
2 2
2
2
°
= + -
°
= + -
+
=
-
Answers: (A), (B), (C), (D) 
3. Let cos x + cos y = a, cos 2x + cos 2y2 = b and cos 3x +
cos 3y = c. Which of the following is (are) true?
(A) cos cos2 2 1
2
x y
b
+ = +
(B)
1
4
2 22( ) cos cosa b x y- - =
(C) 2a3 + c = 3a(1 + b)
(D) abc = 0 for all real x and y
Solution:
cos2x + cos2y + 2 cos x cos y = a2 (1.11)
Multiple Correct Choice Type Questions
These are the regular mul-
tiple choice questions with 
four choices provided. Only
one among the four choices
will be the correctanswer.
SINGLE CORRECT 
CHOICE TYPE 
QUESTIONS
COMPREHENSION-TYPE QUESTIONS
Comprehension-type questions consist
of a small passage, followed by three
multiple choice questions. The ques-
tions are of single correct answer type.
These questions are the 
regular “Match the Follow-
ing” variety. Two columns 
each containing 4 subdivi-
sions or first column with
four subdivisions and sec-
ond column with more sub-
divisions are given and the 
student should match ele-
ments of column I to that 
of column II. There can be 
one or more matches.
MATRIX-MATCH TYPE QUESTIONS
1. Match the items of Column I with those of Column II.
Column I Column II
(A) 
sin
cos
cos
cos
2
1 2 1
q
q
q
q+
æ
èç
ö
ø÷ +
æ
èç
ö
ø÷
is equal to (p) cot
q
2
(B) 
sin
cos
cos
cos
2
1 2
1q
q
q
q-
æ
èç
ö
ø÷
-æ
èç
ö
ø÷
equals (q) tan
q
2
(C) 
cot
cot
q
q
-
+
=
1
1
(r) 
1 2
2
- sin
cos
q
q
(D) secq +q tan q = (s)
1
2
1
2
+
-
tan
tan
q
q
Solution: 
(A)
sin
cos
cos
cos
sin cos
cos
2
1 2 1
2
2
12
2
q
q
q
q
q q
q+
æ
èç
ö
ø÷ +
æ
èç
ö
ø÷
=
æ
èç
ö
ø÷ 22 2
2 2 2
2 2
2
2
cos ( / )
sin( / )cos( / )
cos ( / )
q
q q
q
q
æ
èç
ö
ø÷
=
sec tan
sin
cos
cos sin
cos sin
cos
q q
q
q
q q
q q
q
+ =
+
=
+æèç
ö
ø÷
-
=
1 2 2
2 2
2
2
2 2
++
-
=
+
-
sin
cos sin
tan
tan
q
q q
q
q
2
2 2
1
2
1
2
Answer: (D) Æ (s)
2. Match the items of Column I with those of Column II.
Column I Column II
(A) The value of cos2q + q cos2(60° + q)q +
cos2(60° - q) isq
(p)
3
2
(B) cos 20° cos 40° cos 80° (q) 
1
8
(C) sin sin ( ) sin ( )2 2 2120 120q q q+ ° + + ° -
equals
(r)
3
8
(D) sin 20° sin 40° sin 80° is equal to (s)
3
8
Matrix-Match Type Questions
1. Passage: sin(A ± B) = sinA cos B ± cos A sinB and
cos(A ± B) = cos A cos B ∓ sin A sinB. Based on this 
information, answer the following questions.
 (i) sin A + 2 sin 3A + sin 5A is equal to
(A) 4 sin 3Acos2A2 (B) 4 cos 3A sin2A2
(C) 4 sin 3A sin2A2 (D) 4 cos 3 A cos2A2
 (ii) 
sin sin sin sin
cos cos cos cos
A A A A
A A A A
+ + +
+ + +
=
2 4 5
2 4 5
(A) cot 3A (B) tan 3A
(C) 2 cot 3A (D) 2 tan 3A
(iii)
cos cos cos cos
sin sin sin sin
7 3 5
7 3 5
A A A A
A A A A
+ - -
- - +
=
Comprehension-Type Questions
A22 (B) tan 2A22
(C) 2 cot 2A22 (D) 2 tan 2A22
Solution: From the given information, we have
2
2
2
sin cos sin( ) sin( )
cos sin sin( ) sin( )
cos
A B A B A B
A B A B A B
= + + -
= + - -
AA B A B A B
A B A B A B
cos cos( ) cos( )
sin sin cos( ) cos( )
= + + -
= - - +2
 (i) sin sin sin (sin sin ) sin
sin cos sin
A A A A A A
A A A
+ + = + +
= +
=
2 3 5 5 2 3
2 3 2 2 3
2ssin (cos )
sin cos
3 2 1
4 3 2
A A
A A
+
=
Answer: (A)
 (ii) Numerator = + + +
= +
=
(sin sin ) (sin sin )
sin cos sin cos
sin (
A A A A
A A A A
A
5 2 4
2 3 2 2 3
2 3 ccos cos )2A A+
Denominator = + + +
= +
=
(cos cos ) (cos cos )
cos cos cos cos
cos (
A A A A
A A A A
A
5 2 4
2 3 2 2 3
2 3 ccos cos )2A A+
(ii) (cot cot )(cot cot )(cot cot )B C C A A B+ + +
equals
(A) sin A sin B sin C
(B) secA secB sec C
(C) cosecA cosec B cosec C
(D) cos A cos B cos C
 (iii) 
1
1
+ - +
+ + -
=
cos cos cos
cos cos cos
A B C
A B C
(A) cot tan
B C
2 2
 (B) tan cot
B C
2 2
(C) tan cot
A B
2 2
(D) tan cot
C A
2 2
Solution:
 (i) 1 2 2- +sin sin cos cosB C A A
= + - - - +
= + - - +
1
1
2
2
cos [cos( ) cos( )]cos
cos cos( )cos cos(
A B C B C A
A B C A BB C A
A B C B C A
B C
+
= + + - + -
= + - =
)cos
cos cos( )cos( ) cos
cos sin cos
1
1
2 2
2 2 22 2B C+ cos
Answer: (A)
 (ii) (cot cot )(cot cot )(cot cot )B C C A A B+ + +
INTEGER-TYPE QUESTIONS
The questions in this section are nu-
merical problems for which no choices
are provided. The students are required
to find the exact answers to numerical
problems and enter the same in OMR 
sheets. Answers can be one-digit or
two-digit numerals.
These questions check the 
analytical and reasoning 
skills of the students. Two 
statements are provided – 
Statement I and Statement 
II. The student is expected
to verify if (a) both state-
ments are true and if both 
are true, verify if statement 
I follows from statement
II; (b) both statements are
true and if both are true, 
verify if statement II is not 
the correct reasoning for 
statement I; (c), (d) which
of the statements is untrue.
ASSERTION–REASONING TYPE QUESTIONS
In the following set of questions, a Statement I is given 
and a corresponding Statement II is given just below it. 
Mark the correct answer as:
(A) Both Statements I and II are true and Statement II
is a correct explanation for Statement I.
(B) Both Statements I and II are true but Statement II is 
not a correct explanation for Statement I.
(C) Statement I is true and Statement II is false.
(D) Statement I is false and Statement II is true.
1. Statement I: If 0 < q < p/2 and cos 2pp q =q 2(cos q -q
sin q), then tan q q is equal to 1.q
 Statement II: If 0 < a, aa b < p/2 and sin(pp a +a b)bb = 1, then
a +a b =b p/2. pp
Solution: Statement II is clear. Now
cos (cos sin ) cos sin
(cos sin )
2 2
2
2 2q q q q q
q q
= - Þ -
= -
This is true when cos q =q sin q (i.e.,q q =q p/4) in which case pp
tanq =q 1. Suppose cos q ¹ sin q. Therefore
cos sin
sin
q q
p
q
p
q
p
+ =
+æ
èç
ö
ø÷
=
+ =
2
4
1
4 2
This implies that either
sin
a b+
=
2
0
or cos sin
a b a b-æ
èç
ö
ø÷ = -
-
2
3
2
Therefore
a b p
a b p
+ =
-æ
èç
ö
ø÷
=
-
=
-æ
èç
ö
ø÷
2
2
1
3 6
n or tan tan
a b p
a b
p
p
+ =
-
= -2
2 6
n nor
a b p a b p
p
+ = = + -2 2
3
n nor
Now a + b = 2np impliesp
sin 3a = a sin(6np -p 3b) or sin 3bb a = a sin(6np +p 3b - b p)p
In any case sin 3a = -a sin 3b so that sin 3b a +a sin 3b =b 0.
Answer: (A)
3. Statement I: sin(p /18) is a root of 8 6 1 03x x- + = .
Statement II: For any real a a a a, sin sin sin .3 3 4 3= -
Solution: Statement II is a standard formula. Put a = a p/18. pp
Therefore
3 3a a= =and hence sin
Assertion–Reasoning Type Questions
1. In D ABC, sin(B + C -C A) + sin(C +C A - B) + sin(A +
B - C) = k sinA sin B sinC where C k equals .
Solution: A + B + C = p Þ B + C -C A = p - 2A22 , etc.
Therefore
LHS = - + - + -
= + +
=
sin( ) sin( ) sin( )
sin sin sin
sin s
p p p2 2 2
2 2 2
4
A B C
A B C
A iin sinB C
Answer: 4
2. If A + B + C = p, thenpp
sin sin sin cos cos sin2 2 2
2 2 2 2 2 2
A C B
p q
A B C
- + = -
where p + q is .
Solution:
sin sin sin2 2 2
2 2 2
A B C
+ -
= -1 2
2 2 2
cos cos sin
A B C
Therefore p = 1, q = 2.
Answer: 3
3. If 
1
3
1
3tan tan cot cot
cot
q q q q
q
+
-
+
= k then the value
of k isk .
Solution:
1
3
1
3
1
3
3
3
1
tan tan cot cot
tan tan
tan tan
tan tan
q q q q
q q
q q
q q
+
-
+
=
+
-
+
=
- ttan tan
tan tan tan
cot
3
3
1
4
4
q q
q q q
q
+
= =
Answer: 4
Integer Answer Type Questions
For self-assessment, each chapter has
adequate number of exercise prob-
lems where the questions have been
subdivided into the same categories as
asked in IIT-JEE.
EXERCISES
1. The value of sin 36° =
(A) 
( )5 1 10 2 5
8
+ +
(B) 
5 1
4
+
(C)
10 2 5
8
+
(D)
( )5 1 10 2 5
8
- +
2. If a =
+ +
2
1
sin
cos sin
,
q
q q
 then
1
1
- +
+
cos sin
sin
q q
q
is equal to
(A) 1 - a (B) 1 + a (C) a (D) 1/a
6. Suppose that sin sin cos ,3
0
3x x A xmm
n
=
=å m where A0, 
A1, �, Am are constants and An ¹ 0, then the value of ¹ n is
(A) 8 (B) 6 (C) 4 (D) 9
7. If a, b, c and d are smallest positive angles in thed
ascending order such that the sine of each angle is equal
to a positive constant l, then 4 sin(a/ 2) + 3 sin(b/ 2) + 
2 sin(c/ 2) + sin(d/ 2) is equal todd
(A) 2 1 + l (B) 1 + l
( ) 1 + l ( ) 3 1 + l
 EXERCISES
Single Correct Choice Type Questions
1. 
x
A B
A B
A B
A B
n n
=
+
-
æ
èç
ö
ø÷
+
+
-
æ
èç
ö
ø÷
cos cos
sin sin
sin sin
cos cos
Then
(A) x = 0 if n is an odd positive integer
(B) x = tann(A − B)/ 2 if n is an even positive integer
(C) x = 2 cotn(A − B)/ 2 if nis an even positive integer
(D) x = 0 if n is an even positive integer
2. Which of the following statements are true?
(A) If 2
3. q = (m2 − n2)/(m2 + n2), then
(A) tanq =
-m n
mn
2 2
2
(B) cosecq =
+
-
m n
m n
2 2
2 2
(C) sec q =
+m n
mn
2 2
2
 (D) cotq =
+
2
2 2
mn
m n
4. Which of the following are true?
(A) tan 20°+ tan 72°+ tan 88° = tan 20° · tan 72° · tan 88°
(B) tan 51°+ tan 62° + tan 67° = tan 51° · tan 62° · tan 67°
(C) cot 27° + cot 32°+ cot 31°= cot 27° · cos 32° · cot 31°
(D) 3 + tan 40° + tan 80°= 3 tan 40° tan 80°
5
Multiple Correct Choice Type Questions
In each of the following questions, statements are given 
in two columns, which have to be matched. The state-
ments in Column I are labeled asI (A), (B), (C) and 
(D), while those in Column II are labeled asI (p), (q), 
(r), (s) and (t). Any given statement in Column I can I
have correct matching with one or more statements in 
Column II. The appropriate bubbles corresponding to 
the answers to these questions have to be darkened as 
illustrated in the following example.
Example: If the correct matches are (A) ® (p),(s); 
(B) ® (q),(s),(t); (C) ® (r); (D) ®(r),(t); that is if the 
matches are (A) ® (p) and (s); (B) ® (q),(s) and (t); 
(C) ® (r); and (D) ® (r),(t) then the correct darkening 
of bubbles will look as follows:
A
B
C
D
p q r s t
Matrix-Match Type Questions
1. Passage: acosq +q bsinq = c can be
solved when |c | £ a b2 2+ . Based on this answer the 
following questions (i), (ii) and (iii).
 (i) If 0 £ x £ p and sinp x + cos x = 1, then the number 
of values of x is
(A) 1 (B) 2 (C) 4 (D) 0
 (ii) The maximum and minimum values of 
3 cos q +q 4 sinq -q 5 are respectively (q is real)q
(A) 5, - 5 (B) 4, 3 (C) 0, - 10 (D) 4, - 3
 (iii) The maximum and minimum values of 
5 sin q +q 12 cos q + q 13 are respectively (q is real)q
(A) 13
3. Passage: To eliminate a parameter, we need two 
equations involving the parameter. For example, if 
x = rcos q andq y = rsin q, then by squaring and adding, 
we have x2 + y2 = r2. This shows that q is eliminated q
from the given equations.
Answer the following questions (i), (ii) and (iii).
 (i) If tan q + q sin q = a and tan q - q sin q = b, then after 
eliminating q, (a2 - b2)2 is equal to
(A) 4 ab (B) 4ab (C) 16 ab (D) 16ab
 (ii) Eliminating q from the equationsq x = cot q +q tan q
and y = sec q -q cos q, we have
( ) ( ) ( ) ( 2 )2/3 ( 2)2/3
Comprehension-Type Questions
In the following set of questions, a Statement I is given 
and a corresponding Statement II is given just below it.
Mark the correct answer as:
(A) Both Statements I and II are true and Statement II 
is a correct explanation for Statement I
(B) Both Statements I and II are true but Statement II is 
not a correct explanation for Statement I
(C) Statement I is true and Statement II is false
(D) Statement I is false and Statement II is true
1. Statement I: If x y x y acos sin cos sina a b b+ = + = 2
and 2 2 2 1sin / sin / ,a b = then y a a x2 4= -( )
 Statement II:
qequationequation aa
3. Statement I: For all values of q, 2(sin6q +q cos6q)q -
3(sin4q +q cos4q)q = -1
Statement II: x y x y x xy y3 3 2 2+ = + - +( ) ( )
4. Statement I: If tan(q/2) qq = m, then
1 2 2
1
1
1
2-
+
=
+
-
sin ( / )
sin
q
q
m
m
Statement II: sin
tan
tan
2
2
1 2
q
q
q
=
+
and cos
tan
tan
2
1
1
2
2
q
q
q
=
-
+
5. Statement I:
sin
sin
3 11
25
a
a
=
Assertion–Reasoning Type Questions
The answer to each of the questions in this section is a 
non-negative integer. The appropriate bubbles below
the respective question numbers have to be darkened.
For example, as shown in the figure, if the correct answer
to the question number Y is 246, then the bubbles under Y
Y labeled as 2, 4, 6 are to be darkened.Y
X Y Z
0 0 0 0
1 1 1 1
2 2 2
3 3 3
5 5 5 5
4 4 4
W
3
 6. If 2 tan(A + B) = 3 tan A, then sin(2A22 + B) = ksinB,
where k is equal to .
 7. If 3 cos x = 2 cos(x - 2y2 ), then cot(x - y)coty is equal
to .
 8. If sin 2° + sin 4° + sin 6° + � + sin 178° = n cot 1°, then
n is equal to .
 9. Let 0 < a j < p/2 for pp j = 1, 2, 3, �, 8 and cot a 1 × cot a 2�
cot a 8 = 1. If M is the maximum value of cotM a 1 cot a 2 
� cot a 8, then the value of (32) M is M .
10. The value of (1 + cotA - cosecA)(1 + tanA + sec A)
is .
Integer Answer Type Questions
The Answer key at the end of each chapter contains answers to all exercise problems.
ANSWERS
 1. (D)
 2. (C)
 3. (A)
 4. (B)
 5. (D)
 6. (B)
 7. (A)
 8. (C)
 9. (D)
10. (C)
11. (A)
12. (C)
13. (B)
14. (A)
15. (D)
16. (A)
17. (C)
18. (C)
19. (C)
20. (A)
21. (A)
22. (A)
23. (B)
24. (B)
25. (D)
26. (A)
27. (B)
28. (B)
29. (A)
30. (C)
31. (A)
32. (A)
33. (A)
34. (B)
35. (C)
36. (A)
37. (B)
38. (D)
39. (A)
40. (A)
41. (B)
42. (B)
43. (A)
44. (C)
45. (D)
46. (B)
47. (A)
48. (B)
49. (A)
50. (B)
51. (B)
52. (C)
53. (A)
54. (D)
55. (A)
56. (D)
 ANSWERS
Single Correct Choice Type Questions 
Multiple Correct Choice Type Questions
1. (A), (C)
2. (A), (B), (C)
3. (A), (B), (C)
4. (A), (B), (C), (D)
5. (A), (B), (C), (D)
6. (A), (B), (C)
7. (A), (B), (D)
8. (A), (D)
 9. (B), (C)
10. (A), (C), (D)
11. (A), (B), (C)
12. (B), (C)
13. (A), (D)
14. (A), (C), (D)
15. (A), (B), (C), (D)
16. (A), (B), (C), (D)
Matrix-Match Type Questions
1. (A) ® (p), (s), (B) ® (q), (C) ® (r), (D) ® (t)
2. (A) ® (p), (B) ® (p), (C) ® (p), (D) ® (s)
3. (A) ® (r), (B) ® (s), (C) ® (q), (D) ® (p)
4. (A) ® (s), (B) ® (r), (C) ® (q), (D) ® (p)
5. (A) ® (t), (B) ® (p), (C) ® (r), (D) ® (q)
6. (A) ® (p), (B) ® (p), (C) ® (q), (D) ® (q)
1. (i) (B); (ii) (C); (iii) (C)
2. (i) (A); (ii) (B); (iii) (D)
3. (i) (D); (ii) (B); (iii) (A)
Comprehension-Type Questions
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Contents
1 Trigonometric Ratios and TransformationsTT 1
1.1 Angles and Their Measures ............................................................................................................................ 2
1.2 Trigonometric RatiosTT ....................................................................................................................................... 7
1.3 Periodicity and Variance VV ............................................................................................................................... 19
1.4 Trigonometric Ratios of Compound AnglesTT ................................................................................................. 30
1.5 Trigonometric Ratios of Multiple and Submultiple AnglesTT ........................................................................... 36
1.6 Sum and Product Transformations ................................................................................................................ 47
Worked-Out Problems ..................................................................................................................................51
Summary ....................................................................................................................................................... 87
Exercises ....................................................................................................................................................... 89
Answers ........................................................................................................................................................ 97
2 Inverse Trigonometric Functions 99
2.1 Domains and Ranges of Inverse Trigonometric FunctionsTT .......................................................................... 100
2.2 Properties of Inverse Trigonometric Functions ........................................................................................... 102
Worked-Out Problems ................................................................................................................................ 112
Summary ..................................................................................................................................................... 132
Exercises ..................................................................................................................................................... 135
Answers ...................................................................................................................................................... 139
3 Trigonometric EquationsTT 141
3.1 The Least Positive Angles with a Given Trigonometric RatioTT ..................................................................... 142
3.2 General Solution of Equations of the Form sinx = a .................................................................................. 144
3.3 Solutions of Simple Trigonometric EquationsTT ............................................................................................. 148
Worked-Out Problems ................................................................................................................................ 150
Summary ..................................................................................................................................................... 177
Exercises ..................................................................................................................................................... 178
Answers ...................................................................................................................................................... 184
4 Properties of Triangles 185
4.1 Relations Between the Sides and the Trigonometric Ratios ofTT Angles of a Triangle .................................. 186
4.2 Solution of Triangles TT ................................................................................................................................... 201
4.3 Properties of a Triangle ............................................................................................................................... 209
Worked-Out Problems ................................................................................................................................ 232
Summary ..................................................................................................................................................... 287
Exercises ..................................................................................................................................................... 289
Answers ...................................................................................................................................................... 295
Contentsxxii
5 Addition and Scalar Multiplication of VectorsVV 297
5.1 Definition and Classification of VectorsVV ...................................................................................................... 298
5.2 Addition of VectorsVV ..................................................................................................................................... 300
5.3 Multiplication of a Vector by a Scalar VV ......................................................................................................... 309
5.4 The Division Formula .................................................................................................................................. 312
5.5 Components of a Vector VV ............................................................................................................................. 321
5.6 Vector Equation of a Line and a PlaneVV ........................................................................................................ 333
Worked-Out Problems ................................................................................................................................ 340
 Summary ..................................................................................................................................................... 357
Exercises ..................................................................................................................................................... 362
Answers ...................................................................................................................................................... 364
6 Multiplication of VectorsVV 365
6.1 Scalar or Dot Product ................................................................................................................................. 366
6.2 Vector Equations of a Plane and a Sphere VV ................................................................................................. 382
6.3 Vector or Cross ProductVV .............................................................................................................................. 387
6.4 Vector AreasVV ............................................................................................................................................... 398
6.5 Scalar Triple ProductTT ................................................................................................................................... 404
6.6 Vector Triple Product VV .................................................................................................................................. 415
Worked-Out Problems ................................................................................................................................ 419
Summary ..................................................................................................................................................... 475
Exercises ..................................................................................................................................................... 480
Answers ...................................................................................................................................................... 488
7 Probability 489
7.1 Random Experiments and Events ............................................................................................................... 490
7.2 Classical Definition of Probability ............................................................................................................... 491
7.3 Axiomatic Approach to Probability ............................................................................................................. 494
7.4 Independent and Dependent Events ......................................................................................................... 499
7.5 Random Variables and Probability DistributionsVV ........................................................................................ 504
7.6 Theoretical Discrete Distribution ................................................................................................................ 508
Worked-Out Problems ................................................................................................................................511
Summary ..................................................................................................................................................... 556
Exercises ..................................................................................................................................................... 559
Answers ...................................................................................................................................................... 567
8 Inequalities 569
8.1 Introduction ................................................................................................................................................ 570
 Worked-Out Problems ................................................................................................................................ 571
Exercises ..................................................................................................................................................... 590
 Index 593
Trigonometric Ratios: The 
relationships between the 
angles and the sides of a 
right triangle are expressed 
in terms of trigonometric 
ratios.
Contents
1.1 Angles and Their 
Measures
1.2 Trigonometric Ratios
1.3 Periodicity and
Variance
1.4 Trigonometric Ratios 
of Compound Angles
1.5 Trigonometric Ratios 
of Multiple and 
Submultiple Angles
1.6 Sum and Product 
Transformations
Worked-Out ProblemsWW
Summary
Exercises
Answers
Trigonometric Ratios 
and Transformations 1
T
ri
g
o
n
o
m
e
tr
ic
 R
a
ti
o
s 
a
n
d
T
ra
n
sf
o
rm
a
ti
o
n
s
X
Y
y = sin x
–1
1
X
Y
p p 2p 3pO p/2p/2 3p/2 5p/2
Y
Y
O
1
1
p
XX
y = cos x
2pp/2p/2 3p/2 5p/2
O
X
Y y = tan x
X
pp/2p/2 3p/2
0
X
Y
y = cot x
X
p 2pp/2p/2 3p/2 5p/2
Y
2
1
0
1
2
X
p
y = sec x
X
p/2p/2 3p/2
Y
2
1
O
1
2
X
p
y = cosec x
X
2pp/2 3p/2
2 Chapter 1 Trigonometric Ratios and Transformations
Trigonometry is the study of the relation between angles and sides of a triangle. This is an essential branch of 
mathematics useful for the measurement of areas, distances and heights.
1.1 | Angles and Their Measures
Consider a straight line extending indefinitely in both the directions and let O be a point on this straight line. Then the 
point O divides the line into two parts and each of these two parts is called a ray and O is called the vertex of the two
rays. Usually rays are denoted by OA, where O is the vertex and A is any point on the ray (ray is a part of the straight
line). In Figure 1.1, OA and OB are rays. The vertex of a ray is called the origin of the ray or the initial point of the ray.t
FIGURE 1.1 Rays OA and OB.
DEFINITION 1.1 The figure formed by two rays with the common vertex is called an angle and is denoted by 
AOB or AOB,� where O is the common vertex, A is a point on one ray and B is a point on 
the other ray as shown in Figure 1.2.
O
B
FIGURE 1.2 Angle.
A real number is associated with each angle and this number is called the measure of the angle. There are two impor-
tant systems of measurement of an angle: the sexagesimal system (or British system) and the radian system (or circular 
measure system).
DEFINITION 1.2 Sexagesimal System An angle AOB� is called a right angle if OB is perpendicular to OA, 
geometrically. In sexagesimal system a right angle is divided into 90 equal parts called degrees.
Each degree is divided to 60 equal parts called minutes and each minute is divided into
60 equal parts called seconds. One degree, one minute and one second are denoted by 1°,
1¢ and 1², respectively. Therefore,
and
1 90
1 60
60
right angle
1
=
= ¢
¢ = ¢¢
°
°
Figure 1.3 illustrates the sexagesimal system.
DEFINITION 1.3 Radian System The angle subtended at the centre of a circle by an arc whose length is equal
to the radius of the circle is called a radian and is denoted by 1c.
B
90°
FIGURE 1.3 Sexagesimal system.
31.1 Angles and Their Measures
O
1c
A
B
FIGURE 1.4 Radian system.
Figure 1.4 shows the radian system. In the following theorem, we prove that this angle is independent of the radius of 
the circle, we have considered.
THEOREM 1.1 The radian is constant, in the sense that it is independent of the radius of the circle.
PROOF Consider a circle with centre O and radius r. Let A and B be points on the circle such that the 
length of arc AB is equal to the radius r of the circle (Figure 1.5). Then by definition,
AOB 1 radian c� = = 1
Produce AO to meet the circle at C. Then AC is a diameter of the circle and the length of the arc
ABC is equal to half of the circumference of the circle. Therefore,
AOC 2 right angles� = = °180
We know that the angles subtended at the centre of a circle are proportional to the lengths of the
arcs which subtend them. Therefore
AOC
AOB
arc ABC
arc AB
�
� =
Also, we know that the ratio of the circumference and diameter of a circle is always a constant 
and is denoted by p. Therefore, the circumference of any circle is equal to pp pdpp (= 2prpp ), where d is 
the length of any diameter of the circle. From these, we have
180 2 2° = =
AOB�
( )/p pr
r
AOB� = °180 / ,p which is independent of the radius of the circle.
A
O
1c
B
C
 FIGURE 1.5 Theorem 1.1. ■
4 Chapter 1 Trigonometric Ratios and Transformations
QUICK LOOK 1
1. The value of the constant p mentioned in Theo-p
rem 1.1 is approximately 22/7 (= 3.1415…). In fact
p isp not a rational number; nor it is an alge-
braic number in the sense that p is not a root of p
any nonzero polynomial with rational coefficients. 
However, 22/7 or 3.1415… are only approximate
values of p.pp
2. From Theorem 1.1, it follows that
1
180c = °
p
or p radiansp = 180° = 2 right angles
p c = °180
and therefore
A right angle radians= æèç
ö
ø÷ = °
p
2
90
Also,
1
180
° = p radians
These provide us methods to convert the measure-
ment of an angle in sexagesimal system into that in
radian system and vice-versa.
3. Usually angles are measured in radians or degrees. 
For an angle having degree measure q, we write qq°. 
When we do not mention the superscript °, the 
measure is considered to be in radians.
Example 1.1 
Express 40°36¢20² in radian measure.²
Solution:
20
20
60
20
60 60
1
180
36
36
60
¢¢ =
=
´
= æ
èç
ö
ø÷
°
¢ = æèç
ö
ø÷
 minutes
 degree
°°
= æ
èç
ö
ø÷
°6
10
Therefore
40 36 20 40
6
10
1
180
40
6
10
1
180 180
° ¢ ¢¢ = + +æèç
ö
ø÷
°
= + +æ
èç
ö
ø÷
æ
èç
ö
ø÷
p
raadians
radians=
+ +
´
= æ
èç
ö
ø÷
7200 108 1
180 180
7309
32400
p
p
Example 1.2 
Express (5p/6)pp c in degrees.
Solution:
The given radian measure in degrees is
5
6
5
6
180 150
pæ
èç
ö
ø÷ = ×
æ
èç
ö
ø÷
°
= °
c
THEOREM 1.2 Let q be the angle in radians subtended by an arc of length a at the centre of a circle of radius r.
Then
q = a
r
radians
PROOF We have from the definition of a radian,
1 radian = The angle subtended by an arc of length r at the centre of circler
51.1 Angles and Their Measures
Therefore
1
r
 radian = The angle subtended by an arc of length 1 at the centre
a
r
radian = The angle subtended by an arc of length a at the centre
■
Let us consider a unit circle (a circle of unit radius) having centre at C. Let O be a point on the circle. Let CO be the
initial side of an angle. Since the radius of the circle is one unit, the length of an arc of this circle will be the radian 
measure of the angle subtended by it at the centre C. Consider the line AOB, which is the tangent to the circle at O. 
We can consider AOB as the real line with O representing the real number zero, the points towards OB representing 
positive real numbers and those towards OA representing negative real numbers. Now, if the line OB is roped in 
anticlockwise direction along the circle, then every positive real number will correspond to a radian measure and 
conversely. Similarly, if the line OA is roped in clockwise direction along the circle, then each negative real number
will correspond to a radian measure and conversely.Thus, the radian measures and real numbers are same. However, 
regarding angles, note that two different real numbers may represent the same angle; for example, p/2 and 5pp p/2 reprepp -
sent the same angle. In fact for any angle q, in radians, q + 2np represents the angle p q only for any integer q n.
C
1 unit
FIGURE 1.6 Circle of unit radius.
THEOREM 1.3 For any real number x, there exist unique integer n and a real number q such thatq
x n= + £ <2 0 2p q q pand
PROOF Let n be the integral part of x/2p; that is, n is the largest integer which is less than or equal to x/2p. pp
Then
n
x
n£ < +
2
1
p
Therefore
n x n
x n
( ) ( )2 2 2
0 2 2
p p p
p p
£ < +
£ - <
Now, put q = x - 2np. Thenpp
x n= + £ <2 0 2p q q pand
To prove the uniqueness of n and q, let n and m be integers and q and q f be real numbers such thatf
2 2 0 2 0 2n mp q p f q p f p+ = + £ < £ <, and
6 Chapter 1 Trigonometric Ratios and Transformations
Then
| | | |q f p p- = - ³2 2m n m nif ¹
and
| |q f p- < 2
Therefore m = n and q - f. ■
Note: The integer n above can be considered as the number of completed revolutions of the circle in Figure 1.6 to reach x.
Example 1.3 
Suppose that a clock shows half past 3. Find the angle
between the hours hand and the minutes hand in degrees
and radians.
Solution: From the hypothesis, the hours hand is
exactly in middle of 3 and 4 and minutes hand is at 6 
(Figure 1.7). The 12 divisions in the clock totally repre-
sent 360° and hence each division represents 30°. 
Therefore, the angle between the hours hand and the
minutes hand is 30° + 30° + 30°/2 = 75°, which is equal to
75 ´ p/180pp = 5p/12 radians.pp
12
9 3
4
5
6
FIGURE 1.7 Example 1.3.
Example 1.4 
Determine the radius of the circle in which an arc of 
length 16 cm subtends an angle of 80° at the centre of 
the circle.
Solution: If l is the length of the arc and l q is the angleq
subtended at the centre of the circle of radius r, we know, 
from Theorem 1.2, that
q = l
r
radians
We are given here that q = 80° and l = 16 and we have 
to find r. First we have to convert the measure of q into 
radians. We have
q p p= ° = ´ =80 80
180
4
9
Therefore
4
9
16
9 16
4
36
36
7
22
p
p p
= =
=
´
= = ´æ
èç
ö
ø÷
l
r r
r approx.
Example 1.5 
Let a triangle be given such that its angles are in arithmetic
progression and that the ratio of the greatest to the smallest 
angles, in degrees, is p :60. Find all the angles of the triangle.
Solution: Let the angles be a - d, a and a + d, where
d > 0. The sum of the angles in a triangle is 180°. Therefore
3 180
60
a
a
= °
= °
Since d > 0, a - d < a < a + d, from the hypothesis,
we have
a d
a d
d
d
+
-
= +
-
= =p
60
60
60
180
60
3or
Therefore d = 30° and so 30°, 60° and 90° are all the
angles of the given triangle.
71.2 Trigonometric Ratios
1.2 | Trigonometric Ratios
In this section we will define the trigonometric ratios sine and cosine as real-valued functions in the interval [0, 2p]p
first and later extend the domain of these to the entire real line. Using the definitions of these, we define the other
trigonometric functions such as cosecant, secant, tangent and cotangent on appropriate domains.
DEFINITION 1.4 Sine and Cosine Ratios Let q be a real number such that 0q £ q £ 2p. Consider a rectangularpp
coordinate system OXY. Draw a circle with centre at O and radius r > 0 in the coordinate
plane. Choose a point P on this circle such that OP makes an angle of q radians with OX,
measured in the anticlockwise sense (see Figure 1.8).
With reference to the coordinate system OXY, let (x, y) be the coordinates of the point P.
Then, we define the sine ratio of q as
sinq = y
r
and the cosine ratio of q as
cosq = x
r
First note that these ratios sinq and cosq q are independent of the circle and its radiusq r. Consider Figure 1.9 in which
two circle C1 and C2CC of radius r1 and r2 are given and the points P(x1, y1) and Q(x2, y2) lie on these circles, respectively, 
such that O, P and Q are collinear.
y
xO
X X
X X
q
Y Y
Y Y
P(x, y)
O
q
P(x, y)
Oq
P(x, y)
O
q
P(x, y)
FIGURE 1.8
8 Chapter 1 Trigonometric Ratios and Transformations
X
q
NMO
Q
Y
r2
P
r1
r1 y2
y1
FIGURE 1.9
Let M and N be feet of the perpendiculars drawn from P and Q, respectively, on OX. Then, PM = y1, QN = y2.
The triangles OPM and OQN are similar and hence
PM
OP
QN
OQ
=
Therefore
y
r
y
r
1
1
2
2
=
Thus, sinq is independent of the circle and its radius. Similarly, cosq q is also independent of the circle and its radius.q
QUICK LOOK 2
1. If P(x, y) is a point on the circle of radius r with
centre at the origin O such that OP makes an angle 
of q radians with the ray OX in the anticlockwise q
sense, then we have
x r y r= =cos and sinq q
2. For any 0 £ q £ 2p, we havepp
sin cos2 2 1q q+ =
and hence
and
sin cos
cos sin
2 2
2 2
1
1
q q
q q
= -
= -
where sin2 q = sinq × sinq and cosq 2 q = cos q × cosq.
3. If P(x, y) is a point on the unit circle (circle of 
radius 1) with centre at the origin O and OP makes 
an angle q radians with the X-axis, thenq
x y= =cos andq qsin
4. For any 0 £ q £ 2p,pp
sin (since )
P lies on the X-axis
or radians
q
q p
= Û = >
Û
Û =
0 0 0
0
y r
x y( , )
and
cos
P lies on the Y-axis
or
q
q
p p
= Û =
Û
Û =
0 0
2
3
2
x
x y( , )
DEFINITION 1.5 Let x be a real number. Then there exist unique integer n and a real number q such thatq
x n= + £ <2 0 2p q q pand
91.2 Trigonometric Ratios
(see Theorem 1.3). Now, we define
sin sinx = q
and cos x = cosq
In other words, we are defining
sin sin( )2np q q+ =
and cos(2np + q)q = cos q
for any integer n and real number q such that 0q £ q < 2p.pp
QUICK LOOK 3
1. Sine (sin) and cosine (cos) are real-valued func-
tions defined on the entire real line and both are of 
period 2p, that is,pp
sin sin( )q p q+ =2
and
cos cos( )q p q+ =2
for all real numbers.
2. sin = 0 Û x = np for some integer p n and cos x = 0 Û
x = (2n + 1) p/2 for some integerpp n.
DEFINITION 1.6 Secant and Tangent For any real number x ¹ (2n + 1)p/2, pp n Î�, we define
secant
cos
and tangent
sin
cos
x
x
x
x
x
= =1
and denote secant x and tangent x simply by secx and tan x; that is,
sec
cos
and tan
sin
cos
x
x
x
x
x
= =1
for any real number x which is not of the form (2n + 1)p/2, wherepp n is an integer.
DEFINITION 1.7 Cosecant and Cotangent For any real number x ¹ np, pp n Î�, we define
cosecant
sin
and cotangent
cos
sin
x
x
x
x
x
= =1
and denote cosecant x and cotangent x simply by cosec x and cotx; that is,
cosec
sin
and cot
cos
sin
x
x
x
x
x
= =1
for any real number x which is not of the form np, where pp n is an integer.
QUICK LOOK 4
Unlike the trigonometric functions sine and cosine, the
trigonometric functions secant, tangent, cosecant and
cotangent are not defined on the whole real line. The
functions secx and tanx are not defined at x = (2n + 1)
p/2,pp n �, since cos x = 0 at these x. Similarly cosec x
and cot x are not defined at x = np,pp n �, since sinx = 0 
at these x. The domains of all the trigonometric funct-
ions are collected in Table 1.1 for convenience and for 
quick reference.
10 Chapter 1 Trigonometric Ratios and Transformations
Table 1.1 Domains of all the trigonometric functions
Function Domain
sin x �
cosx �
cosecx � - {np |n �}
secx � �- + Îìí
î
ü
ý
þ
( ) |2 1
2
n n
p
tanx � �- + Îìí
î
ü
ý
þ
( ) |2 1
2
n n
p
cotx � �- Î{ | }n np
We have discussed earlier about all the values of x for which the values of each of the trigonometric ratios sin x, cos x,
tanx, cosecx and cotx are defined. In the following paragraphs, we discuss the values of x for which these are positive
and those values of x for which they are negative.
QUICK LOOK 5
Let q be a real number such that 0q < q < 2p, pp q ¹ p/2,pp
q ¹ p and p q ¹ 3p/2. Consider a rectangular coordinatepp
system OXY, as in Definition 1.4, with O as origin. Let 
r be a positive real number. Choose a pointP in the r
coordinate plane such that OP = r and OP makes an r
angle of q radians with OX measured in the anticlock-q
wise sense. Let (x, y) be the coordinates of the point P. 
We have the following important observations. Recall
that both x and y are nonzero since q Ï{0, p/2,pp p, 3pp p/2}.pp
1. If 0 < q < p/2, then P lies in the first quadrant of the pp
plane and hence x > 0 and y > 0, so that
sin cos
tan sec
cot cosec
q q
q q
q q
= > = >
= > = >
= > = >
y
r
x
r
y
x
r
x
x
y
r
y
0 0
0 0
0
;
;
; 00
tions are positive at q when 0q £ q £ p/2. In this casepp
we say that q lies in the first quadrant.
2. If p/2pp < q < p, then P lies in the second quadrant of pp
the plane and hence x < 0 and y > 0, so that sin q > 0
and cosecq > 0 and all others are negative. In this 
case, we say that q lies in the second quadrant.q
3. If p < q < 3p/2, then P lies in the third quadrant of pp
the plane and hence x < 0 and y < 0, so that tanq > 0 
and cotq > 0 and all others are negative. In this case, 
q is said to be in the third quadrant.q
4. If 3p/2pp < q £ 2p,pp then P lies in the fourth quadrant
of the plane and hence x > 0 and y < 0, so that cosq
and secq are positive and all the others are negative. 
In this case, q is said to be in the fourth quadrant. 
The angles 0, p/2, pp p, 3pp p/2 and 2pp p are called the p
quadrant angles.
5. The signs (positive or negative) of the trigonometric 
ratios in the four quadrants can be remembered as 
shown below.
I II III IV
All Silver Tea Cups
· In quadrant I, all are positive.
· In quadrant II, only sine (and its reciprocal cose-
cant) is positive.
· In quadrant III, only tangent (and its reciprocal 
cotangent) are positive.
· In quadrant IV, only cosine (and its reciprocal
secant) is positive.
Now, we shall prove certain basic identities satisfied by the trigonometric functions. For convenience, we write 
f 2(x) for (f(ff x))2, where f is any function.f
THEOREM 1.4 1. sin2 x + cos2 x = 1 for all x �
2. 1 + tan2 x = sec2 x for all x � - {(2n + 1) p/2pp |n �}
3. 1 + cot2 x = cosec2 x for all x � - {np |n �}
111.2 Trigonometric Ratios
PROOF Let x Î� and x = 2np + q, where n is an integer and q is a real number such that 0q £ q < 2p.pp
Consider a rectangular coordinate system OXY with O as origin. Choose a point P(x, y) in the
coordinate plane such that OP = 1 and OP makes an angle of q radians with OX measured inq
anticlockwise sense (Figure 1.10).
O
Y
P(x, y)
X
y
x
q
1
FIGURE 1.10 Theorem 1.4.
1. By Definition 1.4,
sinq = =y y
OP
and
cos
OP
q = =x x
Since x2 + y2 = (OP)2 = 1, it follows that
sin cos2 2 1q q+ =
Now,
sin cos sin cos
sin cos
2 2 2 2
2 2
2 2
1
q q p q p q
q q
+ = + + +
= +
=
( ) ( )n n
2. Let x Î� - {(2n + 1) p/2pp |n Î�}. Then cosx ¹ 0 and hence tanx and secx are defined
1 1
1
2
2
2
2 2
2
2
2
+ = +
=
+
= =
tan
sin
cos
cos sin
cos
cos
sec
x
x
x
x x
x
x
x
3. Let x Î� - {np |n Î�}. Then sinx ¹ 0 and cosec x and cotx are defined and
1 1
1
2
2
2
2 2
2
2
2
+ = +
=
+
= =
cot
cos
sin
sin cos
sin
sin
cosec
x
x
x
x x
x
x
x
■
COROLLARY 1.1 1. For any x � - {(2n + 1) p/2pp |n �},
( )( )sec tan sec tan sec tanx x x x x x- + = - =2 2 1
that is, secx - tanx and secx + tan x are reciprocal to each other.
12 Chapter 1 Trigonometric Ratios and Transformations
2. For any x � - {np |n �},
( )( )cosec cot cosec tan cosec cotx x x x x x- + = - =2 2 1
that is, cosecx - cot x and cosec x + cot x are reciprocal to each other.
COROLLARY 1.2 1. |sinx| £ 1 and |cosx| £ 1 for all x Î�
2. |cosecx| ³ 1 for all x Î� - {np |n Î�} and |secx| ³ 1 for all x Î� - {(2n + 1) p/2pp |n Î�}
PROOF This is an immediate consequence from the identity sin2 x + cos2 x = 1. ■
Try it out Express all the trigonometric ratios of an angle in terms of parts (1) or (2) of Corollary 1.2.
In the following examples, we have expressed all the trigonometric ratios of an angle in terms of sine and in terms
of cotangent.
Example 1.6 
Express each trigonometric ratios of an angle q in termsq
of sinq.
Solution: Since sin2 q + cos2 q = 1, we have
cos
sec
cos
q q
q
q q
= -
= =
-
1
1 1
1
2
2
sin
sin
cosec
sin
sin
q
q
q q
q
q
q
q q
q
q
=
= =
-
= =
-
1
1
1
2
2
sin
tan
sin
cos
sin
cot
cos
sin ssinq
All the above are valid for approximate values of q.
Example 1.7 
Express all trigonometric ratios of an angle in terms of cotq.qq
Solution: From part (3) of Theorem 1.4, we have, for
all approximate q,
cosec cot
cosec cot
q q
q
q q
= +
= =
+
1
1 1
1
2
2
sin
cos sin
cot
tan
cot
sec
cos
q q
q
q
q
q
q
q
q
= - = -
+
=
+
=
= =
+
1 1
1
1 1
1
1 1
2
2 2cot cot
ccot
cot
2 q
q
Example 1.8 
If cos / ,q = 2 2 3 find the numerical magnitude of all
trigonometric ratios of q.
Solution: Using cos / ,q = 2 2 3 we get
sin
sin
q q
q
q
= - = -
æ
èç
ö
ø÷
=
= =
1 1
2 2
3
1
3
1
3
2
2
cos
cosec
sec
cos
tan
sin
cos
cot
tan
q
q
q q
q
q
q
= =
= =
= =
1 3
2 2
1
2 2
1
2 2
131.2 Trigonometric Ratios
Inter-Relationships Among Trigonometric Ratios
In Table 1.2, we display formulae expressing each of the trigonometric ratios in terms of any other ratio.
Table 1.2 Relationship among trigonometric ratios
Function sin q cos q cosec q sec q tan q cot q
sinq sinq 1 2- sin q
1
sinq
1
1 2- sin q
sinq
q1 2- sin
1 2- sin q
qsin
cosq 1 2- cos q cos q 1
1 2- cos q
1
cosq
1 2- cos q
qcos
cosq
q1 2- cos
cosecq 1
cosecq
cosec
cosec
2 1q
q
- cosecq
cosec
cosec
q
q2 1-
1
12cosec q -
cosec2 1q -
secq sec
2 1q
q
-
sec
1
secq
sec q
qsec2 1-
secq sec2 1q - 1
12sec q -
tanq
tanq
q1 2+ tan
1
1 2+ tan q
1 2+ tan
tan
q
q
1 2+ tan q tanq
1
tanq
cotq
1
1 + 2cot q
cotq
q1 + 2cot
1 + 2cot q 1 +
2cot q
qcot
1
cotq
cotq
Example 1.9 
If tan q + secq = 3/2, evaluate the value of sec q.
Solution: First, express tanq and secq q in terms of sinq q.
We have
tan
sin
secq q
q
q
q
=
-
=
-1
1
12 2sin
and
sin
Put sin q = x. We are given that
tan secq q+ = 3
2
Therefore
x
x x1
1
1
3
22 2-
+
-
=
4 1 9 1
13 8 5 0
13 5 1 0
1
5
13
2 2
2
( ) ( )
( )( )
x x
x x
x x
x
+ = -
+ - =
- + =
= - or
If x = -1, then cos q = 0 (since cos2 q + sin2 q = 1) and
hence tan q and secq q are not defined. Therefore q
x = 5/13; that is, sinq = 5/13.
Now, let us evaluate the trigonometric ratios of certain angles and prepare a table useful for ready reckening. First
recall the following in degrees and radians.
Radians
p
6
p
4
p
3
p
2
2
3
p 3
4
p 5
6
p p 2p
Degrees 30° 45° 60° 90° 120° 135° 150° 180° 360°
14 Chapter 1 Trigonometric Ratios and Transformations
THEOREM 1.5 1. sin sin30
6
1
2
° = =p
2. sin sin45
4
1
2
° = =p
3. sin sin60
3
3
2
° = =p
4. sin sin90
2
1° = =p
5. sin0° = sin0 = 0
PROOF 1. Consider a rectangular coordinate plane OXY with origin O. Let P(a, b) be a point such that
OP makes an angle of 30° with OX in anticlockwise sense. Draw a perpendicular through
P to OX to meet at M. Then PM = b and OM = a. Produce PM to Q such that PM = MQ.
Now, by the properties of triangles, OPQ is an equilateral triangle and PQ = 2b and hence
OP = 2b = OQ. Therefore
sin 30
2
1
2
° = =b
b
This is graphically illustrated in Figure 1.11.
P
2b
30°
2b
M
Q
O
Y
X3
FIGURE 1.11 Theorem 1.5 part (1).
By similar techniques, we can prove (2) through (5) also. ■
Using the values of sine given in Theorem 1.5 and Table 1.2, we can evaluate the values of other trigonometric ratios 
also at these angles.
DEFINITION 1.8 Complementary Angles Two angles are said to be complementary if their sum is a right
angle. Therefore, for 0 £ q < p/2,pp q andq p/2pp - q are complementary.q
THEOREM 1.6 Let q andq y be complementary angles. Theny
1. sinq = cos y and cosy q = siny
2. tanq = cot y and coty q = tany
3. secq = cosecy and cosecy q = secy
PROOF Draw a right-angled triangle OAP with OAP� = °90 and AOP� = q (Figure 1.12). Since the sum of 
the angles of any triangle is180° and since q + y = 90°, it follows that OPA� = y .
O
q
y
P
90°
FIGURE 1.12 Theorem 1.6.
151.2 Trigonometric Ratios
If O is treated as origin and the line OA as the X-axis, then the x-coordinate of P is OA and the 
y-coordinate of P is PA. By the definition of the ratio sin q, we have
sin cosq q= =PA
OP
and
OA
OP
Also, when the angle y is considered, PA becomes the base and OA becomes the perpendicular y
and hence
sin cosy y= =OA
OP
and
PA
OP
And therefore (1) is proved. Similarly (2) and (3) can be proved. ■
COROLLARY 1.3 For any 0 £ q < p/2,pp
1. sin cos and cos sinq p q q p q= -æèç
ö
ø÷ = -
æ
èç
ö
ø÷2 2
2. tan cot and cot tanq p q q p q= -æèç
ö
ø÷ = -
æ
èç
ö
ø÷2 2
3. sec cosec and cosec secq p q q p q= -æèç
ö
ø÷ = -
æ
èç
ö
ø÷2 2
Try it out Prove parts (1)-(3) of Corollary 1.3.
Thus, we have that
1. The sine of any angle is the cosine of its complement.
2. The tangent of any angle is thet cotangent of its complement.t
3. The secant of any angle is thet cosecant of its complement.t
For these reasons, sine and cosine, tangent and cotangent, and secant and cosecant can be remembered as complementary.
From these observations, we can construct Table 1.3 containing the trigonometric ratios of certain important
angles. The reader is advised before proceeding any further to table that if he/she can memorize accurately the portion
of the table included between thick lines, then one should be able to reproduce the rest easily.
Table 1.3 
Ratios
Degree 0° 30° 45° 60° 90° 180° 270° 360°
Radians 0
p
6
p
4
p
3
p
2
p 3
2
p
2p
sin 0
1
2
1
2
3
2
1 0 -1 0
cos 1 3
2
1
2
1
2
0 -1 0 1
cosec Not defined 2 2
2
3
1 Not defined -1 Not defined
sec 1 2
3
2 2 Not defined -1 Not defined 1
tan 0 1
3
1 3 Not defined 0 Not defined 0
cot Not defined 3 1
1
3
0 Not defined 0 Not defined
16 Chapter 1 Trigonometric Ratios and Transformations
Next, we consider variations and periodicity of the trigonometrical ratios. In fact, we have defined sin x and cos x for 
all real numbers x, secx and tanx for all x � - {(2n + 1)p/2, pp n �}, and cosecx and cotx for all x � -{np |n �}.
By the very definition of these functions, it follows that
f x f x( ) ( )2p + =
for all trigonometric functions f. Now, we have the following theorem.ff
THEOREM 1.7 For all real numbers x,
sin sin and cos cos( ) ( )- = - - =x x x x
PROOF The theorem is evidently true when x = 0 or p/2 or pp p or 3p p/2 or 2pp p. Letpp x be any real number. Then 
we can write x = 2np + q, where 0 £ q < 2p andp n ÎZ.
In view of the above remark, we can assume that 0 £ q < 2p andp q ¹ p/2, pp p, 3pp p/2. Consider app
rectangular coordinate system OXY with O as origin. Draw a unit circle in the coordinate plane 
with the origin as the centre. Choose a point P on the unit circle such that XOP� measured in anti-
clockwise sense is q radians. Figureq 1.13 illustrates the cases for each of the four quadrants. Draw
a perpendicular from P to OX to meet the circle at Q.
In each of the four cases, the X-coordinates of P and Q are same and their Y-coordinates are 
same in absolute value but with different sign. Now, we have
sin sin
Y-coordinate of Q
( ) ( )- = +
=
q q p2
P
X
Y
Q
O
q
P
X
Y
Q
O
q
P
X
Y
Q
Oq
X
Y
O
q
P
Q
FIGURE 1.13 Theorem 1.7.
171.2 Trigonometric Ratios
= -
= -
( )
sin
Y-coordinate of P
q
and cos cos
X-coordinate of Q
X-coordinate of P
cos
( ) ( )- = +
=
=
=
q q p
q
2
Next, we have x = 2np + q and 2q np £ x < 2(n + 1)p (recall thatp n is the integral part of x/2p).p
Put f = 2(n + 1)p - x. Then 0 < f £ 2p. From the above argument, we havepp
sin sin and cos cos( ) ( )- = - - =f f f f
Therefore
sin sin
sin
sin
sin
sin
( ) [ ( ) ]
( )
[ ( ) ]
- = + -
=
= - -
= - + - -
= -
x n x
x n
2 1
2 1
p
f
f
p
xx
and
cos cos
cos
cos
cos
cos
( ) [ ( ) ]
( )
[ ( ) ]
- = + -
=
= -
= + - -
=
x n x
x n
x
2 1
2 1
p
f
f
p
■
Argument similar to the one given above can be used to prove the following also.
THEOREM 1.8 The following hold for any real number x.
1. sin(p - x) = sinx and cos(p - x) = -cosx
2. sin(p + x) = -sinx and cos(p + x) = -cosx
3. cos sin and sin cos 
p p
2 2
-æèç
ö
ø÷ = -
æ
èç
ö
ø÷ =x x x x
4. cos sin and sin cos
p p
2 2
+æèç
ö
ø÷ = - +
æ
èç
ö
ø÷ =x x x x
Try it out Prove Theorem 1.8.
DEFINITION 1.9 Supplementary Two angles q and q f are said to bef supplementary if q + f = p.pp
(1)
p
3
and 
2
3
p
 are supplementary angles. (2) 45° and 135° are supplementary angles, as sinq = sinf
and cosq = -cos f.
Examples
18 Chapter 1 Trigonometric Ratios and Transformations
The following are consequences of Theorems 1.7 and 1.8. They can be obtained by using the “Principle of 
Mathematics Induction”.
COROLLARY 1.4 For any real number x and for any integer n, the following are true.
1. sin(np - x) = (-1)n+1 sinx and cos(np - x) = (-1)n cosx
2. sin(np + x) = (-1)n sin x and cos(np + x) = (-1)n cosx
3. sin cos and cos( ) ( ) ( ) ( )2 1
2
1 2 1
2
1n x x n xn n+ +æ
èç
ö
ø÷
= - + +æ
èç
ö
ø÷
= - +
p p 11 sin x
sin cos and cos s( ) ( ) ( ) ( )2 1
2
1 2 1
2
1n x x n xn n+ -æèç
ö
ø÷ = - + -
æ
èç
ö
ø÷ = -
p p
iin x
Try it out Prove parts (1)-(4) of Corollary 1.4.
Table 1.4 contains values of other trigonometric functions also. Here n stands for an integer and x for any real 
number in the domain of corresponding functions.
Table 1.4 Values of other trignometric functions
x sin x cos x tan x cot x cosec x sec x
p - q (-1)n+1 sinq (-1)n cosq -tanq -cotq (-1)n+1 cosecq (-1)n secq
np + q (-1)n sinq (-1)n cosq tan q cotq (-1)n cosecq (-1)n secq
( )2 1
2
n + -p q (-1)n cosq (-1)n sinq cot q tanq (-1)n secq (-1)n cosecq
( )2 1
2
n + +p q (-1)n cosq (-1)n+1 sinq cot q tanq (-1)n secq (-1)n+1 cosecq
DEFINITION 1.10 Coterminal Two angles q andq f are said to bef coterminal if the difference between them isl
an integral multiple of 2p or 360p ° according as the angles are measured in radians or degrees.
The following are pairs of coterminal angles:
(1)
p
2
and 
5
2
p
(2) 40° and 400°
(3) 50° and -310°
Example
If q andq f are coterminal angles, thenf q = 2np + f for some integerf n and hence, for any trigometric function f
f f n f( ) ( ) ( )q p f f= + =2
Example 1.10 
If sin q = -5/13 and q is in the third quadrant, evaluateq
the value of
5 12 132cot tan cosecq q q+ +
Solution: Consider a rectangular coordinate plane 
OXY with O as origin. Let P(x, y) be a point in the plane 
such that OP makes of q in the anticlockwise sense, with q
OX. Since q is given in the third quadrant,q x < 0 and y < 0 
(Figure 1.14). Now, since sin q = -5/13, we get that y = -5, 
OP = 13 and hence x y= - = ± -12 132 2( ). Therefore
tanq = = -
-
=y
x
5
12
5
12
191.3 Periodicity and Variance
cot 
cosec
sin
q
q
q
= =
= =
-
x
y
12
5
1 13
5
Hence
5 12 13
5
12
5
12
5
12
13
13
5
2
2
cot tan cosecq q q+ +
= æ
èç
ö
ø÷
+ æ
èç
ö
ø÷
+
-æ
èç
ö
ø÷÷
= + - =
144
5
5
169
5
0
X
M
N
P(x, y)
Oq
Y
FIGURE 1.14 Example 1.10.
Example 1.11 
Find the value of
sin600° cos330° + cos 120° sin150°
Solution:
sin cos cos sin sin
cos
600 330 120 150 3 60
3
2
60
° ° + ° ° = + °
´ + °æ
èç
ö
ø÷
( )p
p
++ + °æ
èç
ö
ø÷
- °cos sin
p
p
2
30 30( )
= - °× °+ - ° °
= - × + - ×æ
sin sin sin sin (by Table 1.4)60 60 30 30
3
2
3
2
1
2
1
2
( )
èèç
ö
ø÷ = - 1
Example 1.12 
If secq + tanq = 1/5 and 0 < q < 2p, find the quadrant in pp
which q lies and the value of sinq q. Using this, find all the
other trigonometric ratios of q.
Solution: Since sec q + tanq and secq q - tanq are recip-q
rocal to each other, we have
secq - tan q = 5
By adding this to the given equation, we get
2
1
5
5
13
5
sec or secq q= + =
Therefore
cos
sec
tan sec
sin tan cos
q
q
q q
q q q
= = >
= - = - = -
= × = -
1 5
13
0
1
5
1
5
13
5
12
5
12
55
5
13
12
13
0× = - <
Since sin q < 0 and cosq > 0, q lies in the fourth quadrant.q
1.3 | Periodicity and VarianceLet us recall that a function f defined on a subset f E of the real number system E � is called periodic if there exists a
positive real number p such that f(ff x + p) = f(ff x) for all x ÎE whenever E x + p is also in E. In this case, p is called a period
of f. Note that there may be several periods of the same function. For example, if ff p is a period of f, thenff np is also a
period of f for any positive integer f n.
DEFINITION 1.11 Let f be a function defined on a subset f E of E �. If the set all periods of f has minimum, thenf
that minimum period is called the period of f.ff
A function may possess a period, but still it may not possess the minimum period. For example, consider a constant
function f defined on anyf E Í �. Then every positive real number is a period of f and each period is a positive real 
20 Chapter 1 Trigonometric Ratios and Transformations
number and, evidently, there is no minimum positive real number. That is, a function may be periodic without having
the period (i.e., the minimum period).
We know that each trigonometric function is periodic and 2p is a period of each of these. In the following, we p
prove that each trigonometric function has minimum period.
THEOREM 1.9 The sine function is periodic and its minimum period is 2p.pp
PROOF By Definition 1.5 of the sine function, sin(2p + q)q = sinq for all real numbersq q and hence 2q p is p
period of the sine function. We shall prove that any positive real number less than 2p is not a p
period of sine. Suppose, on the contrary, that 0 < p < 2p and p p is a period of the sine function.
Put x0 = 2p - p. Then since p is a period
sin sin( )x p x0 0+ =
Therefore
sin sin sin sin2 2p p= - = - = -( ) ( )p p p
Therefore
sin p = 0
and hence p = p (since 0p < p < 2p). Butp p is not a period of the sine function, sincep
sin sin
p p p
4
1
2
1
2 4
+æèç
ö
ø÷ = - =¹
This implies that there is no p such that 0 < p < 2p andp p is a period of the sine function. Thus,
2p is the minimum period of the sine function. ■
Similar arguments as in Theorem 1.9 yield the following theorem.
THEOREM 1.10 1. The sine, cosine, cosecant and secant functions are all periodic with minimum period 2p.pp
2. The tangent and cotangent functions are periodic with minimum period p.pp
Example 1.13 
Determine the value of the following
(1) tan 780°
(2) sec 840°
(3) cosec 420°
(4) cot 240°
Solution:
(1) tan tan tan780 4 60 60 3° = + ° = ° =( )p
(2) sec sec sec sec
cosec
840 4 120 120
2
30
30 2
° = + ° = ° = + °æèç
ö
ø÷
= - ° = -
( )p p
(3) cosec cosec cosec420 2 60 60
2
3
° = + ° = ° =( )p
(4) cot cot cot240 60 60
1
3
° = + ° = ° =( )p
Next, we discuss the variance of different trigonometric functions; that is, how does the change in value of x affects
each trigonometric ratio at x.
Recall that the domain of each of the sine and cosine is the whole real number system �, while that of secant and
tangent is � - {(2n + 1)p/2,pp n �} and that of cosecant and cotangent is � - {np,pp |n �}. In Table 1.5 we display these 
together with the ranges (or images) of these functions.
211.3 Periodicity and Variance
Table 1.5 Domains and ranges of trigonometric functions
Domain Range
Sine � [-1, 1]
Cosine � [-1, 1]
Tangent � �- + Îìí
î
ü
ý
þ
( ) |2 1
2
n n
p
Cotangent � - {npp | n �} �
Secant � �- + Îìí
î
ü
ý
þ
( ) |2 1
2
n n
p -¥, -1] È [1, +¥)
Cosecant � - {npp | n Î�} (-¥, -1] È [1, ¥)
DEFINITION 1.12 Increasing and Decreasing Functions Let f be a real-valued function defined on a subsetf
of � and E Í Dom f. Thenff f is said to be f increasing on E ifE
x y E x y f x f y, ( ) ( )Î < Þ £and
f is said tof decreasing on E ifE
x y E x y f x f y, ( ) ( )Î < Þ ³and
The following can be easily proved.
Variance and Graph of Sin x
1. The function sinx is increasing on [0, p/2]. Aspp x increases from 0 to p/2, sinpp x increases from 0 to 1.
2. The function sin x is decreasing on [p/2, 3pp p/2]. As pp x increases from p/2 and 3pp p/2 sinpp x decreases from 1 to -1.
3. The function sinx is increasing on [3p/2, 2pp p]. As p x increases from 3p/2 to 2pp p, sinpp x increases from -1 to 0.
We can plot the graph of sinx in the coordinate plane by taking angles x in radian measure on X-axis and the values of 
y = sin x on Y-axis. The graph of sinx is given in Figure 1.15. The sine curve passes through the origin and the values
of sin x vary between -1 and +1 which are, respectively, the minimum and maximum. This curve is in the shape of a
wave whose wavelength is 2p, which is nothing but the period of sinpp x.
X -p -p
2
0
p
2
p
3
2
p
2p
5
2
p
3p
Y 0 -1 0 1 0 -1 0 1 0
X
Y y = sin x
–1
1
X
p p 2p 3pO p/2p/2 3p/2 5p/2
FIGURE 1.15 Graph of sin x.
22 Chapter 1 Trigonometric Ratios and Transformations
Variance and Graph of Cos x
1. The function cosx is decreasing on [0, p]. As p x increases from 0 to p, cospp x decreases from 1 to -1.
2. The function cosx is increasing on [p, 2pp p]. Asp x increases from p to 2p p, cospp x increases from -1 to 1.
By choosing a suitable scale, plot the points (x, cosx) given in the table below and join these points with a smooth 
curve to get the graph of cos x. This cosine curve does not pass through the origin. The maximum and minimum values
of cos x are 1 and -1, respectively. Each real number in [-1, 1] is a value of cos x. The curve also looks like a wave with 
wavelength 2p, which is same as the period of cospp x. The graph of cosx is given in Figure 1.16.
X -p -p
2
0
p
2
p 3
2
p
2p 5
2
p
Y -1 0 1 0 -1 0 1 0
Y
O
1
1
p
XX
y = cos x
2pp/2p/2 3p/2 5p/2
FIGURE 1.16 Graph of cos x.
Variance and Graph of Tan x
1. The function tan x is of period p, the minimum period being pp p.pp
2. The function tan x is not defined at x = [(2n + 1)/2]p, for any integerpp n.
3. The function tanx is increasing in each of the intervals [0, p/2) and (pp p/2,pp p]. Asp x increases from 0 to p/2, tanpp x
increases from 0 to +¥, and as x increases from p/2 to pp p, tanpp x increases from -¥ to 0.
4. The tanx curve passes through the origin and does not intersect the vertical lines at x = [(2n + 1)/2]p, for anypp
integer n rather it looks like touching these lines.
5. The function tan x has no minimum and maximum values.
Figure 1.17 illustrates the graph of tanx.
O
X
Y y = tan x
X
pp/2p/2 3p/2
FIGURE 1.17 Graph of tan x.
231.3 Periodicity and Variance
Variance and Graph of Cot x
1. The function cotx is of period p, the minimum period beingpp p.pp
2. The function cotx is not defined at x = np, for any integerpp n.
3. The function cot x is decreasing on the interval (0, p) and on each of the intervals (p np, (pp n + 1)p), p n �. As x increases
from 0 to p, cotpp x decreases from +¥ to -¥.
4. The cot x curve does not pass through the origin and does not intersect the vertical lines x = np, for any integer pp n,
but looks like touching these lines.
5. The function cotx has no minimum and maximum values.
Figure 1.18 illustrates the graph of cotx.
0
X
Y y = cot x
X
p 2pp/2p/2 3p/2 5p/2
FIGURE 1.18 Graph of cot x.
Variance and Graph of Sec x
1. The function secx is periodic and 2p is its minimum period.p
2. The function secx is not defined at np/2 for any integerpp n.
3. The function secx is increasing on each of the intervals
0
2 2
2 2
1
2
, , , ,
p p
p p pæ
èç
ö
ø÷
æ
èç
ö
ø÷
+æ
èç
ö
ø÷
æ
èç
ö
ø÷
n n and 2
1
2
2 1n n+æ
èç
ö
ø÷
+æ
èç
ö
ø÷
p p,( )
for any integer n. Also, secx is decreasing on each of the intervals
( ) ,2 1 2
3
2
n n+ +æ
èç
ö
ø÷
æ
èç
ö
ø÷
p p and 2
3
2
2 2n n+æ
èç
ö
ø÷
+æ
èç
ö
ø÷
p p,( )
for any integer n.
4. The secx curve does not intersect the X-axis and any line XX x = np/2,pp n � but nearly touches these lines.
5. The function secx has no minimum and maximum values and for no x, -1 < sec x < 1.
Figure 1.19 illustrates the graph of secx.
24 Chapter 1 Trigonometric Ratios and Transformations
Y
2
1
0
1
2
X
p
y = sec x
X
p/2p/2 3p/2
FIGURE 1.19 x.
Variance and Graph of Cosec x
1. The function cosecx is periodic