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MATHEMATICS 61

Notes

MODULE - IVFunctions

Trigonometric Functions-I

16

TRIGONOMETRIC FUNCTIONS - I

We have read about trigonometric ratios in our earlier classes. Recall that we defined theratios of the sides of a right triangle as follows :

c a csin , cos , tan

b b aθ = θ = θ =

and cosec b b a, sec , cot

c a cθ = θ = θ =

We also developed relationships between thesetrigonometric ratios as

2 2sin cos 1θ + θ = , 2 2sec 1 tanθ = + θ, 2 2cosec 1 cotθ = + θ

We shall try to describe this knowledge gained so far in terms of functions, and try todevelop this lesson using functional approach.In this lesson, we shall develop the science of trigonometry using functional approach.We shall develop the concept of trigonometric functions using a unit circle. We shalldiscuss the radian measure of an angle and also define trigonometric functions of the typey = sin x, y = cos x, y = tan x, y = cot x, y = sec x, y = cosec x, y = a sin x, y = b cos x,etc., where x, y are real numbers.

We shall draw the graphs of functions of the type

y = sin x, y = cos x, y = tan x, y = cotx, y = secx, and y = cosecx y = a sin x,y = a cos x.

OBJECTIVES

After studying this lesson, you will be able to :

• define positive and negative angles;

• define degree and radian as a measure of an angle;

• convert measure of an angle from degrees to radians and vice-versa;

Fig.16.1

MATHEMATICS

Notes


62


• state the formula r= θl where r and θ have their usual meanings;

• solve problems using the relation r= θl ;

• define trigonometric functions of a real number;

• draw the graphs of trigonometric functions; and

• interpret the graphs of trigonometric functions.

EXPECTED BACKGROUND KNOWLEDGE• Definition of an angle.

• Concepts of a straight angle, right angle and complete angle.

• Circle and its allied concepts.

• Special products : ( )2 2 2a b a b 2ab± = + ± , ( ) ( )3 3 3a b a b 3ab a b± = ± ± ±

• Knowledge of Pythagoras Theorem and Py thagorean numbers.

16.1 CIRCULAR MEASURE OF ANGLE

An angle is a union of two rays with the common end point. An angle is formed by the rotationof a ray as well. Negative and positive angles are formed according as the rotation is clock-wise or anticlock-wise.

16.1.1 A Unit CircleIt can be seen easily that when a line segment makes one complete rotation, its end pointdescribes a circle. In case the length of the rotating line be one unit then the circle described willbe a circle of unit radius. Such a circle is termed as unit circle.

16.1.2 A RadianA radian is another unit of measurement of an angle other than degree.A radian is the measure of an angle subtended at the centre of a circle by an arc equal in lengthto the radius (r) of the circle. In a unit circle one radian will be the angle subtended at the centreof the circle by an arc of unit length.

Fig. 16.2

Note : A radian is a constant angle; implying that the measure of the angle subtended byan are of a circle, with length equal to the radius is always the same irrespective of theradius of the circle.

MATHEMATICS 63

Notes



16.1.3 Relation between Degree and Radian

An arc of unit length subtends an angle of 1 radian. The circumference 2 ( r 1)π =∵ subtend

an angle of 2π radians.

Hence 2π radians = 360°

π radians = 180°

2π

radians = 90°

4π

radians = 45°

1 radian = 3602

° π

=180 °

π

or 1° = 2

360π

radians =180

π radians

Example 16.1 Convert

(i) 90° into radians (ii) 15° into radians

(iii)6π

radians into degrees. (iv) 10π

radians into degrees.

Solution :

(i) 1° = 2

360π

radians

⇒ 90° = 290

360π

× radiansor 90° = 2π

radians

(ii) 15° = 215

360π

× radians or 15° = 12π

radians

(iii) 1 radian = 3602

° π

6π

radians = 3602 6

°π × π

6π

radians = 30°

(iv)10π

radians = 3602 10

°π × π

10π

radians = 18°

MATHEMATICS

Notes


64


CHECK YOUR PROGRESS 16.11. Convert the following angles (in degrees) into radians :

(i) 60° (ii) 15° (iii) 75° (iv) 105° (v) 270°2. Convert the following angles into degrees:

(i) 4π

(ii) 12π

(iii) 20π

(iv) 60π

(v) 23π

3. The angles of a triangle are 45°, 65° and 70°. Express these angles in radians

4. The three angles of a quadrilateral are 6π

, 3π

, 23π

. Find the fourth angle in radians.

5. Find the angle complementary to 6π

.

16.1.4 Relation Between Length of an Arc and Radius of the CircleAn angle of 1 radian is subtended by an arc whose length is equal to the radius of the circle. Anangle of 2 radians will be substened if arc is double the radius.

An angle of 2½ radians willbe subtended if arc is 2½ times the radius.

All this can be read from the following table :

Length of the arc (l) Angle subtended at thecentre of the circle θ (in radians)

r 1

2r 2

(2½)r 2½

4r 4

Therefore, r

θ =l

or r= θl

where r = radius of the circle,

θ = angle substended at the centre in radians

and l = length of the arc.

The angle subtended by an arc of a circle at the centre of the circle is given by the ratio of thelength of the arc and the radius of the circle.

Note : In arriving at the above relation, we have used the radian measure of the angle

and not the degree measure. Thus the relationr

θ =l is valid only when the angle is

measured in radians.

MATHEMATICS 65

Notes



Example 16.2 Find the angle in radians subtended by an arc of length 10 cm at the centre ofa circle of radius 35 cm.

Solution : l = 10cm and r 35cm.=

rθ =

l radians or 1035

θ = radians

or27

θ = radians

Example 16.3 If D and C represent the number of degrees and radians in an angle prove that

D C180

=π

Solution : 1 radian = 3602

° π

or180 °

π

∴ C radians = 180

C°

× π Since D is the degree measure of the same angle, therefore,

180

D C= ×π

which impliesD C

180=

π

Example 16.4 A railroad curve is to be laid out on a circle. What should be the radius of acircular track if the railroad is to turn through an angle of 45° in a distance of 500m?

Solution : Angle θ is given in degrees. To apply the formula r ,= θ θl must be changed toradians.

45 45180

πθ = °= × radians ....(1)

=4π

radians

l = 500 m ....(2)

l = r θ gives r =θl

∴ 500r m

4

=π

[using (1) and (2)]

= 4

500 m×π

MATHEMATICS

Notes


66


= 2000 ×0.32 m 1

0.32 = π

= 640 mExample 16.5 A train is travelling at the rate of 60 km per hour on a circular track. Through

what angle will it turn in 15 seconds if the radius of the track is 56

km.

Solution : The speed of the train is 60 km per hour. In 15 seconds, it will cover

60 15km

60 60××

= 1

km4

∴ We have, l = 1

km4

and r = 5

km6

∴

14 radians5r6

θ= =l

3

radians10

=

CHECK YOUR PROGRESS 16.21. Express the following angles in radians :

(a) 30° (b) 60° (c) 150°2. Express the following angles in degrees :

(a) 5π

(b) 6π

(c) 9π

3. Find the angle in radians and in degrees subtended by an arc of length 2.5 cm at thecentre of a circle of radius 15 cm.

4. A train is travelling at the rate of 20 km per hour on a circular track. Through what angle

will it turn in 3 seconds if the radius of the track is1

12 of a km?.

5. A railroad curve is to be laid out on a circle. What should be the radius of the circulartrack if the railroad is to turn through an angle of 60° in a distance of 100 m?

6. Complete the following table for l, r, θ having their usual meanings.

MATHEMATICS 67

Notes



l r θ

(a) 1.25m ....... 135°

(b) 30 cm .......4π

(c) 0.5 cm 2.5 m ........(d) ......... 6 m 120°

(e) ......... 150 cm15π

(f) 150 cm 40 m ........

(g) ........ 12 m6π

(h) 1.5 m 0.75 m ........(i) 25 m ........ 75°

16.2 TRIGONOMETRIC FUNCTIONS

While considering, a unit circle you must have noticed that for every real number between 0 and2π , there exists a ordered pair of numbers x and y. This ordered pair (x, y) represents thecoordinates of the point P.

(i) (iii)

(ii) (iv)

Fig. 16.3

MATHEMATICS

Notes


68


If we consider 0θ= on the unit circle, we will have a point whose coordinates are (1,0).

If 2π

θ = , then the corresponding point on the unit circle will have its coordinates (0,1).

In the above figures you can easily observe that no matter what the position of the point,corresponding to every real number θ we have a unique set of coordinates (x, y). The values ofx and y will be negative or positive depending on the quadrant in which we are considering thepoint.Considering a point P (on the unit circle) and the corresponding coordinates (x, y), we definetrigonometric functions as :

sin yθ = , cos xθ =

ytan (forx 0)

xθ = ≠ ,

xcot (for y 0)

yθ = ≠

1sec (for x 0)

xθ = ≠ ,

1cosec (for y 0)

yθ = ≠

Now let the point P move from its original position in anti-clockwise direction. For variouspositions of this point in the four quadrants, various real numbers θ will be generated. Weesummarise, the above discussion as follows. For values of θ in the :

I quadrant, both x and y are positve.II quadrant, x will be negative and y will be positive.III quadrant, x as well as y will be negative.IV quadrant, x will be positive and y will be negative.or I quadrant II quadrant III quadrant IV quadrant

All positive sin positive tan positive cos positivecosec positive cot positive sec positive

Where what is positive can be rememebred by :All sin tan cos

Quardrant I II III IVIf (x, y) are the coordinates of a point P on aunit circle and θ , the real number generatedby the position of the point, then sinθ = y andcosθ = x. This means the coordinates of thepoint P can also be written as (cosθ , sin θ )

From Fig. 16.5, you can easily see that thevalues of x will be between −1 and + 1 as Pmoves on the unit circle. Same will be true fory also.Thus, for all P on the unit circle

P (cos , sin )q q

Fig. 16.4

MATHEMATICS 69

Notes



1 x 1− ≤ ≤ and 1 y 1− ≤ ≤

Thereby, we conclude that for all real numbers θ

1 cos 1− ≤ θ ≤ and 1 sin 1− ≤ θ ≤

In other words, sinθ and cosθ can not be numerically greater than 1

Example 16.6 What will be sign of the following ?

(i) sin718

π (ii) cos

49π

(iii) tan59π

Solution :

(i) Since 718

π lies in the first quadrant, the sign of sin

718

π will be posilive.

(ii) Since 49π

lies in the first quadrant, the sign of cos49π

will be positive.

(iii) Since 59π

lies in the second quadrant, the sign of tan59π

will be negative.

Example 16.7 Write the values of (i) sin 2π

(ii) cos 0 (iii) tan2π

Solution : (i) From Fig.16.5, we can see that the coordinates of the point A are (0,1)

∴ sin2π

=1 , as sin yθ =

Fig.16.5

(ii) Coordinates of the point B are (1, 0)∴ cos 0 1= , as cos xθ =

(iii) sin 12tan

2 0cos2

ππ

= =π

which is not defined

Thus tan2π

is not defined.

MATHEMATICS

Notes


70


Example 16.8 Write the minimum and maximum values of cosθ .

Solution : We know that 1 cos 1− ≤ θ ≤

∴ The maximum value of cosθ is 1 and the minimum value of cosθ is −1.

CHECK YOUR PROGRESS 16.31. What will be the sign of the following ?

(i) cos3

2π(ii)

5tan

6π

(iii)2

sec3π

(iv)35

sec18

π(v)

25tan

18π

(vi)3

cot4π

(vii)8

cosec3π

(viii)7

cot8π

2. Write the value of each of the following :

(i) cos2π

(ii) s i n 0 (iii)2

cos3π

(iv)3

tan4π

(v) sec 0 (vi) tan2π

(vii)3

tan2π

(viii) c o s 2π

16.2.1 Relation Between Trigonometric Functions

By definition x cos= θ

y sin= θ

As ytan

xθ = , ( )x 0≠

sincos

θ=

θ , n2π

θ ≠

and x

coty

θ = , ( )y 0≠

i.e., cos 1

cotsin tan

θθ = =

θ θ( )nθ ≠ π

Similarly, 1

seccos

θ =θ

n2π θ ≠

Fig. 16.6

MATHEMATICS 71

Notes



and 1

cosecsin

θ =θ

( )nθ ≠ π

Using Pythagoras theorem we have, 2 2x y 1+ =

i.e., ( ) ( )2 2cos sin 1θ + θ =

or, 2 2cos sin 1θ + θ =

Note : ( )2cos θ is written as 2cos θ and ( )2sin θ as 2sin θ

Again 2 2x y 1+ =

or 2 2y 1

1 ,x x

+ =

for x 0≠ or, ( ) ( )2 21 tan sec+ θ = θ

i.e. 2 2sec 1 tanθ = + θ

Similarly, 2 2cosec 1 cotθ = + θ

Example 16.9 Prove that 4 4 2 2sin cos 1 2sin cosθ + θ = − θ θ

Solution : L.H.S. 4 4sin cos= θ + θ

4 4 2 2 2 2sin cos 2sin cos 2sin cos= θ + θ + θ θ − θ θ

( )22 2 2 2sin cos 2sin cos= θ + θ − θ θ

( )2 2 2 21 2sin cos sin cos 1= − θ θ θ + θ =∵= R.H.S.

Example 16.10 Prove that 1 sin

sec tan1 sin

− θ= θ − θ

+ θ

Solution : L.H.S.1 sin1 sin

− θ=

+ θ

( ) ( )( ) ( )1 sin 1 sin1 sin 1 sin

− θ − θ=

+ θ − θ

( )2

21 sin

1 sin

− θ=

− θ

MATHEMATICS

Notes


72


( )2

21 sin

cos

− θ=

θ

1 sincos− θ

=θ

1 sincos cos

θ= −

θ θ

sec tan R.H.S.= θ − θ =

Example 16.11 If 21

sin29

θ = , prove that 1

sec tan 22

θ + θ = , given that θ lies in the first

quadrant.

Solution :21

sin29

θ =

Also, 2 2sin cos 1θ + θ =

⇒2

2 2 441 400 20cos 1 sin 1

841 841 29 θ = − θ = − = =

⇒ 20

cos29

θ = (cosθ is positive as θ lies in the first quardrant)

∴ 21tan

20θ =

∴29 21 29 21

sec tan20 20 20

+θ + θ = + =

5 1

2 R.H.S.2 2

= = =

CHECK YOUR PROGRESS 16.4

1. Prove that 4 4 2 2sin cos sin cosθ − θ = θ − θ

2. If 1

tan2

θ = , find the other five trigonometric functions.

3. If b

coseca

θ= , find the other five trigonometric functions, if θ lies in the first quardrant.

4. Prove that 1 cos

cosec cot1 cos

+ θ= θ + θ

− θ

MATHEMATICS 73

Notes



RealNumbers→

5. If cot cosec 1.5θ + θ = , show that 5

cos13

θ =

6. If tan sec mθ + θ = , find the value of cos θ

7. Prove that ( ) ( ) 2tan A 2 2tanA 1 5tanA 2sec A+ + = +

8. Prove that 6 6 2 2sin cos 1 3sin cosθ + θ = − θ θ

9. Prove that cos sin

cos sin1 tan 1 cot

θ θ+ = θ + θ

− θ − θ

10. Prove that tan sin

cot cosec sec1 cos 1 cos

θ θ+ = θ + θ ⋅ θ

+ θ − θ

16.3 TRIGONOMETRIC FUNCTIONS OF SOME SPECIFIC REAL NUMBERS

The values of the trigonometric functions of 0, 6π

, 4π

, 3π

and 2π

are summarised below in the

form of a table :

06π

4π

3π

2π

Function

↓

sin 012

12

32

1

cos 13

2

12

12

0

tan 013

1 3 Not defined

As an aid to memory, we may think of the following pattern for above mentioned values of sinfunction :

04

, 14

, 24

, 34

, 44

On simplification, we get the values as given in the table. The values for cosines occur in thereverse order.Example 16.12 Find the value of the following :

(a) sin sin cos cos4 3 4 3π π π π

− (b) 2 2 24tan cosec cos4 6 3π π π

− −

MATHEMATICS

Notes


74

Trigonometric Functions-ISolution :

(a) sin sin cos cos4 3 4 3π π π π

−

1 3 1 12 22 2

= −

3 12 2

−=

(b) 2 2 24tan cosec cos4 6 3π π π

− −

( ) ( )2

2 2 14 1 2

2 = − −

14 4

4= − −

14

= −

Example 16.13 If A3π

= and B6π

= , verify that

( )cos A B cosAcosB s inAsinB+ = −

Solution : L.H.S. ( )cos A B= +

cos3 6π π = +

cos

2π

= = 0

R.H.S. cos cos sin sin3 6 3 6π π π π

= −

1 3 3 12 2 2 2

= ⋅ − ⋅

3 30

4 4= − =

∴ L.H.S. = 0 = R.H.S. cos (A + B) = cos A cos B − sin A sin B

MATHEMATICS 75

Notes



CHECK YOUR PROGRESS 16.51. Find the value of

(i) 2 2 2sin tan tan6 4 3π π π

+ + (ii) 2 2 2 2sin cosec sec cos3 6 4 3π π π π

+ + −

(iii)2 2

cos cos sin sin3 3 3 3π π π π

− (iv) 2 2 2 24cot cosec sec tan3 4 3 4π π π π

+ +

(v)1

sin sin cos cos6 4 3 4 4π π π π + − +

2. Show that

2 21 tan tan tan tan sec sec6 3 6 3 6 3π π π π π π + + − =

3. Taking A , B3 6π π

= = , verify that

(i) ( ) tan A t a n Btan A B

1 tan A tan B+

+ =−

(ii) ( )cos A B cosAcosB s inAsinB+ = −

4. If 4π

θ = , verify the following :

(i) sin2 2sin cosθ = θ θ (ii) 2 2cos2 cos sinθ = θ − θ

22cos 1= θ −

21 2sin= − θ

5. If A6π

= , verify that

(i) 2cos2A 2cos A 1= − (ii) 22 t a n A

tan2A1 tan A

=−

(iii) sin2A 2 s i n A c o s A=

16.4 GRAPHS OF TRIGONOMETRIC FUNCTIONSGiven any function, a pictorial or a graphical representation makes a lasting impression on theminds of learners and viewers. The importance of the graph of functions stems from the fact thatthis is a convenient way of presenting many properties of the functions. By observing the graphwe can examine several characteristic properties of the functions such as (i) periodicity, (ii)intervals in which the function is increasing or decreasing (iii) symmetry about axes, (iv) maximumand minimum points of the graph in the given interval. It also helps to compute the areas enclosedby the curves of the graph.

MATHEMATICS

Notes


76


16.4.1 Variations of sinθ as θ Varies Continuously From 0 to 2π .

Let X'OX and Y'OY be the axes ofcoordinates.With centre O and radiusOP = unity, draw a circle. Let OPstarting from OX and moving inanticlockwise direction make an angleθ with the x-axis, i.e. ∠ XOP = θ. DrawPM ⊥ X'OX, then sinθ = MP as OP=1.

∴ The variations of sin θ are the same as those ofMP.I Quadrant :

As θ increases continuously from 0 to 2π

PM is positive and increases from 0 to 1.

∴ sin θ is positive.

II Quadrant ,2π π

In this interval, θ lies in the second quadrant.Therefore, point P is in the second quadrant. Here PM= y is positive, but decreases from 1 to 0 as θ varies

from to2π

π . Thus sin θ is positive.

III Quadrant 3

,2π π

In this interval, θ lies in the third quandrant. Therefore,point P can move in the third quadrant only. HencePM = y is negative and decreases from 0 to −1 as θ

varies from π to 32π

. In this intervalsinθ decreases from 0 to −1. In this interval sin θ is

negative.

IV Quadrant 3

, 22π π

In this interval, θ lies in the fourth quadrant. Therefore,point P can move in the fourth quadrant only. Here againPM = y is negative but increases from -1 to 0 as

θ varies from3

to 22π

π . Thus sin θ is negative in this

interval.

Fig. 16.7

Fig. 16.8

Fig. 16.9

Fig. 16.10

MATHEMATICS 77

Notes



16.4.2 Graph of sin θ as θ varies from 0 to 2π .

Let X'OX and Y'OY be the two coordinate axes of reference. The values of θ are to be measuredalong x-axis and the values of sine θ are to be measured along y-axis.

(Approximate value of 1 32 1.41, .707, .87

22= = = )

θ 06π

3π

2π 2

3π 5

6π

π76π 4

3π 3

2π 5

3π 11

6π

2π

sin θ 0 .5 .87 1 .87 .5 0 −.5 −.87 −1 −.87 −.5 0

Fig. 16.11

Some Observations

(i) Maximum value of sin θ is 1.

(ii) Minimum value of sin θ is −1.

(iii) It is continuous everywhere.

(iv) It is increasing from 0 to 2π

and from 32π

to 2π. It is decreasing from2π

to 32π

.

With the help of the graph drawn in Fig. 16.12 we can always draw another graph.y = sin θ in the interval of [2 , 4 ]π π ( see Fig. 16.11)1)

What do you observe ?

The graph of y = sin θ in the interval [2 , 4 ]π π is the same as that in 0 to2π. Therefore, thisgraph can be drawn by using the property sin (2 ) sinπ + θ = θ. Thus, sin θ repeats itself when

θ is increased by2π. This is known as the periodicity of sin θ .

Fig. 16.12

MATHEMATICS

Notes


78

Trigonometric Functions-IWe shall discuss in details the periodicity later in this lesson.Example 16.14 Draw the graph of y = sin 2θ .

Solution :

θ : 06π

4π

3π

2π 2

3π 3

4π 5

6π

π

2 :θ 03π

2π 2

3π

π3

4π 32π 5

3π

2π

sin2 :θ 0 .87 1 .87 0 −.87 −1 −.87 0

Fig. 16.13

The graph is similar to that of y = sin θ

Some Observations

1. The other graphs of sin θ , like a sin θ , 3 sin 2θ can be drawn applying the samemethod.

2. Graph of sin θ , in other intervals namely [ ]4 , 6 ,π π [ ]2 , 0 ,− π [ ]4 , 2 ,− π − π

can also be drawn easily. This can be done with the help of properties of allied

angles: sin ( )2 sin ,θ + π = θ sin ( )2 sin .θ − π = θ i.e.,θ repeats itself when

increased or decreased by 2π .


1. What are the maximum and minimum values of sin [ ]in 0, 2θ π ?

2. Explain the symmetry in the graph of sin [ ]in 0, 2θ π

3. Sketch the graph of y = 2 sin θ , in the interval [ ]0, π

MATHEMATICS 79

Notes



4. For what values of θ in [ ], 2π π , sin θ becomes

(a) 1

2−

(b) 3

2−

5. Sketch the graph of y = sin x in the interval of [ ],−π π

16.4.3 Graph of cosθ as θ Varies From 0 to 2π

As in the case of sinθ , we shall also discuss the changes in the values of cosθ whenθ assumes

values in the intervals 0 , ,2π

, ,

2π π

3,

2π π

and 3

, 2 .2π π

I Quadrant : In the interval 0 ,2π

, point P lies in

the first quadrant, therefore, OM = x is positive but

decreases from 1 to 0 asθ increases from 0 to2π

.

Thus in this interval cosθ decreases from 1 to 0.

∴ cos θ is positive in this quadrant.

II Quadrant : In the interval ,2π π

, point P lies in

the second quadrant and therefore point M lies on thenegative side of x-axis. So in this case OM = x isnegative and decreases from 0 t o 1− asθ increases

from2π

toπ . Hence in this inverval cosθ decreases

from 0 to 1− .

∴ cos θ is negative.

III Quadrant : In the interval 3

,2π π

, point P lies

in the third quadrant and therefore, OM = x remainsnegative as it is on the negative side of x-axis. ThereforeOM = x is negative but increases from−1 to 0 as θ

increases from π to 3

.2π

Hence in this interval cos θ

increases from -1 to 0.

∴ cos θ is negative.

IV Quadrant : In the interval 3, 2

2π π

, point P lies

Fig.16.14

Fig. 16.16

Fig. 16.15

MATHEMATICS

Notes


80


in the fourth quadrant and M moves on the positiveside of x-axis.Therefore OM = x is positive. Also it

increases from 0 to 1 as θ increases from 3

to 22π

π .

Thus in this interval cosθ increases from 0 to 1.

∴ cosθ is positive.

Let us tabulate the values of cosines of some suitablevalues of θ .

θ 06π

3π

2π 2

3π 5

6π

π76π 4

3π 3

2π 5

3π 11

6π

2π

cosθ 1 .87 .5 0 0.5 −.87 −1 −.87 −.5 0 0.5 .87 1

Fig. 16.18

Let X'OX and Y'OY be the axes. Values of θ are measured along x-axis and those of cosθalong y-axis.

Some observations

(i) Maximum value of cos 1.θ =

(ii) Minimum value of cos 1.θ = −

(iii) It is continuous everywhere.

(iv) cos ( )2 cos .θ + π = θ Also( )2 cos .θ − π = θCos θ repeats itself when θ is

increased or decreased by2π. It is called periodicity of cos θ . We shall discuss indetails about this in the later part of this lesson.

(v) Graph of cos θ in the intervals [2 , 4 ] [4 , 6 ] [ 2 , 0]π π π π − π , will be the same as

in [0, 2 ]π .

Example 16.15 Draw the graph of cos θ asθ varies from to− π π . From the graph read

the values of θ when cos 0.5θ = ± .

Fig. 16.17

MATHEMATICS 81

Notes


Trigonometric Functions-ISolution :

θ : −π56

− π 23

− π2

−π3

−π6

−π0

6π

3π

2π 2

3π 5

6π

π

cos θ : −1.0 −0.87 −0.5 0 .50 −.87 1.0 0.87 0.5 0 −0.5 −0.87 −1

cos 0.5θ =

when ,3 3π −π

θ =

cos 0.5θ = −

when 2 2,

3 3π − π

θ =

Fig. 16.19

Example 16.16 Draw the graph of cos 2θ in the interval 0 toπ .

Solution :

θ 06π

4π

3π

2π 2

3π 3

4π 5

6π

π

2θ 03π

2π 2

3π

π3

4π 32π 5

3π

2π

c o s 2θ 1 0.5 0 −0.5 −1 −0.5 0 0.5 1

Fig. 16.20

MATHEMATICS

Notes


82



1. (a) Sketch the graph of y = cos θ asθ varies from to4 4

−π π .

(b) Draw the graph of y = 3 cos θ asθ varies from 0 to 2π .

(c) Draw the graph of y = cos 3θ from to− π π and read the values of θ whencos 0.87θ = and cos 0.87.θ = −

(d) Does the graph of y = cos θ in 3

,2 2π π

lie above x-axis or below x-axis?

(e) Draw the graph of y = cos θ in [2 , 4 ]π π

16.4.4 Graph of tan θ as θ Varies from 0 to 2π

In I Quadrant : tan θ can be written as sincos

θθ

Behaviour of tan θ depends upon the behaviour of sin θ and 1

cos θ

In I quadrant,sin θ increases from 0 to 1,cos θ decreases from 1 to 0

But 1

cos θ increases from 1 indefintely (and write it as increasses from 1 to ∞ ) tan 0θ >

∴ tan θ increases from 0 to ∞ . (See the table and graph of tan θ ).

In II Quadrant :sin

tancos

θθ =

θsinθ decreases from 1 to 0.cos θ decreases from 0 to 1− .tan θ is negative and increases from to 0−∞

In III Quadrant :sin

tancos

θθ =

θsinθ decreases from 0 to 1−cos θ increases from 1 t o 0−

∴ tan θ is positive and increases from 0 to ∞

In IV Quadrant :sin

tancos

θθ =

θsinθ increases from 1to0−cos θ increases from 0 to 1tan θ is negative and increases form to0−∞

Graph of tan θ

θ 0 6π

3π

02π

− ° 02π

+ °23π 5

6π

π76π 4

3π 3

02π

− °3

02π

+ °53π

11

6π

2π

tan θ 0 .58 1.73 +∞ -1.73 -.58 0 .58 1.73 +∞ −∞ -1.73 -.58 0 0

MATHEMATICS 83

Notes



Fig. 16.21

Observations

(i) tan(180 ) tan°+ θ = θ . Therefore, the complete graph of tan θ consists ofinfinitely many repetitions of the same to the left as well as to the right.

(ii) Since tan( ) tan−θ = − θ , therefore, if ( ),tanθ θ is any point on the graph then

( , tan )−θ − θ will also be a point on the graph.

(iii) By above results, it can be said that the graph of y tan= θ is symmetrical inopposite quadrants.

(iv) tan θ may have any numerical value, positve or negative.

(v) The graph of tan θ is discontinuous (has a break) at the points 3,

2 2π π

θ = .

(vi) As θ passes through these values, tan θ suddenly changes from to+ ∞ − ∞.

16.4.5 Graph of cot θ as θ Varies From 0 to 2π

The behaviour of cot θ depends upon the behaviour of cos θ and 1

sinθ as 1

cot cossin

θ= θθ

We discuss it in each quadrant.

I Quadrant :1

cot cossin

θ = θ×θ

cos θdecreases from 1 to 0

sin θ increases from 0 to 1

∴ cot θ also decreases from + ∞to 0 but cot 0.θ >

II Quadrant :1

cot cossin

θ = θ×θ

cos θ decreases from 0 t o 1−

sinθ decreases from 1 to 0

MATHEMATICS

Notes


84


⇒ cot 0θ < or cot θ decreases from 0 to − ∞

III Quadrant :1

cot cossin

θ = θ×θ

cos θ increases from 1 t o 0−

sin θ decreases from 0 to 1−

∴ cot θ decreases from to 0.+ ∞

IV Quadrant :1

cot cossin

θ = θ×θ

cos θ increases from 0 to 1

sin θ increases from 1 t o 0−

∴ cot θ< 0

cot θ decreases from 0 to − ∞

Graph of cot θ

θ 06π

3π

2π 2

3π 5

6π

0π − 0π +76π 4

3π 3

2π 5

3π 11

6π

2π

cot θ ∞ 1.73 .58 0 -.58 -1.73 −∞ + ∞ 1.73 .58 0 -.58 -1.73 −∞

Fig 16.22

Observations

(i) Since cot ( ) cotπ + θ = θ, the complete graph of cot θconsists of the portion from

0 toθ = θ= π or 3to

2 2π π

θ = θ= .

MATHEMATICS 85

Notes



Fig. 16.23

Fig. 16.24

Fig.16.25

(ii) cot θ can have any numerical value - positive or negative.

(iii) The graph of cot θ is discontinuous, i.e. it breaks at 0 , , 2 ,π π .

(iv) Asθ takes values 0, , 2 ,π π cot θ suddently changes from to− ∞ + ∞

CHECK YOUR PROGRESS 16.81. (a) What is the maximum value of tan θ ?

(b) What changes do you observe in 3

tan at , ?2 2π π

θ

(c) Draw the graph of y tan= θ from −π toπ . Find from the graph the value of θ for

which tan 1.7.θ =

2. (a) What is the maximum value of cot θ?

(b) Find the value of θ whencot 1θ = − , from the graph.

16.4.6 To Find the Variations And Draw The Graph of sec θ As θ Varies From 0 to 2π .

Let X'OX and Y'OY be the axes of coordinates. Withcentre O, draw a circle of unit radius.

Let P be any point on the circle. Join OP and drawPM ⊥ X'OX.

OP 1sec

OM OMθ = =

∴ Variations will depend upon OM.

I Quadrant : sec θ is positive as OM is positive.

Also sec 0 = 1 and sec 2π

= ∞ when we approach 2π

from the right.

∴ As θ varies from 0 to 2π

, secθ increases from 1 to

∞ .

II Quadrant : secθ is negative as OM is negative.

sec2π

=−∞ when we approach 2π

from the left. Also sec

1.π = −

∴ As θ varies from 2π

to π , sec θ changes from

MATHEMATICS

Notes


86


to 1.− ∞ −

It is observed that asθ passes through 2π

, secθ changes

from to+ ∞ − ∞ .

III Quadrant : secθ is negative as OM is negative.

sec 1π = − and 3

sec2π

= − ∞ when the angle approaches

32π

in the counter clockwise direction. As θ varies from

π to32π

, sec θ decreases from 1to− − ∞ .

IV Quadrant : secθ is positive as OM is positive. when θ

is slightly greater than 32π , sec θ is positive and very large.

Also sec 2 1.π = Hence sec θ decreases from ∞ to 1 as

θ varies from 3to 2

2π

π .

It may be observed that as θ passes through

3; sec

2π

θ changes from to .− ∞ + ∞

Graph of sec θ asθ varies from 0 to 2π

θ 0 6π

3π

02π

− 02π

+23π 5

6π

π76π 4

3π 3

02π

−3

02π

+53π 11

6π

2π

cot θ 1 1.15 2 + ∞ −∞ -2 -1.15 -1 -1.15 -2 −∞ + ∞ 2 1.15

1

Fig. 16.28

Fig.16.26

Fig.16.27

MATHEMATICS 87

Notes



Fig. 16.29

Fig. 16.30

Fig. 16.31

Observations

(a) secθ cannot be numerically less than 1.

(b) Graph of sec θ is discontinuous, discontinuties (breaks) occuring at 2π

and 32π

.

(c) As θ passes through 2π

and 32π

, see θ changes abruptly from + ∞ to − ∞ and then

from − ∞ to + ∞ respectively..

16.4.7 Graph of cosec θ as θ Varies From 0 to 2π

Let X'OX and Y'OY be the axes of coordinates. Withcentre O draw a circle of unit radius. Let P be anypoint on the circle. Join OP and draw PMperpendicular to X'OX.

OP 1cosec

MP MPθ = =

∴ The variation ofcosec θ will depend upon MP.

I Quadrant : cosec θ is positive as MP is positive.

cosec 12π

= when θ is very small, MP is also small and

therefore, the value of cosec θ is very large.

∴ As θ varies from 0 to 2π

, cosec θ decreases from

∞ to 1.

II Quadrant : PM is positive. Therefore, cosec θ is

positive. cosec 12π

= and cosec π = ∞ when the

revolving line approaches π in the counter clockwisedirection.

∴ As θ varies from2π

to π , cosec θ increases from

1 to ∞ .

III Quadrant :PM is negative

∴ cosec θ is negative. When θ is slightly greater than π ,

MATHEMATICS

Notes


88


cosec θ is very large and negative.

Also cosec 3

12π

= − .

∴ As θ varies from 3

to2π

π , cosecθ changes from

to 1− ∞ − .

It may be observed that as θ passes through π , cosec θchanges from to+ ∞ − ∞ .

IV Quadrant :

PM is negative.

Therefore, cosec θ = − ∞ as θ approaches 2π.

∴ as θ varies from 32π

to 2π, cosecθ varies from −

1 to−∞ .

Graph of cosec θ

θ 06π

3π

2π 2

3π 5

6π

0π − 0π +76π 4

3π 3

2π 5

3π 11

6π

2π

cosec θ ∞ 2 1.15 1 1.15 2 + ∞ −∞ −2 −1.15 −1 −1.15 −2 −∞

Fig. 16.34

Observations

(a) cosecθ cannot be numerically less than 1.

(b) Graph of cosecθ is discountinous and it has breaks at 0, , 2θ = π π.

(c) As θ passes through π , coses θ changes from + ∞ to −∞ . The values at 0 and2π are+ ∞ and −∞ respectively..

Fig. 16.32

Fig. 16.33

MATHEMATICS 89

Notes



Example 16.17 Trace the changes in the values of sec θ as θ lies in −π to π .Soluton :

Fig. 16.35


1. (a) Trace the changes in the values of sec θ when θ lies between 2− π and 2π anddraw the graph between these limits.

(b) Trace the graph of cosec θ ,when θ lies between 2− π and 2π.

16.5 PERIODICITY OF THE TRIGONOMETRIC FUNCTIONS

From your daily experience you must have observed things repeating themselves after regularintervals of time. For example, days of a week are repeated regularly after 7 days and monthsof a year are repeated regularly after 12 months. Position of a particle on a moving wheel isanother example of the type. The property of repeated occurence of things over regular intervalsis known as periodicity.

Definition : A function f (x) is said to be periodic if its value is unchanged when the value of thevariable in increased by a constant, that is if f (x + p) = f (x) for all x.

If p is smallest positive constant of this type, then p is called the period of the function f (x).

If f (x) is a periodic function with period p, then ( )1

f x is also a periodic function with period p.

16.5.1 Periods of Trigonometric Functions

(i) ( )sinx sin x 2n= + π; n = 0, ± 1, ± 2, .....

(ii) ( )cosx cos x 2n= + π; n = 0, ± 1, ± 2 ,.....

Also there is no p, lying in 0 to 2π, for which

( )sinx sin x p= +

( )cosx cos x p= + , for all x

MATHEMATICS

Notes


90


∴ 2π is the smallest positive value for which

( )sin x 2 sinx+ π = and ( )cos x 2 cosx+ π =⇒ sin x and cos x each have the period 2π.

(iii) The period of cosec x is also2π because cosec x 1

sin x= .

(iv) The period of sec x is also 2π as sec x 1c o s x

= .

(v) Also ( )tan x t a n x+ π = .

Suppose ( )p 0 p< < π is the period of tan x, then

( )tan x p tanx,+ = for all x.

Put x = 0, then tan p = 0, i.e., p = 0 or π .

⇒ the period of tan x is π .

∴ p can not values between 0 and π for which ( )tanx tan x p= +

∴ The period of tanx is π

(vi) Since 1

cot xt a n x

= , therefore, the period of cot x is also π .

Example 16.18 Find the period of each the following functions :

(a) y = 3 sin 2x (b) x

y cos2

= (c) y =x

tan4

Solution :

(a) Period is 22π

, i.e., π .

(b)1

y cos x2

= , therefore period 2

412

π= = π

(c) Period of y = x

tan 4144

π= = π


1. Find the period of each of the following functions :

(a) y = 2 sin 3x (b) y = 3 cos 2x

(c) y = tan 3x (d) 2y sin 2x=

MATHEMATICS 91

Notes



WH

l An angle is generated by the rotation of a ray.l The angle can be negative or positive according as rotation of the ray is clockwise or

anticlockwise.l A degree is one of the measures of an angle and one complete rotation generates an

angle of 360°.

l An angle can be measured in radians, 360° being equivalent to 2π radians.

l If an arc of length l subtends an angle of θ radians at the centre of the circle with radiusr, we have l = rθ .

l If the coordinates of a point P of a unit circle are (x, y) then the six trigonometric functions

are defined as sin yθ = , cos xθ = , ytan

xθ = , x

coty

θ = , 1sec

xθ = and

1cosec

yθ = .

The coordinates (x, y) of a point P can also be written as ( )cos , sinθ θ .

Hereθ is the angle which the line joining centre to the point P makes with the positivedirection of x-axis.

l The values of the trigonometric functions sin θ and cos θ when θ takes values 0,

, , ,6 4 3 2π π π π

are given by

06π

4π

3π

2π

sin 012

12

32

1

cos 13

2

12

12

0

l Graphs of sin θ , cos θ are continous every where

— Maximum value of both sin θ and cos θ is 1.

— Minimum value of both sin θ and cos θ is -1.

— Period of these functions is 2π.

LET US SUM UP

MATHEMATICS

Notes


92


l tan θ and cot θ can have any value between −∞ and + ∞.

— The function tan θ has discontinuities (breaks) at 2π

and 32π

in ( )0, 2π .

— Its period is π .— The graph of cot θ has discontinuities (breaks) at 0, π ,2π. Its period is π .

l sec θ cannot have any value numerically less than 1.

(i) It has breaks at 2π

and 32π

. It repeats itself after2π.

(ii) cosecθ cannot have any value between −1 and +1.

It has discontinuities (breaks) at 0, π , 2π.It repeats itself after 2π.

l http://www.wikipedia.orgl http://mathworld.wolfram.com

TERMINAL EXERCISE ONS

1. A train is moving at the rate of 75 km/hour along a circular path of radius 2500 m.Through how many radians does it turn in one minute ?

2. Find the number of degrees subtended at the centre of the circle by an arc whose lengthis 0.357 times the radius.

3. The minute hand of a clock is 30 cm long. Find the distance covered by the tip of the handin 15 minutes.

4. Prove that

(a) 1 sin

sec tan1 sin

− θ= θ − θ

+ θ(b)

1sec tan

sec tan= θ − θ

θ + θ

(c) 2 2tan cot

2sin cos1 tan 1 cot

θ θ− = θ θ

+ θ + θ(d) ( )21 sin

tan sec1 sin

+ θ= θ + θ

− θ

(e) ( )( )8 8 2 2 2 2sin cos sin cos 1 2sin cosθ − θ = θ − θ − θ θ

(f) 2 2sec cosec tan cotθ + θ = θ + θ

5. If 4π

θ = , verify that 3sin3 3sin 4sinθ = θ − θ

SUPPORTIVE WEB SITES

MATHEMATICS 93

Notes



6. Evaluate :

(a)25

sin6

π(b)

21sin

4π

(c) 3tan

4π

(d)17

sin4

π (e)19

cos3

π

7. Draw the graph of cos x from x2π

= − to 3

x2π

= .

8. Define a periodic function of x and show graphically that the period of tan x is π , i.e. theposition of the graph from x = π to 2π is repetition of the portion from x = 0 toπ .

MATHEMATICS

Notes


94


ANSWERS


1. (i) 3π

(ii) 12π

(iii) 512

π(iv)

712

π(v)

32π

2. (i) 45° (ii) 15° (iii) 9° (iv) 3° (v) 120°

3. 4π

, 1336

π,

1436

π4.

56π

5. 3π


1. (a) 6π

(b) 3π

(c) 56π

2. (a) 36° (b) 30° (c) 20°

3.16

radian; 9.55° 4. 15

radian 5. 95.54 m

6. (a) 0.53 m (b) 38.22 cm (c) 0.002 radian(d) 12.56 m (e) 31.4 cm (f) 3.75 radian(g) 6.28 m (h) 2 radian (i) 19.11 m.

CHECK YOUR PROGRESS 16.31. (i) − ive (ii) − ive (iii) − ive (iv) + ive

(v) + ive (vi) − ive (vii) + ive (viii) − ive

2. (i) zero (ii) zero (iii) 12

− (iv) − 1

(v) 1 (vi) Not defined (vii) Not defined (viii) 1


2.1

sin5

θ = , 2

cos5

θ = ,cot 2θ = , cosec 5θ = , 5

sec2

θ =

3.a

sinb

θ = , 2 2b a

cosb−

θ = , 2 2

bsec

b aθ =

−,

2 2

atan

b aθ =

−,

2 2b acot

a−

θ = 6. 22m

1 m+

MATHEMATICS 95

Notes




1. (i) 1

44

(ii) 1

62

(iii) 1− (iv) 223

(v) Zero


1. 1, 1− 3. Graph of y = 2 sin θ , [ ]0, π

Fig. 16.36

4. (a) 7 11

,6 6π π

(b) 4 5

,3 3π π

5. y sinxfrom to= − π π

Fig. 16.37


1. (a) y cos , to4 4π π

= θ −

Fig. 16.38

MATHEMATICS

Notes


96

Trigonometric Functions-I(b) y 3cos ; 0 to 2= θ π

Fig. 16.39

(c) y cos3 , to= θ −π π

cos 0.87θ =

,6 6π π

θ = −

cos 0.87θ = −

5 5

,6 6π π

θ = −

(d) Graph of y cos= θ in 3

,2 2π π

lies below the x-axis.

(e) y cos= θ

θ lies in 2π to 4π

Fig. 16.41


1. (a) Infinite (b) At 2π

, 32π

there are breaks in graphs.

Fig. 16.40

MATHEMATICS 97

Notes



(c) y tan2 , to= θ − π π

At ,tan 1.73π

θ = θ =

2. (a) Infinite (b) 3

cot 1 at4π

θ = − θ =


1. (a) y sec= θ

Fig. 16.42

Points of discontinuity of sec2θ are at 3

,4 4π π

in the interval [ ]0, 2π .

(b) In tracing the graph from 0 to 2− π , use cosec ( ) cosec −θ = − θ.


1. (a) Period is 23π

(b) Period is 22π

= π (c) Period of y is 3π

(d) 2 1 cos4x 1 1y sin 2x cos4x

2 2 2−

= = = − ; Period of y is 2

i.e4 2π π

(e) x 1

y 3cot3+ =

, Period of y is 3

13

π= π

TERMINAL EXERCISE

1.1

radian2

2. 20.45° 3. 15 cmπ

6. (a) 12

(b) 12

− (c) 1− (d) 12

(e) 12

MATHEMATICS

Notes


98


7.

Fig. 16.43

8.

y sec= θ

Fig. 16.44

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