Download Basic Trigonometric results and trigonometric ratios of arbitrary

Basic Trigonometric results and
trigonometric ratios of arbitrary angles
INTRODUCTION :
The word Trigonometry in its primary sense signifies the measurement of triangles. From
ancient times the Trigonometry also included the establishment of the relations which
subsist between the sides , angles and area of a triangle; but now it has a much wider
scope and embraces all manner of geometrical and algebraical investigations carried on
through the medium of certain quantities called trigonometrical ratios. In every branch of
Higher Mathematics, whether Pure or Applied, a knowledge of Trigonometry is of the
greatest value.
1.
DEFINITION OF ANGLE : Suppose that the straight line in the figure is capable of
revolving about the point O, and suppose that in this way it has passed successively from
the position OA to the position occupied by OB, OC, OD, …., then the
angle between OA and any position such as OC is measured by the
which the line OP has undergone in passing from its initial position
OA into its final position OC. Moreover the line OP may make any
number of complete revolutions through the original position OA
before taking up its final position. In Trigonometry angles are not
restricted as Geometry, but may be of any magnitude. The point O
is called the origin, and OA the initial line; the revolving line OP is known as generating
line or the radius vector.
2.
MEASUREMENT OF ANGLES : There are several ways of measuring angles in trigonometry.
The student is already familiar with the Sexagesimal Measure in which a right angle is
equal to 90. There is an absolute measure of an absolute angle which is called Circular
or Radian Measure. In this system one unit of angle (one radian) is the angle subtended
at the centre of a circle by an arc whose length is equal to the radius. Since the half
circumference of a circle of unit radius is . We must have c = 180 where the index c
 180 
signifies the circular measure. This is loosely written as  = 180. Thus 1c  
  and
  
  
10  
 .
 180 
The student is advised to remember the following angles in radians.
c
Deg
30
45
60
90
120
135
150
Rad

6

4

3

2
2
3
3
4
5
6
180

210
225
240
270
300
315
330
360
7
6
5
4
4
3
3
2
5
3
7
4
11
6
2
We will be using Sexagesimal system (angles in degrees) in evaluation of trigonometric
ratios initially only after which angles will be in radians unless the contrary is specified.
3. DEFINITIONS OF TRIGONOMETRIC FUNCTIONS
For an angle  between 0 and 90 we define
several trigonometric ratios to describe size or magnitude of the angle. Let the initial position
1
acquires a position OB given that BOA = .
to OA, we define
of the revolving line be OA and the revolving line
Choose a point P on OB and draw PM perpendicular
MP
OP
sin  
, cos ec 
OP
MP
OM
OP
cos 
, sec 
OP
OM
PM
OM
tan  
, cot  
OM
PM
We can easily note that
(i) the values of the above six ratios do not depend upon the choice of P (If P′ is some other
point and M′ is the foot of  then  OPM and OP′M′ are similar)
1
1
1
sin 
cos
(ii)
, tan  
cos ec 
,sec 
, cot  
, cot  
sin 
cos
tan 
cos
sin 
(iii)
sin 2   cos2   1, 1  tan 2   sec2  , 1 + cot2  = cosec2 
(follow by Pythogorous theorem)
Section –I Trigonometric transformations
1.
We are already familiar with transformations like
sin(90   )  cos ,
cos(90   )  sin 
, tan(90   )  cot  etc.
Let us now show that if  is any arbitrary angle then
(i)
sin(  )   sin 
(ii)
cos(  )  cos
(iii)
sin( 180   )   sin 
(iv)
cos(180   )   cos 
(v)
sin( 90   )  cos ,
(vi)
cos(90   )   sin 
We will give circle proof for above transformations formulas
We will first of all assume that  is a positive acute angle i.e. 0    90 or 0   

2
and consider
the circle with centre O and radius r . Let A' OA is the horizontal diameter of the circle. Let the
revolving line make an angle  with positive directions of
x  axis .
Draw PM perpendicular to x  axis and produce it to a point Q
on the circle. Then PM  MQ and Q
2
is the point (r cos ,r sin  ) . This is symbolically written as
r cos(  )  r cos , r sin(  )  r sin  on
Cancelling r we get cos(  )  cos , sin(  )   sin 
we can also get tan(  )   tan  ,
cot(  )   cot  , sec(  )  sec 
cos ec( )   cos ec from the second formula .
For instance tan(  ) 
sin(  )  sin 

  tan  etc.
cos(  ) cos 
Now to prove (iii) and (iv) we again consider the same circle (see figure) produce the radius PO in the
opposite direction to reach the point Q on the circle
Note that OAQ  180   then OM '  OM , M ' Q  MP . But co-ordinate sign convection wise Q
will have co-ordinates (r cos  ,r sin  ) .
This is symbolically written as r cos(180   )  r cos
r sin(180   )   sin  . Thus cos(180   )   cos
Also tan(180   ) 
sin( 180   )  sin 

 tan  .
cos(180   ) cos
From these relations we at once get
sin( 180   )  sin(   180)
  sin(  )  sin  , cos(180   )  cos(   180)
  cos( )   cos . Also tan(180   )   tan 
To prove (v) and (vi) we again consider the circle let
AOP   , OP  r
POQ  90
Then OM  r cos
PM  r sin 
OM '  PM  r sin 
QM '  OM  r cos 
Following sign conventions of co-ordinate geometry co-ordinates of Q are (r sin  , r cos  ) giving us
cos(90   )   sin  , sin( 90   )  cos 
3
Using above six transformations formula’s we can derive all types of transformations formula’s.
For example (270   )  sin( 180  90   )   sin( 90   )   cos 
cos( 270   )  cos(180  90   )
  cos(90   )  sin 
tan( 270   )   cot 
We will now show that all above results are true for arbitrary  i.e.  may be any angle.
If  lies between 90 and 180 then   90  
Now sin( 90   )  sin( 90  90   ) = cos(90   )   sin 
  sin(   90)
 sin( 90   )  cos 
if  lies between 180 and 270 then   180  
sin( 90   )  sin( 90  180   )
 sin( 270   )
  cos 
  cos(  180)   cos 
Thus sin( 90   )  cos in all cases geometrically it is easily seen as sin(   90)  cos
We can show the validity of transformation formulas for an arbitrary angle θ directly also
We will first prove it for 90  θ  180 . If 90  θ  180 then the point P  cos θ,sin θ  lies in second
quadrant (see figure and note that cos θ  0,sin θ  0 )
If P moves in counter clockwise direction for an angle 180 then P reaches a point P  in the fourth
quadrant. P  has co-ordinates   cos θ ,  sin θ  . Note that  cos θ  0,  sin θ  0
cos  θ  180   cos θ  sin  θ  180   sin θ as before
4
Again if P lies in third quadrant then P is  cos θ ,sin θ  while P  is   cos θ ,  sin θ  in first quadrant
Finally if P lies in fourth quadrant then P is   cos θ ,  sin θ 
In all the cases we have
cos  θ  180   cos θ , sin  θ  180   sin θ
We can similarly show this fact for other transformation formulas. If an angle φ is more than 360 then
φ  360n  θ where 0  θ  360 and n is a positive integer. Now T-ratio of φ will be same as
corresponding T-ratios of θ . For example
sin 180  φ  sin 180  360n  θ   sin 180  θ    sin θ
Thus sin 180  θ    sin θ for arbitrary θ
The other transformation formulas are also true for all θ  R . See more illustrations in 2(d)
2.
How to remember all transformation formulas easily ?:
Though we have proved all transformation formulas but have not discussed the way to remember all the
formulas. Let us discuss four important points in detail.
IInd
IIIrd
90  θ
θ
180  θ
90  θ
first
270  θ
180  θ
360  θ
270  θ
IVth
θ
(a).
Quadrant determination by assuming that θ is an acute angle. This is shown in the figure. As an
example if an angle is of the type 180  θ or 270  θ then it belongs to IIIrd quadrant.
(b).
Sign of Trigonometric functions : It is necessary to know the sign of signs of sines, cosines and
tangents only in various quadrants. The signs of cosec, sec and cot wil be same as that of
sin, cos and tan respectively.
sin +
All +
tan +
cos +
These signs convention is based up on co-ordinate geometry sign convention for above. In the
first quadrant all trigonometric functions are positive since x and y co-ordinates are both
positive. In the second sin θ is positive since y co-ordinate is positive. In the third tan θ is
positive since both x and y are negative and tan θ 
sin θ
. In the fourth cossinθ is positive as
cos θ
x-coordinate is positive. We can remember it by recalling the AFTER SCHOOL TO COLLEGE
A of AFTER stands for all , S of school stands for sin θ
T of To stands for tan θ , C of College stands for cosθ
5
Thus, sin 210  0, cos 300  0
Since 210 belongs to third quadrant while 300 belongs to fourth. We can also say
sin 180  θ   0,cos  θ   0
tan  360  θ   0 , sec  270  θ   0,cot  90  θ   0 etc.
( c).
The transformation formulas do not change the type of trigonometric function on both side of it
if transformation is being applied on an angle of the type θ ,180  θ ,360  θ . But it changes
from sin to cos, cos to sin, tan to cot, cot to tan, sec to cosec, cosec to sec
If transformation is being applied on an angle of the type 90  θ , 270  θ
Example :
If we want to simplify sin 180  θ  we note that
(i) 180  θ belongs to third quadrant
(ii) sin 180  θ   0
(iii) No change in T-ratio will occur  sin 180  θ    sin θ
Again cos  270  θ   sin θ . Since cos will change to sin and 270  θ belongs to fourth
quadrant where cos is positive. Further note cot  θ    cot θ . Thus we can remember all
formulas easily. We have the following long list of formulas.
SET I
SET II
sin  θ    sin θ
sin 180  θ   sin θ
cos  θ   cos θ
cos 180  θ    cos θ
tan  θ   tanθ
tan 180  θ    tan θ
cot  θ    cot θ
cot 180  θ    cot θ
sec  θ   sec θ
sec 180  θ    sec θ
co sec  θ   co sec θ
co sec 180  θ   co sec θ
SET III
SET IV
sin 180  θ    sin θ
sin  360  θ    sin θ
cos 180  θ    cos θ
cos  360  θ   cos θ
tan 180  θ   tan θ
tan  360  θ    tan θ
6
cot 180  θ   cot θ
cot  360  θ    cot θ
sec 180  θ    sec θ
sec  360  θ   sec θ
cosec 180  θ   co sec θ
cos ec  360  θ    cos ecθ
SET V
SET VI
SET VII
sin  90  θ   cos θ
sin  270  θ    cos θ
sin  270  θ    cos θ
cos  90  θ    sin θ
cos  270  θ    sin θ
cos  270  θ   sin θ
tan  90  θ    cot θ
tan  270  θ   cot θ
tan  270  θ    cot θ
cot  90  θ    tan θ
cot  270  θ   tan θ
cot  270  θ    tan θ
sec  90  θ    cos ecθ
sec  270  θ    cos ecθ
sec  270  θ   cos ecθ
co sec  90  θ   sec θ
co sec  270  θ    sec θ
co sec  270  θ    sec θ
TRIGONOMETRIC RATIOS OF ANY ARBITRARY ANGLE: If the angle is greater than 360, then the
T-ratios can be found by using the fact that,
sin(360 n + ) = sin ; cos(360 n + ) = cos ; tan(360 n + ) = tan ,
where “n” is an integer. The following steps may be used :Step I: Remove negative sign by using transformations (i)
Step II: Divide the angle by 360 and consider the T-ratio of the remainder.
Step III : Express the angle as 180   or 360 - , if it is not acute (use of 90  , 270   may
be avoided since in 180  , 360   and - cases, the T-ratio remains same).
We will illustrate it by examples
(i)
sin  900  θ   sin  2  360  180  θ 
 sin 180  θ    sin θ
cos 1360  θ   cos  3 360  270  θ 
 cos  270  θ    sin θ
(ii)
sin 840 = sin(360  2 + 120) = sin 120 = sin(180 – 60) = sin 60 =
(iii)
tan(-1200) = -tan 1200 = -tan(3  360 + 120) = -tan 120
= -tan(180 – 60) = -(-tan 60) = 3
3
2
(iv)
tan  7π  θ   tan  6π  π  θ   tan  π  θ    tan θ
(v)
π


 2016π  π

tan  2015  θ   tan 
 θ  (Bringing a multiple of 4 near 2015)
2
2




7
π


 π

 tan 1008π   θ   tan    θ  ( 1008π are 504 resolutions of line)
2


 2

 π

π

 tan     θ     tan   θ    cot θ

2

 2
Section –II Given one T-ratio with location of angle how to find other T-ratios
In earlier classes we have done that, given sin θ  3/ 5. Find cos θ , tan θ , cot θ ,sec θ etc. We used to
solve this problem quickly by constructing a triangle only.
And used to determine other T-ratios easily. This used to be a simple and routine problem in earlier
classes but at an advanced level it is to be supplemented by more concepts and rigour. For instance if
3
sin θ  and 90  θ  180 then as cos θ  1  sin 2 θ
5
We have  cos θ  1  sin 2 θ

 cos θ   1  sin 2 θ   1 
cos θ  0  cos θ   cos θ 
sin θ
3/ 5
9
4

  Also tan θ 
cos θ 4 / 5
25
5
TRIGONOMETRIC RATIOS OF 0, 90 AND SOME POSITIVE ACUTE ANGLES :
0
30
45
60
1
1
3
sin
0
2
2
2
1
1
3
cos
1
2
2
2
1
tan
0
1
3
3
Trigonometric ratios of angles which are multiples of
90
1
0
not defined

: If n is any integer then the angles
2
n
are of great importance in mathematic since sine and cosine become  1 or 0 at these
2
angles.
(i)
sin n = 0
(ii)
cos n = (1)n (iii)
tan n = 0
(iv)
sin(4n + 1) /2 = 1
(v)
sin (4n – 1) /2 = -1
(vi)
cos(2n + 1) /2 = 0
(vii)
tan (2n + 1) /2 is not defined.
8
Domain, Range of trigonometric functions:
T – Function
Domain
Range
sin x
R
[-1, 1]
cos x
R
[-1, 1]
tan x
R – {x : x = (2n + 1)/2} R
cot x
R – {x : x = n}
sec x
R – {x : x = (2n + 1)/2} (-, -1]  [1, )
cosec x
R – {x : x = n }
R
(-, -1]  [1, )
SOLVED EXAMPLES
1.
Show that
SOLUTION:-
sin( 90   ) cos(180   ) sec(180   )
1
tan( 270   ) sin( 360   )
,
cos(180   )   cos 
sec(180   )   sec ,
tan( 270   )   cot 
sin( 90   )  cos
sin( 360   )   sin 
Thus LHS of above expression 
2.
Show that (i)
cos  ( cos  )(  sec  )
1
 cot  ( sin  )
sin( 2015   )   sin( 35   ) (ii)
sin( 2015   )  sin( 1800  215   )
SOL:- (i)
 sin( 215   )  sin( 180  35   )
(ii)
sin( 600   )
 tan( 60   )
cos(1680   )
600    360  240  

LHS

  sin( 35   )
1680    360  4  240  
sin( 240   )
 tan( 240   )
cos( 240   )
 tan(180  60   )  tan( 60   )
3.
 3

cos
   sin( 2   )
 2

Show that
 3

cos ec
   sec(8   )
 2

 sin 2  cos 2 
9
SOL:-
 3

cos
    sin 
 2

sin( 2   )   sin 


cos ec      sec 
2

sec(8   )  sec(  )  sec 


LHS


sin  ( sin  )
( sec  ) sec 
 sin 2  cos 2 


     cos
2

4.
Show that sin  2015
SOL:-
The integer just less than 2015 which is divisible by 4 is 2012 . We write sin  2015


 2012  3

 sin 
  
2


3


 sin 1006 
 
2




 
2

 3

 sin 
     cos 
 2

EXERCISE
1.
2.
3.
4.
5.
6.
7.
8.
9.
Find the radian measures corresponding to the following degree measures
15
240 (4)
530
(1)
(2)
(3)
3730
Find the degree measures corresponding to the following radian measures
3
5
7
(1)
(2)
-4
(3)
(4)
4
3
3
A wheel makes 360 revolutions in one minute .Through how many radians does it turn in one
second?
Find the degree measure of the angle subtended at the centre of a circle diameter 200 cm
by an are of length 22cm.
In a circle of diameter 40cm , the length of chord is 20cm. Find the length of minor arc of the
chord.
If, in two circles, arcs of the same length subtend angles of 60 and 75 at the centre , find the
ratio of their radii:
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip
describes an arc of length:
(1)
10cm
(2)
15cm
(3)
21cm
Find the values of the other five trigonometric functions in each of the following problems:
1
3
(1)
(2) sin   , lies in second quadrant.
cos   , lies in third quadrant
2
5
4
13
(3)
(4)
tan   , lies in third quadrant.
sec  , lies in fourth quadrant.
3
5
show that
(1)
sin 3060  0 (2)
cos570  
10
1
2
(3) sin 1845 
1
2
(4)
(7)
(10)
(13)
(16)
10.
1
2
1
sin(330) 
2
1
sin 765 =
(11)
2
1
cos 855  
2
cos 1755 
sin 4530  
1
2
(5)
cos ec(1200) 
(8)
sin 510 
2
3
1
2
 15 
cot  
 1
 4 
(14)
cos(1125) 
(17)
tan
(6)
tan 315  1
(9)
cos ec(1410)  2
1
11

6
3
(12)
tan
1
2
(15)
tan(585)  1
(18)
sin
13
. 3
3
3
5

3
2
Prove that :
1
2
(1)
sin 780 sin120  cos 240 sin 390
(2)
sin 300 cos 210  cos 300 sin 210  1
(3)
tan 720  cos 270  sin150 cos120 
(4)
(5)
(6)
1
4
sin 600 cos390  cos 480 sin150  1
tan(225)cot (405)  tan (765)cot 675 0
sin
8
23
13
35 1
cos
 cos
sin

3
6
3
6 2
(7)
cos 24  cos55  cos125  cos 204  cos300 
(8)
tan 225 cot 405  tan 765 cot 675 0
(9)
cos570 sin510  sin (330 cos(390)  0.
(10)
2sin 2
(12)
2sin 2
(14)
tan
(15)
3


 2cos2  sec2  6
4
4
3

6
 cos ec
7

cos2  0
6
3
1
2
(11)
sin 2
(13)
cot 2

6

6
11
2 3

17 3  4 3
 2sin
 cos ec 2  4cos 2

.
3
3 4
4
6
2
cos(  x)cos(  x)
 cot 2 x


sin(  x)cos   x 
2

11
 cos2

3
 cos ec
 tan 2

4

1
2
5

 3tan 2  6
6
6
11.
(16)
sin 180  A cot  90  A cos  360  A


 sin A
tan 180  A tan  90  A
sin   A
(17)
sin(180   )cos(90   ) tan(270   )cot (360   )
1
sin(360   )cos(360   )cos ec( )sin(270   )
(18)
cos ec(90   )  cot (450   )
tan(180   )  sec(180   )

2
cos ec(90   )  tan (180   )
tan (360   )  sec( )
(19)


 3


  3
 

sin     cos     cot 
    sin     sin 
   cot    
2
2
2
2
2





 
 



cos(2   )cos ec(2   ) tan    
2

 1.
(20).


sec     cos cot    
2

  3

 3


(21).
cos 
 x  cos(2  x) cot 
 x   cot(2  x)   1
 2


  2

In a ABC, prove that:
C
 A B 
(1)
cos(A + B) + cos C = 0
(2)
cos 
  sin
2
 2 
A B
C
(3)
tan
 cot
2
2
Answers
1.
2.
(1)
(1)
3.
12 
7.
(1)
8.
(1)
(2)
(3)
(4)
(5)
(10)
(15)
/12
42 57 16
2/15
(2)
(2)
- 5/24
-229 5′ 27
4.
12 36
5.
(2)
1/5
(3)
(3)
4/3
(3)
300
20
cm
3
7/25
(4)
(4)
53/18
420
6.
5:4
3
2
1
,cos ec  
,sec  2, tan   3,cot  
2
3
3
5
4
5
3
4
cos ec  ,cos   ,sec   , tan    ,cot   
3
5
4
4
3
4
5
3
5
3
sin    ,cos ec   ,cos   ,sec   ,cot  
5
4
5
3
4
12
13
5
12
5
sin    ,cos ec   ,cos  , tan    ,cot   
13
12
13
5
12
1
1
3


(6)
0
(7)
(8)
2
2
3
2

½
(11)
-½
(12)
½
(13)
3
1
1
1

(16)
(17)
(18)
2
2
2
sin   
12
(9)
-1
(14)
-1
-½
Section –III
TRIGONOMETRIC RATIOS OF COMPOUND ANGLES
When we learn a new function, we come to know about various peculiar properties of these functions.
Trigonometric functions in general do not satisfy the relations
sin  A  B  sin A  sin B
tan  A  B  tan A  tan B
The actual and mathematically correct expansion of sin  A  B, tan  A  B initially look very
astonishing but later we realize that almost entire trigonometry run around them, one single
formula
sin  A  B  sin A cos B  cos Asin B
(*)
is capable of generating almost all formulas of trigonometry. If we assume the validity of (*)
(Proof later) then
sin  A  B  sin A cos  B  cos Asin  B  sin A cos B  cos Asin B






cos A  B   sin   A  B   sin   A  cos B  cos  A  sin B
2
2
2






 cos A cos B  sin Asin B
cos A  B  cos A cos  B  sin Asin  B  cos A cos B  sin Asin B
sin A cos B cos A sin B

sin A sin B  cos A sin B
tan A  tan B
sin  A  B 

 cos A cos B cos A sin B 
tan  A  B  
cos A cos B sin A sin B 1  tan A tan B
cos A  B  cos A cos B  sin A sin B

cos A cos B cos A cos B
tan  A  B  
tan A  tan  B 
1  tan A tan  B 
cot  A  B  
cot  A  B  
cot A cot B  1
1
1  tan A tan B


cot B  cot A
tan  A  B  tan A  tan B
cot A cot  B   1 cot A cot B  1

cot B  cot A
cot  B   cot A
In the next two chapters we will observe that by adding, subtracting two A  B formulas of same kind
we can express sums as products and products as sums. We will further observe that main
trigonometric formulas sin 2 A, cos 2 A, sin 3 A are also derived from these A  B formulas only.
In this chapter we will be discussing questions on following six formulas :-
sin  A  B  sin A cos B  cos Asin B
13
sin  A  B  sin A cos B  cos A sin B
cos A  B  cos A cos B  sin Asin B
cos A  B  cos A cos B  sin Asin B
tan  A  B  
tan A  tan B
tan A  tan B
, tan  A  B  
1  tan A tan B
1  tan A tan B
In addition to these there are two ‘surprise formulas’
sin  A  B . sin  A  B   sin 2 A  sin 2 B and cos A  B . cos A  B   cos 2 A  sin 2 B which
are easily proved by expanding and multiplying. We call them surprise formulas because the
pattern is false for other trigonometric functions. For example
tan  A  B . tan  A  B   tan 2 A  tan 2 B .
We will now give formal proof of these formulas
To prove the formulae
sin  A  B   sin A cos B  cos A sin B,
cos  A  B   cos A cos B  sin A sin B , see figure
Let XOX '  A, and X ' OX ''  B; then XOX ''  A  B.
On OX '' , the final position of initial line of the compound angle A  B , take any point P , and draw
PQ and PR perpendicular to OX and OX ' respectively; also draw RS and
RT perpendicular to OX and PQ respectively.
Now,
Since
PQ RS  PT RS PT



OP
OP
OP OP
RS OR PT PR

.

.
 sin A.cos B  cos A.sin B
OR OP PR OP
TPR  90  TRP  TRO  ROS  A;
sin  A  B  
Thus sin  A  B   sin A cos B  cos A sin B
14
Also
cos  A  B  
OQ OS  TR OS TR



OP
OP
OP OP

OS OR TR PR
.

.
OR OP PR OP
 cos A.cos B  sin A.sin B
Thus cos  A  B   cos A cos B  sin A sin B
Finally tan  A  B  
But
RS PT
RS PT


OS
OS
OS
OS


TR
TR TP
1
1
.
OS
TP OS
PQ RS  PT

OQ OS  TR
RS
TR
 tan A, and
 tan A;
OS
TP
TP PR

 tan B
OS OR
tan A  tan B
Thus tan  A  B  
1  tan A tan B
also the triangles ROS and TPR are similar 
We can similarly derive tan  A  B  
tan A  tan B
1  tan A tan B
To prove the formulas
sin  A  B   sin A cos B  cos A sin B
cos  A  B   cos A cos B  sin A sin B
Let XOX '  A, and X ' OX ''  B; then XOX ''  A  B.
On OX ' take any point P , and draw PQ and PR perpendicular to OL and OM respectively.
Draw RS and RT perpendicular to OL and QP respectively.
Now,
PQ RS  PT RS PT



OP
OP
OP OP
RS OR PT PR

.

.
OR OP PR OP
sin  A  B  
15
 sin A.cos B  cos A.sin B
Since
TPR  90  TRP  X '' RT  X ' OX  A;
sin  A  B   sin A cos B  cos A sin B.
Also
OQ OS  RT OS RT



OP
OP
OP OP
OS OR RT RP

.

.
OR OP RP OP
cos  A  B  
 cos A.cos B  sin A.sin B
SOLVED EXAMPLES
1.

 


 

   cos     sin     sin     = sin(    )
4
 4

4
 4

Show that cos
Sol : If we examine carefully LHS  cos A cos B  sin A sin B where A 
4
 , B 





LHS  cos( A  B)  cos        cos         sin    
4
4

2


2.

cos 2 33  cos 2 75
 2
2 69
2 21
sin
 sin
2
2
Show that

 
Sol : cos 2 33  cos 2 57  1  sin 2 33  1  sin 2 57

 sin 2 57  sin 2 33
 sin 57  33sin 57  33  sin 90 sin 24  sin 24




 69 
 21 
 69 21 
 69 21 
sin    sin 2   = sin    . sin     sin 45 sin 24
2
2
 2 
 2
 2
 2
2
LHS 

3.
Show that
RHS
=

sin 24

sin 45 sin 24
2
cos65  sin 65
  cot 70
cos65  sin 65
1  tan 65
1  tan 65
tan 45  tan 65
1  tan 45  tan 65
( on dividing by cos 65 )
= tan 45  65
16
tan( 20) =  tan 20   cot 70

4

4.
Show that in a triangle whose angles are A, B, C
tan A  tan B  tan C  tan A.tan B.tan C
Sol :
A  B 180  C



tan( A  B)  tan(180  C )
tan A  tan B
tan C

1  tan A  tan B
1
tan A  tan B   tan C  tan A tan B tan C

tan A  tan B  tan C  tan A tan B tan C
NOTE: In general , If ' tan' of both sides is taken we get a result of the type
Sum of tangents = Product of tangents
For example tan 10  tan 7    

5.
tan 10 
tan 7  tan 
1  tan 7 tan 
tan 10  tan 7  tan   tan 10 tan 7 tan 
If  and  are the solutions of the equation a tan   b sec   c, then show that
tan     
Sol :

2ac
a  c2
2
a tan   b sec  c 
c  a tan  2  b 2 sec 2 
On arranging

c  a tan   b sec

(*)
c 2  a 2 tan 2   2ac tan   b 2 (1  tan 2  )
tan 2  (a 2  b 2 )  2ac tan   (c 2  b 2 )  0
(**)
Since  and  are the solutions of (*) . Therefore , tan  and tan  are roots of
equation (**).


tan   tan  
2ac
c2  b2
and
tan

tan


a 2  b2
a 2  b2
2ac
2ac
tan   tan 
(a  b 2 )
=
= 2
tan(   ) 
2
2
c b
a  c2
1  tan  tan 
1 2
a  b2
2
EXERCISE
17
1.
2.
3.
24
3
3
3
, cos   where    
and     2 .
25
5
2
2
3
4
Show that (i) sin     
(ii) cos      
5
5
3

12
3
16
sin   ,0    ,cos    ,    
show that tan     
5
2
13
2
63
If cos   
Prove the following identities:
(1)
cos  A  B  cos B  sin  A  B  sin B  cos A
(2)
sin 3A cos A – cos 3A sin A = sin 2A
(3)
cos(30 + A) cos(30 - A) – sin(30 + A)sin(30 - A) = ½
(4)
sin(60 - A) cos(30 + A) + cos(60 - A)sin(30 + A) = 1
(5)
sin 2 cos  + cos 2 sin  = sin 4 cos  - cos 4 sin 
(6)
cos 2 cos  + sin 2 sin  = cos 
(7)
cos 4 cos  + sin 4 sin  = cos 2 cos  - sin 2 sin 
(8)
cos 4 cos  - sin 4 sin  = cos 3 cos 2 - sin 3 sin 2
(9)
tan      tan 
1  tan     tan 
 tan 
(10)
tan 4 A  tan 3 A
 tan A
1  tan 4 A tan 3 A
(12)
sin(150 + x) + sin(150 - x) = cos x
(15)
cos2 2x – cos2 6x = sin 4x sin 8x
(11)
cot(   ) cot   1
 cot 
cot   cot(   )
(13)
 3

 3

cos 
 x   cos 
 x    2 sin x
 4

 4

(14)




cos   x   cos   x    2 sin x
4
4




(16)
tan 3x tan 2x tan x = tan 3x – tan 2x – tan x
(17)
sin(60 + A) – sin(60 – A) = sin A
(18)
cos  30  A  cos  30  A   3 cos A (19)




cos      cos       2 sin 
4

4

(20)
sin2 5A – sin2 2A = sin 7A sin 3A
cos 2A cos 5A = cos2
(22)
(23)
sin     
sin  sin 

sin  B  C 
cos B cos C
sin    
sin  sin 


(21)
sin    
sin  sin 
sin  C  A 
cos C cos A

0
sin  A  B 
cos A cos B
18
0
7A
3A
 sin 2
2
2
(24)
tan( - ) + tan( - ) + tan( - ) = tan( - )tan( - )tan( - )
(25)
cos2  + cos2  - 2cos  cos  cos( + ) = sin2 ( + )
(26)
sin2  + sin2  + 2sin  sin  cos( + ) = sin2( + )
(27)
cos ec     
(29)

 


 

cos   x  cos   y   sin   x  sin   y   sin( x  y)
4
4
4
4

 


 

(30)


tan   x 
2
4
   1  tan x 
 1  tan x 



tan   x  
4

(31)
sin(n  1) x sin(n  2) x  cos(n  1) x cos(n  2) x  cos x
(32)
 3

 3

cos 
 x   cos 
 x    2 sin x
 4

 4

(33)
sin 2 6 x  sin 2 4 x  sin 2 x sin10 x (34)
(35)
cot x cot 2 x  cot 2 x cot 3x  cot 3x cot x 1
(36)
cos2 cos2  sin 2 (   )  sin 2 (   )  cos2(   )
cos ec cos ec
cot   cot 
(28)
cos     
1  tan  tan 
sec  sec 
cos2 2 x  cos2 6 x  sin 4 x sin8 x
4.
If 2 tan   cot   tan  , prove that cot   2 tan(    )
5.
If tan  
6.
If A  B 
7.
If A + B = 225, prove that
8.
Prove that :
9.
If sin(    )  1 and sin(    ) 
n sin  cos 
, show that tan(    )  (1  n) tan 
1  n sin 2 

4
, prove that : (i)
(1  tan A)(1  tan B)  2 (ii) (cot A  1)(cot B  1)  2
cot A
cot B
1


1  cot A 1  cot B 2
tan 2 2 x  tan 2 x
 tan 3x tan x
1  tan 2 2 x tan 2 x
1

, where 0   ,   , then find the values of
2
2
tan(   2 ) and tan( 2   ) .
19
10.
Prove that
1
cot( x  a)  cot( x  b)

sin( x  a) sin( x  b)
sin( a  b)
11.
If tan  
m
1
, tan  
, show that  +  = /4.
m 1
2m  1
Section –IV Expressing product as sum and sum as products
Product as sum : In the last section we have learnt four important addition and subtraction formulas. If
we add or subtract two such formulas of same kind we get formulas expressing product as a sum and
sum as a product. Indeed by adding sin  A B and sin  A B formulas we easily get
sin  A  B  sin  A  B  2 sin Acos B . If we write it in the reverse order we get a formula expressing
the product of the type 2 sin  cos  as a sum of two sines.
For example, 2 sin 65 cos 25  sin 65  25  sin 65  25  sin 90  sin 30  1  1/ 2  3 / 2 .
Similarly, we can have formulas 2 cos Asin B  sin  A  B  sin  A  B
2 cos Acos B  cos A  B  cos A  B
2 sin Asin B  cos A  B  cos A  B
One has to be careful in applying last formula in which   comes at first place
Sum as product : - If in all above four formulas we put A  B  C , A  B  D then by writing in the
CD
CD
cos
reverse direction we easily get sin C  sin D  2 sin
2
2
CD
CD
sin C  sin D  2 cos
sin
2
2
CD
CD
cos C  cos D  2 cos
cos
2
2
CD
CD
cos C  cos D  2 sin
sin
2
2
Once again we have to be careful in applying formula number 4. The intricacies of these formulas will be
more clear when we do sufficient number of problems. For instance the formula
2 cos Asin B  sin  A  B  sin  A  B may be avoided if we write
2 cos Asin B  2 sin B cos A  sin B  A  sin B  A  sin  A  B  sin  A  B
A   will never appear in cos C  cos D because of cos    cos . Let us illustrate usage of these
formulas by means of examples.
SOLVED EXAMPLES
1.
1
Show that sin140 cos 40 cos10
2
Sol :
sin 140 cos 40 

1
2 sin 140 cos 40
2
1
sin( 140  40)  sin( 140  40)
2
20
1
sin 180  sin 100  1 sin100  1 cos10
2
2
2
1
Show that sin 10 sin 50 sin 70 
8
1
1
LHS  sin 102 sin 50 sin 70  sin 10cos (50  70)  cos(50  70)
2
2
( on applying 2 sin A sin B  cos( A  B)  cos( A  B)]
1
= sin 10 cos( 20 )  cos120
2
1
1
1

 sin 10cos 20  
(  cos( 20)  cos 20 and cos 120   )
2
2
2

1
1
1
1
 sin 10 cos 20  sin 10
 2 sin 10 cos 20  sin 10
2
4
4
4
1
1
 sin (10  20)  sin( 10  20)  sin 10
4
4
1
1
sin 30  sin 10  sin 10  1 sin 30  1 sin 10  1 sin 10  1
=
4
4
4
4
4
8

2.
Sol :
3.
1
Show that sin A sin(60  A)sin(60  A)  sin 3 A
4
Sol :
LHS 
4.
1
1
1
1

sin A  cos 2 A    sin A cos 2 A  sin 3 A
2
2
2
4

1
1
1
  sin 3 A  sin A  sin A  sin 3 A
4
4
4
Show that sin( B  C ) cos( A  D)  sin( C  A) cos( B  D)  sin( A  B) cos(C  D)  0
1
sin A 2 sin( 60  A) sin( 60  A) 
2

1
sin A  cos( 2 A)  cos120
2

1
[ sin( B  C  A  D)  sin( B  C  A  D)  sin( C  A  B  D)  sin( C  A  B  D)
2
 sin( A  B  C  D)  sin( A  B  C  D)]
= 0 ( sin( B  C  A  D) gets cut with sin( C  A  B  D) etc )
Show that cos 4  cos10  2 cos 7 cos 3
CD
CD
cos
By applying cos C  cos D  2 cos
2
2
4  10
4  10
cos 4  cos10  2 cos
cos
 2 cos 7 cos 3
2
2
Show that cos18  sin 18  2 cos 63
LHS  cos18  sin 18  sin 72  sin 18
72  18
72  18
 2 cos
sin
 2 cos 45 sin 27  2 cos 63
2
2
LHS 
5.
Sol:
6.
Sol:
21
7.
Sol:
8.
Sol:
sin 5 x  sin 3 x
 tan 4 x
cos 5 x  cos 3x
5 x  3x
5 x  3x
2 sin
cos
2 sin 4 x cos x
2
2
 tan 4 x
=
LHS 
5 x  3x
5 x  3x
2 cos 4 x cos x
2 cos
cos
2
2
Show that
Show that sin x  sin 3x  sin 5x  sin 7 x  4 cos x cos 2 x cos 4 x
We will use sin C  sin D formula twice on correct pairs. The correct pairs are
sin x , sin 7 x ; sin 3 x and sin 5x
LHS   sin x  sin 7 x    sin 3x  sin 5 x 
 2 sin 4 x cos 3x  2 sin 4 x cos x
 2sin 4 x  cos3x  cos x   2 sin 4 x.2 cos 2 x cos x = 4 cos x cos 2x sin 4x
9.
Show that
Sol:
LHS 
sin 8 A cos 5 A  cos12 A sin 9 A
 cot 4 A
cos 8 A cos 5 A  cos12 A cos 9 A
sin 13 A  sin 3 A  sin 21A  sin 3 A
2sin8 A cos5 A  2cos12 A sin 9 A
=
2cos8 A cos5 A  (2cos12 A cos9 A)
cos13 A  cos 3 A  (cos 21A  cos 3 A)
sin 13 A  sin 21A
2 sin 17 A cos 4 A
CD
CD

cos
( Applying sin C  sin D  2 sin
cos13 A  cos 21A 2 sin 17 A sin 4 A
2
2
CD
CD
sin
and cos C  cos D   2 sin
)  cot 4 A
2
2
tan( A  B) k  1
If sin 2 A  k sin 2 B prove that
.

tan( A  B) k  1
sin 2 A k
sin 2 A  sin 2 B k  1


We
(applying componendo & divenendo )

sin 2 B 1
sin 2 A  sin 2 B k  1

10.
Sol:


2 A  2B
2 A  2B
cos
sin( A  B) cos( A  B) k  1
2
2
=

2 A  2B
2 A  2B
cos( A  B) sin( A  B) k  1
2 cos
sin
2
2
2 sin
tan( A  B) k  1
.

tan( A  B) k  1
EXERCISE
5
 1
sin

12
12 2
1.
Show 2 sin
2.
Show that sin 25 cos115 
1
(sin 40  1)
2
22
1
8
3.
Show that cos 40 cos 80 cos160  
4.
Show that cos
5.
4 cos12 cos 48 cos 72  cos 36
6
cos 3A sin 2A – cos 4A sin A = cos 2A sin A
7.
Prove the following identities:
5

1
cos

12
12 4
(1)
sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
(2)
sin  + sin 3 + sin 5 + sin 7 = 4cos  cos 2 sin 4
(3)
cos 7x + cos 5x + cos 3x + cos x = 4 cos x cos 2x cos 4x
(4)
cot 4x(sin 5x + sin 3x) = cot x(sin 5x – sin 3x)
(5)
cos9 x  cos5 x
sin 2 x

sin17 x  sin 3x
cos10 x
(6)
sin 5 x  sin 3x
 tan 4 x
cos5 x  cos3x
(7)
sin x  sin y
x y
 tan
cos x  cos y
2
(8)
sin x  sin 3x
 tan 2 x
cos x  cos3x
(9)
sin x  sin y
x y
 tan
cos x  cos y
2
(10)
tan 5  tan 3
 4cos 2 cos 4
tan 5  tan 3
(11)
sin x  sin 3x
 2sin x
sin 2 x  cos 2 x
(12)
sin 5x  2sin 3x  sin x
 tan x
cos5 x  cos x
(13)
(sin 7 x  sin 5 x)  (sin 9 x  sin 3x)
 tan 6 x
(cos7 x  cos5x)  (cos9 x  cos3x)
(14)
cos 4 x  cos3x  cos 2 x
 cot 3x
sin 4 x  sin 3x  sin 2 x
(15)
sin 3x + sin 2x – sin x = 4 sin x cos
(16)
cos  cos3
 tan 2
sin 3  sin 
(17)
sin 2  sin 3

 cot
cos 2  cos3
2
(18)
cos 4  cos
5
 tan
sin   sin 4
2
(19)
cos 2  cos12
 tan 5
sin12  sin 2
(20)
(22)
cos  2  3   cos3
sin  2  3   sin 3
sin      sin 4
cos      cos 4
x
cos
2
 cot 
 tan
(21)
  3
2
23
cos   3   cos  3   
sin  3     sin   3 
 tan   2 
8.
(23)
cos 3A + sin 2A – sin 4A = cos 3A(1 – 2sin A)
(24)
sin 3 - sin  - sin 5 = sin 3(1 – 2cos 2)
(25)
cos  + cos 2 + cos 5 = cos 2(1 + 2cos 3)
(26)
sin  - sin 2 + sin 3 = 4sin /2 cos  cos 3/2
(27)
sin 3 + sin 7 + sin 10 = 4sin 5 cos 7/2 cos 3/2
(28)
sin A + 2sin 3A + sin 5A = 4sin 3A cos2 A
(29)
sin 2  sin 5  sin 
 tan 2
cos 2  cos5  cos 
(31)
cos7  cos3  cos5  cos
 cot 2
sin 7  sin 3  sin 5  sin 
(32)
sin( +  + ) + sin( -  - ) + sin( +  - ) + sin( -  + ) = 4sin  cos  cos 
(33)
cos( +  - ) – cos( +  - ) + cos( +  - ) – cos( +  + ) = 4sin  cos  sin 
(34)
sin 2 + sin 2 + sin 2 - sin 2( +  + ) = 4sin ( + )sin( + )sin( + )
(35)
cos  + cos  + cos  + cos( +  + ) = 4cos
(36)
cos( + )sin( - ) + cos( + )sin( - ) + cos( + )sin( - ) + cos( + )sin( - ) = 0
(37)
sin  cos( + ) – sin  cos( + ) = cos  sin( - )
(38)
cos  cos( + ) – cos  cos( + ) = sin  sin( - )
(39)
sin( - ) + sin( - ) + sin( - ) + 4sin
(40)
sin (sin 3 + sin 5 + sin 7 + sin 9) = sin 6 sin 4
(41)
sin   sin 3  sin 5  sin 7
 tan 4
cos  cos3  cos5  cos7
(42)
sin x  sin3x  sin5x  sin 7 x  4cos x cos 2 x sin 4 x
(43)
2 cos
(44)
tan 60    tan 60    
(45)
cos 20 cos100  cos100 cos140  cos140 cos 200  
(46)
sin

2

13
cos
.sin
(30)
sin   sin 2  sin 4  sin 5
 tan 3
cos  cos 2  cos 4  cos5
 
2
 
2
sin
cos
 
2
 
2
sin
cos
 
 
2
2
0
9
3
5
 cos
 cos
0
13
13
13
2 cos 2  1
2 cos 2  1
3
4
7
3
11
 sin
sin
 sin 2 sin 5
2
2
2
If cos( A  B).sin C  D  cos( A  B) sin( C  D) prove that tan A tan B tan C  tan D  0
24
9.
Show that sin x sin y sin(x – y) + sin y sin z sin(y – z) + sin z sin x sin(z – x) + sin(x – y) sin(y – z)
sin(z – x) = 0 for all x, y, z.
Section –V Trigonometric Functions of multiple angles
Almost all trigonometric formula have been generated by sin( A  B ) formula
Now we will observe that most useful formulas being generated by these formula’s.
The formulas express Trigonometric functions of the angle n A in terms of Trigonometric
Functions of angles A where n is an integer ( in general n can be positive rational ) we start deriving
these formula’s.We will put frequently used formulas in a block.
sin 2 A  sin( A  B)  sin A cos A  cos A sin A  2 sin Acos A
Thus sin 2 A  2 sin A cos A
We can similarly derive,
sin 4 A  2 sin 2 A cos 2 A, sin 8 A  2 sin 4 A cos 4 A, sin A  2 sin
A
A
cos
2
2
A
A
A
 2 sin n1 cos n1
n
2
2
2
Again cos 2 A  cos ( A  A)  cos A cos A  sin A sin A  cos 2 A  sin 2 A
sin
But cos 2 A  sin 2 A  cos 2 A  (1  cos 2 A)  2 cos 2 A  1
 1 2 sin 2 A
Thus cos 2 A  cos 2 A  sin 2 A  2 cos 2 A  1  1  2 sin 2 A
We can similarly derive cos 4 A  cos 2 2 A  sin 2 2 A  2 cos 2 2 A  1  1  2 sin 2 2 A
A
A
A
A
cos A  cos 2  sin 2  2 cos 2  1 1  2 sin 2
2
2
2
2
tan A  tan A
2 tan A

Again tan 2 A  tan( A  A) 
.
1  tan A tan A 1  tan 2 A
A
2 tan
2 tan A
2 tan 2 A
2
Thus tan 2 A 
Also tan 4 A 
or tan A 
2
1  tan A
1  tan 2 2 A
2 A
1  tan
2
2
2 tan A
1  tan A
we can easily get sin 2 A 
, cos 2 A 
2
1  tan A
1  tan 2 A
Or cos 2 A  sin 2 A  (1  sin 2 A)  sin 2 A
3A formulas :-
sin 3 A  sin(2 A  A)
 sin A cos 2 A  cos A sin 2 A = sin A (1  2 sin 2 A)  cos A.2 sin A cos A
= sin A (1  2 sin 2 A)  2 sin A(1  sin 2 A) = 3sin A  4 sin 3 A
cos3 A  cos( A  2 A)  cos A cos 2 A  sin Asin 2 A
=
cos A (2 cos 2 A  1)  2 sin 2 A cos A = cos A(2 cos 2 A  1)  2 (1  cos 2 A) cos A
= 4cos3 A  3cos A
tan 3 A  tan( A  2 A) 
tan A  tan 2 A
1  tan A tan 2 A
25
2 tan A
3
1  tan 2 A = 3tan A  tan A
=
2 tan A
1  3tan 2 A
1  tan A.
1  tan 2 A
sin 3 A  3sin A  4 sin 3 A
tan A 
Thus
cos 3 A  4 cos 3 A  3 cos A
tan 3 A 
3 tan A  tan 3 A
1  3 tan 2 A
(i)
(ii)
The sine, cosine, tangents of angle of n A when n  4 can also be found.
We will observe that if n is a positive integer.
cos nA will be a polynomial of degree n in cos A for all n .
sin nA will be a polynomial of degree n in sin A if n is odd .
(iii)
tan nA 
n
C1 tan A  n C3 tan 3 A  n C5 tan 5 A.....
1  n C2 tan 2 A  n C4 tan 4 A.....
The proof of (i), (ii), (iii) involve three branches of algebra which are complex numbers, binomial
theorem and polynomial theory. We will limit our solves upto n  5 . The general proof is beyond the
scope of this book.


2
For example : cos 4 A  2 cos 2 2 A  1  2 2 cos 2 A  1  1  8 cos 4 A  8 cos 2 A  1
or cos 4 A  Real part of cos 4 A  i sin 4 A
4
 Real part of cos A  i sin A
 Real part of
cos A 
4
4
C1 cos A i sin A  4C2 cos A i sin A  4C3 cos A i sin A  4C4 i sin A
3
2
 cos 4 A  6 cos 2 A sin 2 A  sin 4 A

 
2
1
3

2
 cos 4 A  6 cos 2 A 1  cos 2 A  1  cos 2 A
 8 cos 4 A  8 cos 2 A  1
5
sin 5 A  Imaginary part of cos 5 A  i sin 5 A  I.P of cos A  i sin A
This time writing only Imaginary part we get
i sin 5 A  5C1 cos A i sin A  5C3 cos A i sin A  5C5 i sin A
4
1
2
3
5
 sin 5 A  5 sin Acos A  10 sin 3 A cos 2 A  sin A
4

5

2

 

 5sin A 1  sin 2 A  10sin3 A 1  sin2 A  sin5 A
 16 sin 5 A  20 sin 3 A  5 sin A
(*)
The above discussion may lead to something very basic in mathematics that sin x , or
cos x are solutions of polynomial equations with integer coefficients.
For example If A 18 then 5A  90 from (*)
 1  16 x 5  20 x 3  5 x
where x  sin A
5
3
or 16 x  20 x  5 x  1  0
 sin 18 is a solution of a polynomial equation with integer coefficients. Actually numbers
which are solutions of polynomial equations with integer coefficients are called algebraic
numbers. The number  is not algebraic but cos 20 is algebraic since
26
4

cos 60  4 cos3 20  3 cos 20 (by formula cos 3 A  4 cos3 A  3 cos A )
 1 / 2  4 x 3  3x
where x  cos 20
3
 8x  6 x  1  0
 cos 20 is algebraic
We will later see that sin 18 is a solution of a polynomial equation with much lesser degree.
SQUARE, CUBE FROM POWER REMOVAL : From 2 A formula’s we can get rid of power 2,3,4
easily. For example :
1  cos 2 A
1  cos 2 A
, sin 2 A 
2
2
cos 3 A  3 cos A
3 sin A  sin 3 A
cos3 A 
, sin 3 A 
4
4
cos 2 A 
1  2 cos 2 A  cos 2 2 A
4
1  cos 4 A
1  2 cos 2 A 
3 1
1
2
  cos 2 A  cos 4 A

8 2
8
4
2
 1  cos 2 A 
cos 4 A  cos 2 A  

2




2

(**)
3 1
1
Similarly sin 4 A   cos 2 A  cos 4 A etc. From these fourth power formula we easily get
8 2
8
cos 4

8
 cos 4
3
5
7 3
 cos 4
 cos 4
 by using (**) several times.
8
8
8
2
SOLVED EXAMPLES
Example 1 :
Prove that cot  - cot 2 = cosec 2
Solution :
cot   cot 2 

cos  cos 2 sin 2 cos   cos 2 sin 


sin  sin 2
sin  sin 2
sin 2   
sin 

 cos ec 2  RHS
sin  sin 2 sin  sin 2
Example 2 :
Prove that 1 + cot 2 cot  = cosec 2 cot 
Solution :
1  cot 2 cot   1 

cos 2 cos 
sin 2 sin 
sin 2 sin   cos 2 cos  cos2    cos 
1


.
sin 2 sin 
sin 2 sin  sin  sin 2
cot  cos ec2  RHS
27
Example 3 :
tan  + cot  = 2cosec 2
Solution :
LHS  tan   cot  
2
2 sin  cos 
sin  cos  sin 2   cos 2 
1



cos  sin 
sin  cos 
sin  cos 

2
 2 cos ec 2
sin 2
2
Example 4 :
A
A

 sin  cos   1  sin A
2
2

Solution :
LHS  sin 2
Example 5 :
Prove that
Solution :
LHS 
A
A
A
A
 cos 2  2 sin cos  1  sin A
2
2
2
2
cos 2
 tan  45   
1  sin 2
cos 2
cos 2   sin 2   cos   sin   cos   sin  


2
1  sin 2  cos   sin  2
 cos  sin  
cos   sin  1  tan 
tan 45  tan 


 tan 45   
cos   sin  1  tan  1  tan 45 tan 
OR



 

sin   2 
2 sin     cos   
cos 2
2
 
4
 4
  tan     



4
1  sin 2



2 


1  cos  2 
2 cos    
2
4




(Applying sin A  2 sin
1  cos A  2 cos 2
A
A

cos where role of A is being played by  2 and
2
2
2

A
with A as  2 )
2
2
Example 6 :
Prove that  cos  cos     sin   sin    4cos2
Solution :
On squaring
2
LHS
2
 
2
 cos 2   cos 2   2 cos  cos   sin 2   sin 2   2 sin  sin 
 2  2cos  cos   sin  sin  
 2  2 cos     21  cos   
 2.2 cos 2
 4 cos 2
 
2
 
2
(Applying 1  cos A  2 cos 2
 RHS
28
A
with A as     )
2
Example 7 :
Prove that cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Solution :
cos 6x  2 cos 2 3x  1

( cos 2 A  2 cos 2 A  1 with A as 3x )

2
 2 4cos3 x  3cos x  1  32cos6 x  48cos4 x  18cos2 x  1
Example 8 :
A

Prove that sin A  1  2sin 2  45  
2

Solution :
A

RHS  1  2 sin 2  45    1  2 sin 2 
2

A

  45  
2

 cos 2  cos90  A  sin A
Example 9 :




Prove that cos2      sin 2      sin 2
4

4

Solution :
On removing squares


1  cos 2   


4

cos 2     
4
2




1  cos  2 
2
  1  sin 2

2
2


1  cos  2 


2
  1  sin 2
sin 2     
2
2
4

On subtracting we get the desired result
Example 10 :
Prove that
Solution :
LHS
1  sin 2  cos 2
 tan 
1  sin 2  cos 2

1  cos 2  sin 2 2sin 2   2sin  cos

1  cos 2  sin 2 2cos 2   2sin  cos

2 sin  sin   cos  
 tan   RHS
2 cos  cos   sin  
Example 11 :
Prove that sin 2 x  2sin 4 x  sin 6 x  4cos 2 x sin 4 x
Solution :
LHS
 sin 2 x  sin 6 x  2sin 4 x
 2 sin 4x cos 2x  2 sin 4x
 2 sin 4 x1  cos 2 x   2 sin 4 x.2 cos 2 x
 4 cos 2 x sin 4 x  RHS
29
Example 12 :
Prove that 2 cos  
Solution :
1  cos 4  2 cos 2 2
2  2  2 cos 4
 2  2 cos 4  4 cos 2 2
On taking square root  2  2 cos 4  2 cos 2
On adding 2 on both sides we get
2  2  2 cos 4  2  2 cos 2  21  cos 2   2.2 cos 2 
On taking the square root again we get
2  2  2 cos 4  2 cos  as desired
Example 13 :
Prove that tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution :
LHS
 tan A  2 tan 2 A  4 tan 4 A  8cot 8 A
  cot A  tan A  2 tan 2 A  4 tan 4 A  8cot 8 A  cot A
(Subtracting and adding cot A )
Now
 cot A  tan A  

cos A sin A sin 2 A  cos 2 A


sin A cos A
sin A cos A
 cos 2 A
.2  2 cot 2 A
2 sin A cos A
(*)
 LHS  2cot 2 A  2 tan 2 A  4 tan 4 A  8cot 8 A  cot A
 4cot 4 A  4 tan 4 A  8cot 8 A  cot A (using (*) for A replaced by 2 A )
 8cot 8 A  8cot 8 A  cot A
(Again using (*))
cot A  RHS
Example 14 :
Prove that
Solution :
sec 8 A  1
sec 4 A  1
sec8 A  1 tan 8 A

sec 4 A  1 tan 2 A
1
1
1  cos 8 A cos 4 A
 cos 8 A

.
1
cos
8
A
1  cos 4 A
1
cos 4 A

2 sin 2 4 A cos 4 A 2 sin 4 A cos 4 A sin 4 A
,

.
2 sin 2 2 A cos 8 A
cos 8 A
2 sin 2 2 A
 tan 8 A .
2 sin 2 A cos 2 A tan 8 A

2 sin 2 2 A
tan 2 A
30
A
if tan A  3 / 4
2
Example 15 :
Find the values of tan
Solution :
A
2  2 x where x  tan A
tan A 
A 1  x2
2
1  tan 2
2
2 tan
Now tan A  3 / 4  A is either in first quadrant (or 2n to 2n 
quadrant
(i.e. from 2n   to 2n 
If A is from 2n to 2n 

2
then
If A is from 2n   to 2n 
 tan
2.
2
) OR in the third
3
)
2

A
A 1
A
 tan  0  tan 
is from n to n 
4
2
2 3
2

3
A
3
then is from n  to n 
2
2
4
2
A
A
 0  tan  3
2
2
EXERCISE
1.

x
x
x
Find sin ,cos and tan in each of the following :
2
2
2
4
1
(1).
(2).
tan x   , x in quadrant II
cos x   , in quadrant III
3
3
1
(3).
tan x  , in quadrant II
4
Prove the following identities:
(1)
cosec 2 - cot 2 = tan 
(2)
1 + tan 2 tan  = sec 2 
(3)
2 cosec 2 = sec  cosec 
(4)
cos4  - sin4  = cos 2
(5)
cot  - tan  = 2cot 2
(6)
cot 2 A 
(7)
cot A  tan A
 cos 2 A
cot A  tan A
(8)
1  cot 2 A
 cos ec 2 A
2cot A
(9)
cot 2 A  1
 sec 2 A
cot 2 A  1
(10)
1  sec

 2cos2
sec
2
(11)
2  sec2 
 cos 2
sec2 
(12)
cos ec 2  2
 cos 2
cos ec 2
31
cot 2 A  1
2cot A
cos 2
 cot  45   
1  sin 2
2
(13)
A
A

 sin  cos   1  sin A
2
2

(15)
sin 8A = 8sin A cos A cos 2A cos 4A
(17)
cos 2 sin 2

 cos3
sec cosec
2
(14)
2
(16)
1 + cos 4A = 2cos 2A(1 – 2sin2 A)
tan 4 

(18)
cos 4x = 1 – 8 sin x cos x
(20)
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
(21)
cot  
(22)
cosec  (sec  - 1) – cot (1 – cos ) = tan  - sin 
(23)
tan(45 + A)– tan(45 - A) = 2tan 2A
(24)
tan (45 + A) + tan(45 - A) = 2 sec 2A
(26)
cot 3 A 
(28)
sin 3  sin 3 
 cot 
cos3   cos3
(30)
cos 5A cos 2A – cos 4A cos 3A = -sin 2A sin A
(31)
sin 4 cos  - sin 3 cos 2 = sin  cos 2
(32)
4sin A sin(60 + A)sin(60 - A) = sin 3A
(33)
 2
  2

4cos cos 
   cos 
    cos3
 3
  3

(34)
 2

 2

cos  cos 
    cos 
   0
3
3




(35)
cos2 A + cos2 (60 + A) + cos2 (60 – A) = 3/2
(36)
sin2 A + sin2(120 + A) + sin2(120 - A) = 3/2
(37)
(38)
(39)
(19)
4 tan  1  tan 2 

1  6 tan   tan 
2
4
sin 
 cos ec
1  cos 
cot 3 A  3cot A
3cot 2 A  1
sin   sin   sin    
sin   sin   sin    
 cot

2
cot
(25)
sin 3 A cos3 A

2
sin A cos A
(27)
3cos   cos3
 cot 3 
3sin   sin 3
(29)
cos3   cos3 sin 3   sin 3

3
cos 
sin 

2
(cos A – sin A)(cos 2A – sin 2A) = cos A – sin 3A
sin 2 A  cos 2 A 
1  tan A2  2 tan 2 A
1  tan A
2
(40)
sin 4 A 

4 tan A 1  tan 2 A
1  tan A
2
32
2

4cos 2 A
1  2sin 2 A
(41)
cot 15  A  tan 15  A 
(42)
cot 15  A   tan 15  A  
(43)
tan  A  30 tan  A  30  
(44)
(2cos A + 1)(2cos A – 1)(2cos 2A – 1) = 2cos 4A + 1
(45)
cos2( - ) + cos2( - ) + cos2( - ) = 1 + 2cos ( - )cos( - )cos( - )
(46)
2cos2 (45 - A) = 1 + sin 2A
(47)
2cos 2x cosec 3x = cosec x – cosec 3x
(49)
2tan 2x =
(50)
(cos x + cos y)2 + (sin x – sin y)2 = 4cos2
x y
2
(51)
(cos x – cos y)2 + (sin x – sin y)2 = 4sin 2
x y
2
(52)
1  sin 2 x


 tan 2   x 
1  sin 2 x
4

(54)
(sin3x  sin x)sin x  (cos3x  cos x)cos x  0
(55)
1
1
sin A  2sin 3 A  sin 5 A sin 3 A

 cot 2 A. (56)

tan 3 A  tan A cot 3 A  cot A
sin 3 A  2sin 5 A  sin 7 A sin 5 A
(57)
sin 6
(58)
3A
A
  A   A
4 cos    sin     sin
cos ec
2
2
6 2 3 2
(59)
cos 5 = 16 cos5  - 20 cos3  + 5 cos 
(60)
cos A cos (60 – A) cos(60 + A) = ¼ cos 3A
(61)
cos8 A cos 5 A  cos12 A cos 9 A
 tan 4 A
sin 8 A cos 5 A  cos12 A sin 9 A
(62)
cos3 A  cos3 120  A   cos3  240  A 
4
cos 2 A  3 sin 2 A
1  2cos 2 A
1  2cos 2 A
(48)
cos A cos 2A cos 4A cos 8A =
sin16 A
16sin A
cos x  sin x cos x  sin x

cos x  sin x cos x  sin x
(53)
cos3 sin 3

 2cot 2
sin 
cos 
A
A 3
 cos 6   sin 2 A  4  cos A.
2
2 4
33
3
cos 3 A
4
(63)
2sin A cos3 A – 2sin3 A cos A =
1
sin 4A
2
(64)
sin 3A cos3 A + cos 3A sin3 A =
3
sin 4A
4
(65)
tan 3 
cot 3 
1  2sin 2  cos 2 


sin  cos 
1  tan 2  1  cot 2 
(66)
2cos  cos cos ( + ) = cos2  + cos2  - sin2 ( + ).
sin 2 3 cos 2 3

 8cos 2
sin 2 
cos 2 
(67)
 3

1  sin 4 A  cot   2 A  cos 4 A  0
 4

(69)
2cos  - cos 3 - cos 5 = 16cos3  sin2 
(70
cos3 ( + ) sin3 ( - ) + cos3 ( + ) sin3 ( - ) + cos3 ( + ) sin3 ( - )
(68)
= 3cos ( + ) cos ( + ) cos( + ) sin( - ) sin( - ) sin ( - )
2.
(71)
cos2 A + cos2 B – 2cos A cos B cos(A + B) = sin2 (A + B).
(72)
tan 3 A
cot 3 A

 sec A cos ec A  sin 2 A
1  tan 2 A 1  cot 2 A
(73)
tan  tan    60  tan  tan    60  tan    60  tan    60   3
(74)
2sin4 - sin 10  +sin2 = 16sin  cos  cos 2 sin 2 3 .
Prove the following numerical equalities (based on last three sections)
(1)
cos 80 + cos 40 – cos 20 = 0
(2)
cos 130 cos 40 + sin 130 sin 40 = 0
(3)
cos 5 – sin 25 = sin 35
(4)
cos 70 cos 10 + sin 70 sin 10 = ½
(5)
sin 65 + cos 65 = 2 cos 20
(6)
sin 78 – sin 18 + cos 132 = 0
(7)
cot 15 + cot 75 + cot 135 – cosec 30 = 1
(8)
(9)
sin 10 sin 30 sin 50 sin 70 = 1/16
1
sin 20 sin 40 sin 80 =
3
8
(10)
tan 40 + cot 40 = 2sec 10
2
(11)
tan 70 + tan 20 = 2 cosec 40
(12)
(13)
tan 6 tan 42 tan 66 tan 78 = 1
(14)
cos 33  cos 57
1
(16)

sin 21  cos 21
2
cos 20  2sin 2 55  1  2 sin 65
2
(15)
(17)
2
34
 1  cot 60  1  cos30

 
 1  cot 60  1  cos30
cosec 10 - 3 sec10 = 4
1
cos ec10  2sin 70  1
2
3.
(18)
sin 70  8cos 20 cos 40 cos80  2 cos 2 10
(19)
cos 2 73  cos 2 47  cos 73 cos 47 
(20)
tan10  tan 70  tan 50  3
(21)
tan 20 + tan 40 +
3
4
3 tan 20tan 40   3
If sin A + sin B + sin C = cos A + cos B + cos C = 0, prove that
sin2 A + sin2 B + sin2 C = cos2 A + cos2 B + cos2 C = 3/2
ANSWERS
1.
(1).
2 5
5
,
,2
5
5
(2).
6
3
,
, 2
3
3
(3).
8  2 15
8  2 15
,
, 4  15
4
4
Section –VI Periodicity, Graphs of Trigonometric functions
All basic trigonometric functions are periodic. The functions sin x, cos x, sec x and cosec x have periods
2n (n is any integer) with fundamental period as 2 since
sin (x + 2n) = sin x
cos(x + 2n) = cos x
sec(x + 2n) = sec x
cosec(x + 2n) = cosec x
The periods of tan x and cot x are n with fundamental period as . Note that
1.
The period of sin 2x is 2/2 =  by using the fact that if period of & f(x) is T then period of f(x) is
T/ (  0).
2
 6 .
1
3
2.
Using (1) period of tan 3x is /3 while period of cot x/3 is
3.
Fundamental period of sin
4.
5.
6.
S1 = 6, 12, 18, 24, 30, ………………….
(periods of sin x/3)
and
S2 = 8, 16, 24, 32, ………………… is 24
Fundamental period of sin x, cos(sin x) and sin2 x are 
Period of sin x + cos x is /2
Functions of the type sin x2, sin x , sin 1/x are non periodic.
x
x
 cos is 24 since least common element in the two series
3
4
Graphs
Since trigonometric functions are periodic we need to trace their graphs only within periods. Let us
observe the graph of sin x we observe that
(1)
sin x increases from 0 to 1 when x increases from 0 to /2.
(2)
sin x decreases from 1 to 0 when x increases from /2 to .
35
(3)
sin x further decreases from 0 to -1 when x increases from  to 3/2.
(4)
sin x finally increases from -1 to 0 when x increases from 3/2 to 2.
The same pattern repeats from 2 to 4 and so on.
The following graphs can easily be traced by using the graph of sin x
(a)
sin 3x (or sin x in general)
(b)
2 sin 3x (or  sin x)
(c)
2sin 3x + 5
To trace y = sin 3x we will draw the graph of sin x and will divide each of the located point on x-axis by 3.
Again to draw y = 2 sin 3x we will draw the graph of sin x then
(a)
Change (0, 1) and (0, -1) to (0, 2), (0, -2)
(b)
divide each point marked on x-axis by 3.
Finally to draw y = 2 sin 3x + 5 we will raise the last graph by 5 units (Translation along positive y-axis)
Note that graph of sin (x + ) can also be traced by graph of sin x. We just have to subtract  from each
point located on x-axis.
The graph of cos x is easily seen to as shown in the figure. As in the case of sin x the graphs of
associated functions can also be traced.
Graphs of other trigonometric functions
Since tan x, cot x, sec x, cosec x may approach - or  their graphs have tangent lines meeting them
(called asymptotes) at infinity. These functions have infinite discontinuities at n (In case of cot x and
cosec x) and at (2n + 1)/2 (In case of tan x and sec x)
Tan x
Cot x
36
Graph of sec x
Graph of cosec x can similarly be traced
Solved Examples
1.
Compare the graphs of y = sin 2x and y 
sin x
1  tan 2 x

cos x
1  cot 2 x
.
Solution: The graph of y1 = sin 2x is continuous and smooth through out the interval
[0, 2] (Indeed over the entire number line). On the contrary, the function
y2 
sin x

1  tan 2 x
n
not defined at
)
2
Also
cos x
1  cot 2 x
1  tan 2 x  sec x
Thus, if x 
n
then
2
and
is not defined at x 
n
(because tan x, cot x are
2
1  cot 2 x  cos ecx
y2 = sin x sec x + cos x cosec x
Observe the following table for the function y2 (Note sin x cos x + sin x cos x = sin 2x)
37
Interval
sec x
cosec x
y2
 
 0, 
 2
sec x
cosec x
sin 2x
 
 ,
2 
-sec x
cosec x
0
 3 
 , 
 2 
-sec x
-cosec x
-sin 2x
 3 
 ,
 2 
sec x
-cosec x
0
Exercise
1.
From the graph of y  sin x draw the graph of following functions :
(ii)
(iv)
y  sin 3x
y  3 sin 5 x
(v)
y  2 sin x
(iii)
y  3 sin 5 x (vi)
(vii)
y  cos x
(viii)
y  sin 2 x
(i)
2.
Draw the graphs of sin x + cos x [Hint: sin x + cos x =
38
y  sin x  2
y  3 sin 5x  3

2 sin  x   ]

4