Trigonometry (Theory)

Trigonometry
When writing about transcendental issues, be transcendentally clear....... Descartes, Rene
The word 'trigonometry' is derived from the Greek words 'trigon' and ' metron' and it means 'measuring the sides
and angles of a triangle'.
Angle :
Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial
side and the final position of the ray after rotation is called the terminal side of the angle. The point of
rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive
and if the direction of rotation is clockwise, then the angle is negative.
B
Initial Side
e Vertex O A
l Sid
na
mi Te
Ter rm
ina
lS
Vertex O A ide
Initial Side
B
(i) Positive angle (ii) Negative angle
(anticlockwise measurement) (clockwise measurement)
Systems For Measurement of Angles :

An angle can be measured in the following systems.
One complete rotation is equal to 360 degree = 400 grade = 2  radian
Relation between radian, degree and grade :
To Sexagesimal System Centesimal System (French Circular System

(British system) system) (Radian Measurement)
From
Sexagesimal 400 
System 1 degree = grade 1 degree (1°) = radian
360 180
(British system)
1
1min(1)= degree (1°=60)
60
1
1 sec(1) = min (1 = 60)
60
Centesimal 360
System (French 1 grade = degree
400
system)
Circular System 180 200

(Radian 1 radian = degree 1 radian = grade
 
Measurement)
1 degree = 60 min (1°=60) 1 grade=100 min(1g = 100')
1 min = 60 sec (1 = 60) 1 min = 100 sec(1' =100'')
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ADVTRI- 1
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
Trigonometry
Note : # The minutes and seconds in the Sexagesimal system are different with the minutes and
seconds respectively in the Centesimal System. Symbols in both systems are also different.
# If no symbol is mentioned while showing measurement of angle, then it is considered to be
measured in radians.
e.g.  = 15 implies 15 radian
Arc length AB = = r

1 2
Area of circular sector = r  sq. units
2
Trigonometric Ratios for Acute Angles :
Let a revolving ray OP starts from OA and revolves into the position OP, thus tracing out the angle
AOP.
In the revolving ray take any point P and draw PM perpendicular to the initial ray OA.
In the right angle triangle MOP, OP is the hypotenuse, PM is the perpendicular, and OM is the base.
The trigonometrical ratios, or functions, of the angle AOP are defined as follows :
sin(AOP) cos(AOP) tan(AOP) cot(AOP) sec(AOP) cosec(AOP)
Perp MP Base OM Perp MP Base OM Hyp OP Hyp OP

     
Hyp OP Hyp OP Base OM Perp MP Base OM Prep MP
It can be noted that the trigonometrical ratios are all real numbers.
Trigonometric ratios for angle  R :

We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and
study them as trigonometric functions. (also called circular functions) Consider a unit circle (radius 1
unit) with centre at origin of the coordinate axes. Let at origin of the coordinate axes. Let P(a, b) be any
point on the circle with angle AOP = x radian, i.e., length of arc AP = x We define
cos x = a and sin x = b Since  OMP is a right triangle, we have OM2 + MP2 = OP2 or a2 + b2 =1 Thus,
for every point on the unit circle, we have a2 + b2 = 1 or cos2x + sin2 x = 1

Since one complete revolution subtends an angle of 2 radian at the centre of the circle,  AOB = ,
2
3 
 AOC =  and AOD = . All angles which are integral multiples of are called quadrantal angles.
2 2
The coordinates of the points A, B, C and D are, respectively, (1, 0), (0, 1), (–1, 0) and (0, –1).
Therefore, for quadrantal angles, we have
ADVTRI- 2
Trigonometry
cos 0 = 1 sin 0 = 0,
 
cos =0 sin =1
2 2
cos  = –1 sin  = 0
3 3
cos =0 sin =–1
2 2
cos 2 = 1 sin 2 = 0
Now if we take one complete revolution from the position OP, we again come back to same position
OP. Thus, we also observe that if x increases (or decreases) by any integral multiple of 2 , the values
of sine and cosine functions do not change. Thus, sin (2n + x) = sin x , n  Z, cos (2n + x) = cos x,
n  Z. Further, sin x = 0, if x = 0, ±  , ± 2 , ± 3 ....., i.e., when x is an integral multiple of  and
 3 5 
cos x = 0, if x = ± , ± ,± , .....i.e., cos x vanishes when x is an odd multiple of . Thus sin x =
2 2 2 2

0 implies x = n , where n is any integer cos x = 0 implies x = (2n + 1) , where n is any integer.
2
We now define other trigonometric functions in terms of sine and cosine functions :
1
cosec x = , x  n , where n is any integer.
sin x
1 
sec x = , x  (2n + 1) , where n is any integer.
cos x 2
sin x 
tan x = , x  (2n + 1) , where n is any integer.
cos x 2
cos x
cot x = , x  n , where n is any integer.
sin x
We have shown that for all real x, sin2x + cos2x = 1

It follows that 1 + tan2x = sec2x (Think ! ) {x  (2n + 1) ; n  Z}
2
1 + cot2x = cosec2x (Think !) {x  n ; n  Z}
Sign of The Trigonometric Functions

(i) If is in the first quadrant then P(a, b) lies in the first quadrant. Therefore a > 0, b > 0
and hence the values of all the trigonometric functions are positive.
(ii) If is in the quadrant then P(a, b) lies in the quadrant. Therefore a < 0, b > 0 and
hence the values sin, cosec are positive and the remaining are negative.
(iii) If is in the quadrant then P(a, b) lies in the quadrant. Therefore a < 0, b < 0 and
hence the values of tan, cot are positive and the remaining are negative.
(iv) If is in the V quadrant then P(a, b) lies in the IV quadrant. Therefore a > 0, b < 0 and
hence the values of cos, sec are positive and the remaining are negative.
sin cos tan cot sec cosec
st Quadrant + + + + + +
nd Quadrant + – – – – +
rd Quadrant – – + + – –
Vth Quadrant – + – – + –
ADVTRI- 3
Trigonometry
Values of trigonometric functions of certain popular angles are shown in the following table :
   
0
6 4 3 2
0 1 1 2 1 3 3 4
sin 0    1
4 4 2 4 2 4 2 4
3 1 1
cos 1 0
2 2 2
1
tan 0 1 3 N.D.
3
N.D. implies not defined
The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cosx and tan x,
respectively.
Trigonometric Ratios of allied angles

 3
If  is any angle, then  ± ,  ± , ± , 2 ±  etc. are called allied angles.
2 2
–   – + 3 3 2 –  2 + 
– + – +
2 2 2 2
sin –sin cos sin sin –sin –cos –cos –sin sin
cos cos sin –cos –cos –cos –sin sin cos cos
tan –tan cot –tan –tan tan cot –cot –tan tan
cot –cot tan –cot –cot cot tan –tan –cot cot
sec sec cosec –sec –sec –sec –cosec cosec sec sec
cosec –cosec sec cosec cosec –cosec –sec –sec –cosec cosec
Think, and fill up the blank blocks in following table.
    2 5 7 4 3 5 11
0  2
6 4 3 2 3 6 6 3 2 3 6
1 1 3
sin 0 1
2 2 2
3 1 1
cos 1 0
2 2 2
1
tan 0 1 3 N.D.
3
ADVTRI- 4
Trigonometry
Trigonometric functions :
Domain Range Graph
y
y = sinx R [–1, 1]
1
0
3    3
x
–2 – 2
2 2 2 2
–1
y
y = cosx R [–1, 1]
1
0

x
–2 3 –   3 2
2 2 2 2
–1
y
y = tanx R– R
  
(2n  1) , n     
 2  – 2 0 2 
x
– 32 3
2
y = cotx R – {n , n   } R y
–    3 2 x
0
2 2 2
y = secx R– (,  1]  [1,

   )
(2n  1) , n   
 2 
y=cosecx R – {n, n } (,  1]  [1,

)
ADVTRI- 5
Trigonometry
Trigonometric functions of sum or difference of two angles:


(a) sin (A ± B) = sinA cosB ± cosA sinB

(b) cos (A ± B) = cosA cosB sinA sinB
(c) sin²A  sin²B = cos²B  cos²A = sin (A+B). sin (A B)
(d) cos²A  sin²B = cos²B  sin²A = cos (A+B). cos (A  B)
tan A  tan B cot A cot B 1
(e) tan (A ± B) = (f) cot (A ± B) =
1 tan A tan B cot B  cot A
(g) sin (A + B + C) = sin A cos B cos C + sin B cos A cos C + sin C cos A cos B – sin A sin B sin C
(h) cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C
tan A  tan B  tanC  tan A tan B tan C
(i) tan (A + B + C) = .
1  tan A tan B  tan B tan C  tan C tan A
S1  S3  S5  ......
(j) tan (1 + 2 + 3 + ....... + n) =
1  S2  S4  .......
where Si denotes sum of product of tangent of angles taken i at a time
Example # 1 : Prove that
(i) sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B)
   3 
(ii) tan     tan     = –1
 4   4 
Solution : (i) Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B)
= sin (45º + A + 45º – B) = sin (90º + A – B) = cos (A – B)
   3  1  tan  1  tan 
(ii) tan     × tan    = × =–1
4   4  1  tan  1  tan 
Self practice problems :
2 2 4
(1) If cos  =, sin  = , then find cos ( + ) (2) Find the value of cos 375º
3 5
A A
(3) Prove that 1 + tan A tan = tan A cot – 1 = sec A
2 2
6 2  4 3 1
Answers : (1) (2)
15 2 2
Transformation formulae :
CD CD
(i) sin(A+B) + sin(A  B) = 2 sinA cosB (a) sinC + sinD = 2 sin cos
2 2
CD CD
(ii) sin(A+B)  sin(A  B) = 2 cosA sinB (b) sinC  sinD = 2 cos sin
2 2
CD CD
(iii) cos(A+B) + cos(A  B) = 2 cosA cosB (c) cosC + cosD= 2 cos cos
2 2
CD D C
(iv) cos(A  B)  cos(A+B) = 2 sinA sinB (d) cosC  cosD = 2 sin sin
2 2
 15A  A
Example # 2 : Prove that cos7A + cos8A = 2cos   cos  
 2  2
 15A   
A
Solution : L.H.S. cos7A + cos8A = 2cos   cos  2 
 2   
CD C D
[ cos C+ cos D = 2 cos cos ]
2 2
Example # 3 : Find the value of 2sin3 sin – cos2+ cos4

Solution : 2sin3 sin – cos2+ cos4 = 2 sin 3 sin  – 2sin3 sin = 0
ADVTRI- 6
Trigonometry

sin8 cos   sin6 cos3
(i) = tan 2
cos2 cos   sin3 sin4
(ii) If A + B = 45º then prove that (1 + tanA) (1 + tanB) = 2
2sin8 cos   2sin6 cos3 sin9  sin7  sin9  sin3  2sin2 cos5 
Solution : (i) = = = tan 2
2cos2 cos   2sin3 sin4 cos3  cos   cos   cos7 2cos5 cos2 
(ii) A + B = 45º
tan A  tanB
tan (A + B) = 1  =1
1– tan A tanB
tanA + tanB = 1 – tanA tanB  tanA + tanB + tanA tanB + 1 = 2
(1 + tanA) (1 + tanB) = 2
Self practice problems
(4) Prove that

13x 3x
(i) cos 8x – cos 5x = – 2 sin sin
2 2
cos A  cos3A
(ii) = – tan2A
sin A  sin3A
sin2A  sin4A  sin6A  sin8A
(iii) = tan 5A
cos2A  cos 4A  cos6A  cos8A
sin A  2sin3A  sin5A sin3A
(iv) =
sin3A  2sin5A  sin7A sin5A
sin A  sin5A  sin9A  sin13A
(v) = cot 4A
cos A  cos5A  cos9A  cos13A
 7 3 11
(5) Prove that sin sin + sin sin = sin 2 sin 5
2 2 2 2
(6) Prove that cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B) = 0
 9 3 5
(7) Prove that 2 cos cos + cos + cos =0
13 13 13 13
Multiple and sub-multiple angles :

 
(a) sin 2A = 2 sinA cosA Note : sin  = 2 sin cos etc.
2 2
(b) cos 2A = cos²A  sin²A = 2cos²A  1 = 1  2 sin²A
 
Note : 2 cos² = 1 + cos , 2 sin² = 1  cos .
2 2

2 tan A 2 tan
(c) tan 2A = Note : tan  = 2
2 
1  tan2 A 1  tan 2
2 tan A 1  tan2 A
(d) sin 2A = , cos 2A =
1  tan A
2
1  tan2 A
(e) sin 3A = 3 sinA  4 sin3A
(f) cos 3A = 4 cos3A  3 cosA
3 tan A  tan3 A
(g) tan 3A =
1  3 tan2 A
ADVTRI- 7
Trigonometry

sin2A
(i) = tan A (ii) tan A + cot A = 2 cosec 2 A
1  cos2A
1  cos A  cosB  cos(A  B) A B
(iii) = tan cot
1  cos A  cosB  cos(A  B) 2 2
sin2A 2sin A cos A
Solution : (i) L.H.S. = = tan A
1  cos2A 2cos2 A
1  tan2 A  1  tan2 A  2
(ii) L.H.S. tan A + cot A = =2   = = 2 cosec 2 A
tan A  2 tan A  sin2A
A A A 
2sin2  2sin sin   B 
1  cos A  cosB  cos(A  B) 2 2  2 
(iii) L.H.S. =
1  cos A  cosB  cos(A  B) 2 A A A 
2cos  2cos cos   B 
2 2  2 
 A A    A B B 
 sin  sin   B    2sin cos   
= tan
A  2 2   = tan A  2  2   = tan A cot B
2  A A  2  A B B  2 2
 cos 2  cos  2  B    2sin 2 sin  2  
       

sin 4  sin2 2 tan 
(8) Prove that =
1  cos 4  cos2 1  tan2 
 3 5 7 1
(9) Prove that sin sin sin sin =
18 18 18 18 16
(10) Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A

 A
(11) Prove that tan  45º   = sec A + tan A
 2
Important trigonometric ratios of standard angles :

(a) sin n  = 0 ; cos n  = (1)n ; tan n  = 0, where n  
 3 1 5
(b) sin 15° or sin = = cos 75° or cos ;
12 2 2 12
 3 1 5
cos 15° or cos = = sin 75° or sin ;
12 2 2 12
3 1 3 1
tan 15° = = 2  3 = cot 75° ; tan 75° = = 2  3 = cot 15°
3 1 3 1
 5 1
(c) sin or sin 18° = = cos 72°
10 4
 5 1
cos 36° or cos = = sin 54°
5 4
ADVTRI- 8
Trigonometry
Conditional identities:
If A + B + C =  then :
(i) sin2A + sin2B + sin2C = 4 sinA sinB sinC
A B C
(ii) sinA + sinB + sinC = 4 cos cos cos
2 2 2
(iii) cos 2 A + cos 2 B + cos 2 C =  1  4 cos A cos B cos C
A B C
(iv) cos A + cos B + cos C = 1 + 4 sin sin sin
2 2 2
(v) tanA + tanB + tanC = tanA tanB tanC
A B B C C A
(vi) tan tan + tan tan + tan tan =1
2 2 2 2 2 2
A B C A B C
(vii) cot + cot + cot = cot . cot . cot
2 2 2 2 2 2
(viii) cot A cot B + cot B cot C + cot C cot A = 1
Example # 6 : If A + B + C = 90°, Prove that, tan A tan B + tan B tan C + tan C tan A = 1
Solution : A + B = 90º – C
tan A  tanB
 cot C
1– tan A tanB
tan A tan B + tan B tan C + tan C tan A = 1
2x 2y 2z 2x 2y 2z
Example # 7 : If x + y + z = xyz, Prove that + + = . . .
1 x 1 y
2 2
1 z 2
1 x 2
1 y 2
1  z2
Solution : Put x = tanA, y = tanB and z = tanC,
so that we have
tanA + tanB + tanC = tanA tanB tanC  A + B + C = n where n  
Hence L.H.S.
2x 2y 2z 2 tan A 2 tanB 2 tanC
 + + = + + .
1 x 2
1 y 2
1 z 2
1  tan A
2
1  tan B 1  tan2 C
2
= tan2A + tan2B + tan2C [ A + B + C = n ]

2x 2y 2z
= tan2A tan2B tan2C = . .
1  x2 1 y 2
1  z2
Self practice problem
(12) If A + B + C = 180°, prove that
BC CA A B
(i) sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = 4sin sin sin
2 2 2
sin2A  sin2B  sin2C A B C
(ii) = 8 sin sin sin .
sin A  sinB  sinC 2 2 2
(13) If A + B + C = 2S, prove that
(i) sin(S – A) sin(S – B) + sinS sin (S – C) = sinA sinB.
A B C
(ii) sin(S – A) + sin (S – B) + sin(S – C) – sin S = 4sin sin sin .
2 2 2
Sine and Cosine series:
n
sin 2  n  1 
(i) sin  + sin (+) + sin ( + 2 ) +...... + sin   (n  1)  = sin    
  2 
2
sin
n
sin 2  n  1 
(ii) cos  + cos (+) + cos ( + 2 ) +.... + cos   (n  1)  = cos    
  2 
sin 2
where :   2m, m  
ADVTRI- 9
Trigonometry
sin2 n
Example # 8 : (i) Prove that sin + sin3 + sin5 + .... + sin(2n–1)=
sin 
(ii) Find the average of sin2º, sin4º, sin6º , ...... , sin180º
 3 5 7 9 1
(iii) Prove that cos + cos + cos + cos + cos =
11 11 11 11 11 2
 2     (2n – 1) 
sinn   sin  
 = sin n
2
Solution : (i) sin + sin3 + sin5 + .... + sin(2n–1) =  2   2
 2  sin 
sin  
 2 
sin2º  sin4º ....  sin180º sin90º(sin91º ) cos1º cot1º
(ii) = = = =
90 90sin1º 90 sin1º 90
10 5 10
cos sin sin
 3 5 7 9 22 11 = 11 = 1
(iii) cos + cos + cos + cos + cos =
11 11 11 11 11   2
sin 2 sin
11 11
Self practice problem
Find sum of the following series :
 3 5
(14) cos + cos + cos + ...... up to n terms.
2n  1 2n  1 2n  1
(15) sin2 + sin3 + sin4 + ..... + sin n, where (n + 2) = 2

1
Answers : (14) – (15) 0.
2
Product series of cosine angles

sin2n 
cos  . cos 2 . cos22 . cos23 ...... cos2n–1 =
2n sin 
Range of trigonometric expression:
E = a sin  + b cos 

 a b 

 E = a2  b2  sin   cos 

 a b
2 2
a b
2 2


b a
Let = sin  & = cos 
a b
2 2
a  b2
2
b
 E = a2  b2 sin ( + ), where tan  =
a
Hence for any real value of ,
 a2 + b2  E  a2 + b2
Example # 9 : (i) If +  = 90º then find the maximum value of sin sin
(ii) Find maximum and minimum value of 1 + 2sinx + 3cos 2x
1
Solution : (i) sinsin(90º – ) = sincos = × sin2
2
1
maximum value =
2
ADVTRI- 10
Trigonometry
2
 2sin x   1  13
(ii) 1 + 2sinx + 3cos x = – 3sin x + 2sinx + 4=– 3  sin2 x   +4=–3  sin x  3  + 3
2 2
 3   
2 2
 1 16 16  1
Now 0   sin x     – –3  sin x  3   0
 3 9 9  
2
 1 13 13
– 1  – 3  sin x   + 
 3 3 3

(16) Find maximum and minimum values of following
(i) 3 + (sinx – 2)2
(ii) 9cos2x + 48sinx cosx – 5sin2x – 2
   
(iii) 2 sin     + 3 cos    
 6   6 
Answers : (i) max = 12, min = 4. (ii) max = 25, min = –25
(iii) max = 13 , min = – 13
Trigonometric Equation :
An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric
equation.
Solution of Trigonometric Equation :

A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.
1  3  9  11
e.g. if sin =  = , , , , ...........
2 4 4 4 4
Thus, the trigonometric equation may have infinite number of solutions (because of their periodic
nature) and can be classified as :
(i) Principal solution (ii) General solution.
Principal solutions :
The solutions of a trigonometric equation which lie in the interval [0, 2) are called Principal solutions.
1
e.g. Find the Principal solutions of the equation sinx = .
2
Solution :
1
 sinx =
2
 there exists two values
 5 1
i.e. and which lie in [0, 2) and whose sine is
6 6 2
1  5
 Principal solutions of the equation sinx = are ,
2 6 6
General Solution :
The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called
General solution. General solution of some standard trigonometric equations are given below.
ADVTRI- 11
Trigonometry
General Solution of Some Standard Trigonometric Equations :

  
(i) If sin  = sin      = n  + (1)n  where     ,  , n  .
 2 2
(ii) If cos  = cos      = 2n ±  where   [0, ], n  .
  
(iii) If tan  = tan      = n +  where    ,  , n  .
 2 2
(iv) If sin²  = sin²     = n  ± , n  .
(v) If cos²  = cos²     = n  ± , n  .
(vi) If tan²  = tan²     = n  ± , n  . [Note:  is called the principal angle ]
Some Important deductions :

(i) sin = 0   = n, n 

 (ii) sin = 1   = (4n + 1) , n
2

(iii) sin = – 1   = (4n – 1) , n 
2

(iv) cos = 0   = (2n + 1) , n
2
(v) cos = 1   = 2n,n 
(vi) cos = – 1  = (2n + 1), n

(vii) tan = 0   = n, n 
1
Example # 10: Solve cos  =
2
1  
Solution :  cos  =  cos = cos   = 2n ± , n 
2 3 3
2
Example # 11 : Solve : sec 2 = –
3
2 3
Solution :  sec 2 = –  cos2 = –
3 2
5 5 5
 cos2 = cos  2 = 2n ± , n   = n ± , n 
6 6 12
3
Example # 12 : Solve tan =
4
3
Solution :  tan = ............(i)
4
3
Let = tan  tan = tan
4
3
    = n + , where  = tan–1   , n
4
Self Practice Problems :
3
(17) Solve cot = – 1 (18) Solve cos4 = –
2
 n 
Answers : (17)  = n – , n (18)  , n  
4 2 24
ADVTRI- 12
Trigonometry
Example # 13 : Solve tan2 = 1

Solution :  tan2 = 1  tan2 = (1)2
 
  tan2 = tan2    = n ± , n 
4 4
Example # 14 : Solve 4 sec2 = 5 + tan2

Solution :  4 sec2 = 5 + tan2 .............(i)

For equation (i) to be defined  (2n + 1) , n 
2
 equation (i) can be written as:
  
   4(1 + tan2) = 5 + tan2
3tan2= 1
tan2= tan2/6

 = n ± , n 
6



tan3x  tan2x
(19) Solve =1 (20) Solve 2 cos2x + sin22x = 2
1  tan3x tan2x

Answers : (19) no Solution  (20) n, n  or n ± , n
4

Types of Trigonometric Equations :


Type -1
Trigonometric equations which can be solved by use of factorization.

x x
sin3 – cos3
Example # 15 : 2 2  cos x
2  sin x 3
x x  x x  x x x x
sin3 – cos3  sin – cos  sin2  cos2  sin cos 
Solution : 2 2  cos x   2 2  2 2 2 2  cos x
=
2  sin x 3 2  sin x 3
 x x
 sin 2 – cos 2   2  sin x  cos x
   x x  x x
 =  3  sin – cos  – 2  cos2 – sin2   0
2  2  sin x  3  2 2  2 2
 x x  x x x x x
 sin 2 – cos 2  3  2sin 2  2cos 2   0  sin 2  cos 2  tan 2  1
  
x  
 n  ,n   x  2n  ,n   
2 4 2


x
(21) Solve cos3x + cos2x – 4cos2 =0 (22) Solve tan2 + 3sec + 3 = 0
2
Answers : (21) (2n + 1), n  
2
(22) 2n ± , n  or (2n + 1) , n 
3
Type - 2
Trigonometric equations which can be solved by reducing them in quadratic equations.


ADVTRI- 13
Trigonometry
cos x 1
Example # 16 : Solve sin2x – =
4 4
cos x 1
Solution : sin2x – =
4 4
4(1 – cos2x) – cosx = 1
4cos2x + cosx – 3 = 0
(cosx + 1) (4cosx –3) = 0
3
cosx = – 1 , cosx =
4
3
x = (2n+1) , x = (2m±) where  = cos–1 , m, n 
4

(23) Solve 4sin2+ 2sin  
3 1 – 3 = 0
(24) Solve 4cos – 3sec = tan
   
Answers : (23) n + (–1)n , n  or n + (–1)n   , n 
6  3 
 1  17 
(24) n + (– 1)n  where  = sin–1   , n 
 8
 
 1  17 
or n + (–1)n  where  = sin–1   , n 
 8
 
Type - 3

Trigonometric equations which can be solved by transforming a sum or difference of trigonometric

ratios into their product.

Example # 17 : Solve cosx + cos3x – 2cos2x = 0
Solution : cosx + cos3x – 2cos2x = 0
2cos2x cosx – 2cos2x = 0
2cos2x (cosx–1) = 0
cos2x = 0, cosx = 1

x = (2n + 1),  x = 2m , m, n 
2



(25) Solve sin7= sin3 + sin (26) Solve 1 + cos3x = 2cos 2x
sin6x
(27) Solve 8cosx cos 2x cos4x =
sin x
n n 
Answers : (25) , n  or ± , n 
3 2 12
 n 
(26) n  , n  or 2n, n (27)  , n 
6 7 14
Type - 4


Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their
sum or difference.
ADVTRI- 14
Trigonometry
Example # 18 : Solve sec4 – sec2 = 2

1 1
Solution : – =2
cos 4 cos2
cos2 – cos4 = 2 cos4 cos2
cos2 – cos4 = cos6 + cos2
cos6 + cos4 = 0
2cos5 cos = 0
cos5 = 0 or cos = 0 
 
  5= (2n + 1)  = (2m + 1) m, n

2 2
Type - 5
Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c  R, can be solved by dividing
both sides of the equation by a2  b2 .

Example # 19 : Solve sinx + 2cosx = 5
Solution :  sinx + 2cosx = 5 ..........(i)
Here a = 1, b = 2.
 divide both sides of equation (i) by 5 , we get
1 1
sinx . + 2cosx. =1  sinx.sin + cosx.cos = 1  cos (x– )= 1
5 5
 x –  = 2n, n   x = 2n + , n 
 1
 Solution of given equation is 2n+ , n  where  = tan–1   
2

Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosx
into their corresponding tangent of half the angle.
Example # 20 : Solve 3cosx + 4sinx = 5
Solution :  3cosx + 4sinx = 5 .........(i)

x x
1  tan2 2 tan
  cosx = 2 & sinx = 2
2 x 2 x
1  tan 1  tan
2 2
  equation (i) becomes
 2 x   x 
 1  tan 2   2 tan 2 
  3   +4  =5 ........(ii)
 1  tan2 x   1  tan2 x 
 2  2
x
Let tan =t
2
 1  t2   2t 
 equation (ii) becomes 3 
 1  t 2 
+4   =5
   1  t2 
 4t2 – 4t + 1 = 0  (2t – 1)2 = 0
1 x
   t=  t = tan
2 2
x 1 x 1
 tan =  tan = tan, where tan =
2 2 2 2
x  1
  = n +    x = 2n + 2 where  = tan–1   , n 
2 2
ADVTRI- 15
Trigonometry

(28) Solve 2 2 cosx + sinx = 3
x
(29) Solve sinx + tan =0
2
 1 
Answers : (28) 2n + , n where  = tan–1  
2 2 
(29) x = 2n, n 
Type - 6
Trigonometric equations of the form P(sinx ± cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can be
solved by using the substitution sinx ± cosx = t.
Example # 21 : Solve sin2x + 3sinx = 1 + 3 cosx

Solution : sin2x + 3sinx = 1 + 3 cosx
sin2x + 3(sinx – cosx) = 1 ......... (i)
Let sinx – cosx = t
  sin2x + cos2x – 2 sinx.cosx = t2  sin2x = 1– t2
Now put sinx – cosx = t and sin2x = 1– t in (i)2
1 – t2 + 3t = 1
t2 – 3t = 0
t =0 or t=3 (not possible)
sinx – cosx = 0

tanx = 1  x = n + , n 
4
Self Practice Problems:
(30) Solve 1– sin2x + 2sinx – 2cosx = 0 (31) Solve 2cosx + 2sinx + sin3x – cos3x = 0
(32) Solve (1 – sin2x) (cosx – sinx) = 1 – 2sin2x.

  n 
Answers : (30) n + , n  (31) n –  or  (–1)n , n
4 4 2 12
 
(32) 2n + , n  or 2n, n  or n + , n
2 4
Type - 7
Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios
sinx and cosx.
Example # 22 : Solve sin2x + cos4x = 2

Solution : sin2x + cos4x = 2
Now equation will be true if sin2x = 1 and cos4x = 1


 2x = (4n + 1) , n  and 4x = 2m, m
2
 m  m 4n  1
  x = (4n + 1) , n  and x= , m (4n + 1) = m=
4 2 4 2 2
Which is not possible for m, n

(33) Solve cos50x – sin50x = 1
(34) Solve 12 sin x + 5cosx = 2y2 – 8y + 21 for x & y
 5 
Answers : (33) n, n  (34) x = 2nwhere  = cos–1   , n y = 2
 13 
ADVTRI- 16
Trigonometry
IMPORTANT POINTS :
1. Many trigonometrical equations can be solved by different methods. The form of solution obtained in
different methods may be different. From these different forms of solutions, the students should not
think that the answer obtained by one method are wrong and those obtained by another method are
correct. The solutions obtained by different methods may be shown to be equivalent by some
supplementary transformations.
To test the equivalence of two solutions obtained from two methods, the simplest way is to put values
of
n = .......–2, –1, 0, 1, 2, 3....... etc. and then to find the angles in [0, 2]. If all the angles in both solutions
are same, the solutions are equivalent.
2. While manipulating the trigonometrical equation, avoid the danger of losing roots. Generally, some
roots are lost by cancelling a common factor from the two sides of an equation. For example, suppose
1
we have the equation tanx = 2 sinx. Here by dividing both sides by sinx, we get cosx = . This is not
2
equivalent to the original equation. Here the roots obtained by sinx = 0, are lost. Thus in place of
dividing an equation by a common factor, the students are advised to take this factor out as a common
factor from all terms of the equation.
3. While equating one of the factors to zero, take care of the other factor that it should not become infinite.
For example, if we have the equation sinx = 0, which can be written as cos x tan x = 0. Here we cannot
put
cosx = 0, since for cos x = 0, tanx = sinx/ cosx is infinite.
4. Avoid squaring : When we square both sides of an equation, some extraneous roots appear. Hence it is
necessary to check all the solutions found by substituting them in the given equation and omit the
solutions not satisfying the given equation.
For example : Consider the equation,
sin  + cos = 1 .....(1)
Squaring we get
1 + sin 2= 1 or sin 2= 0 .....(2)
i.e. 2= n or = n/2,
 3
This gives = 0, ,, , ......
2 2
3
Verification shows that and do not satisfy the equation as sin  + cos= –1, 1
2
3 3
and sin + cos = – 1, 1.
2 2
The reason for this is simple.
The equation (2) is not equivalent to (1) and (2) contains two equations : sin + cos  = 1
and sin+ cos  = – 1. Therefore we get extra solutions.
Thus if squaring is must, verify each of the solution.
5. Some necessary restrictions :
If the equation involves tanx, secx, take cosx  0. If cot x or cosec x appear, take sinx  0.
If log appear in the equation, i.e. log [f()] appear in the equation, use f() > 0 and base of log > 0, 1.
Also note that [f()] is always positive, for example sin2  = |sin |, not ± sin .
6. Verification : Student are advice to check whether all the roots obtained by them satisfy the equation
and lie in the domain of the variable of the given equation.
Trigonometric Inequalities :
To solve a trigonometric inequality, transform it into many basic trigonometric inequalities. The
transformation process proceeds exactly the same as in solving trigonometric equations. The common
period of a trigonometric inequality is the least common multiple of all periods of the trigonometric
functions presented in the inequality. For example : the trigonometric inequality
sinx + sin2x + cos x/2 < 1 has 4 as common period. Unless specified, the solution set of a
trigonometric inequality must be solved, at least, within one whole common period.
ADVTRI- 17
Trigonometry
Example : Find the solution set of inequality sinx > 1/2.

Solution : When sinx = 1/2, the two values of x between 0 and 2 are /6 and 5/6.
0 
3 x
–2 3 – 
   5 2
 2 6 2
2 6 2
–1
y = sinx
From, the graph of y = sinx, it is obvious that, between 0 and 2sinx > 1/2  /6 < x < 5/6.
Hence sinx >1/2  2n + /6 < x < 2n + 5/6, n  
  The required solution set is (2n + /6, 2n + 5/6)

nI

(35) Solve the following inequations
(i) (sinx – 2) (2sinx–1) < 0 (ii) sinx + 3 cosx  1
 5   
Ans. (i) x  6  2n, 6  2n  (ii) x [– +2n, 2n ]
n   nI 6 2
Heights and distances :
Angle of elevation and depression :
Let OX be a horizontal line and P be a point which is above point O. If an observer (eye of observer) is
at point O and an object is lying at point P then XOP is called angle of elevation as shown in figure. If
an observer (eye of observer) is at point P and object is at point O then QPO is called angle of
depression.
Q P
O X
ADVTRI- 18

Trigonometry (Theory)

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Trigonometry

When writing about transcendental issues, be transcendentally clear....... Descartes, Rene

Systems For Measurement of Angles :

Relation between radian, degree and grade :

To Sexagesimal System Centesimal System (French Circular System

Circular System 180 200

1 min = 60 sec (1 = 60) 1 min = 100 sec(1' =100'')

Arc length AB = = r

sin(AOP) cos(AOP) tan(AOP) cot(AOP) sec(AOP) cosec(AOP)

Perp MP Base OM Perp MP Base OM Hyp OP Hyp OP

Trigonometric ratios for angle  R :

Sign of The Trigonometric Functions

sin cos tan cot sec cosec

N.D. implies not defined

Trigonometric Ratios of allied angles

Think, and fill up the blank blocks in following table.

y = secx R– (,  1]  [1,

y=cosecx R – {n, n } (,  1]  [1,

Trigonometric functions of sum or difference of two angles:

(a) sin (A ± B) = sinA cosB ± cosA sinB

Example # 3 : Find the value of 2sin3 sin – cos2+ cos4

Example # 4 : Prove that

Self practice problems

(4) Prove that

Multiple and sub-multiple angles :

Example # 5 : Prove that

Self practice problems

(10) Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A

Important trigonometric ratios of standard angles :

= tan2A + tan2B + tan2C [ A + B + C = n ]

(15) sin2 + sin3 + sin4 + ..... + sin n, where (n + 2) = 2

Product series of cosine angles

Range of trigonometric expression:

Self practice problems

Solution of Trigonometric Equation :

General Solution of Some Standard Trigonometric Equations :

Some Important deductions :

(vi) cos = – 1  = (2n + 1), n

Example # 13 : Solve tan2 = 1

Example # 14 : Solve 4 sec2 = 5 + tan2

Self Practice Problems :

Types of Trigonometric Equations :

Trigonometric equations which can be solved by use of factorization.

Self Practice Problems :

Trigonometric equations which can be solved by reducing them in quadratic equations.

Self Practice Problems :

Trigonometric equations which can be solved by transforming a sum or difference of trigonometric

Self Practice Problems :

Example # 18 : Solve sec4 – sec2 = 2

Example # 20 : Solve 3cosx + 4sinx = 5

Solution :  3cosx + 4sinx = 5 .........(i)

Self Practice Problems :

Example # 21 : Solve sin2x + 3sinx = 1 + 3 cosx

(32) Solve (1 – sin2x) (cosx – sinx) = 1 – 2sin2x.

Example # 22 : Solve sin2x + cos4x = 2

Now equation will be true if sin2x = 1 and cos4x = 1

Example : Find the solution set of inequality sinx > 1/2.

  The required solution set is (2n + /6, 2n + 5/6)

Self practice problems