Visual-fidelity edition: source-page equations, diagrams and charts are preserved as page images. Text transcripts are provided where extractable. The original source may contain markings within the artwork.
MAASTrigonometry
© MAAS. All rights reserved.Page 1 / 47
Extracted text · page 1
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
Side Vertex O A
Terminal Terminal
Vertex O A Side 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
1 degree = grade 1 degree (1°) = radian System 360 180
(British system)
1
1min(1)= degree (1°=60)
60
1
1 sec(1) = min (1 = 60)
60
Centesimal 360
1 grade = degree
System (French 400
system)
Circular System 180 200
1 radian = degree 1 radian = grade
(Radian
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'')
ADVTRI- 1
MAASTrigonometry
© MAAS. All rights reserved.Page 2 / 47
Extracted text · page 2
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
Area of circular sector = r2 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
MAASTrigonometry
© MAAS. All rights reserved.Page 3 / 47
Extracted text · page 3
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.
sinx
1
sec x = , x (2n + 1) , where n is any integer.
cosx 2
sinx
tan x = , x (2n + 1) , where n is any integer.
cosx 2
cosx
cot x = , x n , where n is any integer.
sinx
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
MAASTrigonometry
© MAAS. All rights reserved.Page 4 / 47
Extracted text · page 4
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
MAASTrigonometry
© MAAS. All rights reserved.Page 5 / 47
Extracted text · page 5
Trigonometry
Trigonometric functions :
Domain Range Graph
y = sinx R [–1, 1] y
1
0
–2 3 – 3 2 x
2 2 2 2
–1
y = cosx R [–1, 1] y
1
0 x
–2 3 – 3 2
2 2 2 2
–1
y = tanx R– R y
(2n 1) , n 2 – 2 0 2 x 3 3
– 2 2
y = cotx R – {n, n} R y
– 0 3 2 x
2 2 2
y = secx R– (, 1] [1,
)
(2n 1) , n
2
y=cosecx R – {n, n } (, 1] [1,
)
ADVTRI- 5
MAASTrigonometry
© MAAS. All rights reserved.Page 6 / 47
Extracted text · page 6
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 S 4 .......
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
MAASTrigonometry
© MAAS. All rights reserved.Page 7 / 47
Extracted text · page 7
Trigonometry
Example # 4 : Prove that
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
tanA tanB
tan (A + B) = 1 = 1
1– tanA 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
cosA cos3A
(ii) = – tan2A
sinA sin3A
sin2A sin4A sin6A sin8A
(iii) = tan 5A
cos2A cos4A cos6A cos8A
sinA 2sin3A sin5A sin3A
(iv) =
sin3A 2sin5A sin7A sin5A
sinA sin5A sin9A sin13A
(v) = cot 4A
cosA 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 2 (c) tan 2A = Note : tan =
1 tan2 A 1 tan 2 2
2 tan A 1 tan2 A
(d) sin 2A = , cos 2A =
1 tan2 A 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
MAASTrigonometry
© MAAS. All rights reserved.Page 8 / 47
Extracted text · page 8
Trigonometry
Example # 5 : Prove that
sin2A
(i) = tan A (ii) tan A + cot A = 2 cosec 2 A
1 cos2A
1 cosA cosB cos(A B) A B
(iii) = tan cot
1 cosA cosB cos(A B) 2 2
sin2A 2sinAcosA
Solution : (i) L.H.S. = = tan A
1 cos2A 2cos 2 A
1 tan2 A 1 tan2 A 2
(ii) L.H.S. tan A + cot A = = 2 = = 2 cosec 2 A tanA 2tanA sin2A
2 A A A
2sin 2sin sin B 1 cosA cosB cos(A B) 2 2 2 (iii) L.H.S. =
1 cosA 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 A 2 2 A 2 2 A B = tan = tan = tan cot
2 A A 2 A B B 2 2
cos cos B 2sin sin
2 2 2 2
Self practice problems
sin4 sin2 2tan
(8) Prove that 1 cos4 cos2= 1 tan 2
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
MAASTrigonometry
© MAAS. All rights reserved.Page 9 / 47
Extracted text · page 9
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
tanA tanB
cotC
1– tanA 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 2 1 y 2 1 z 2 1 x 2 1 y 2 1 z 2
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 2tanA 2tanB 2tanC
+ + = + + .
1 x 2 1 y 2 1 z 2 1 tan 2 A 1 tan 2 B 1 tan2 C
= tan2A + tan2B + tan2C [ A + B + C = n ]
2x 2y 2z
= tan2A tan2B tan2C = . .
1 x 2 1 y 2 1 z 2
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 .
sinA 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 sin 2
n sin 2 n 1 (ii) cos + cos (+) + cos ( + 2 ) +.... + cos (n 1) = cos 2 sin 2
where : 2m, m
ADVTRI- 9
MAASTrigonometry
© MAAS. All rights reserved.Page 10 / 47
Extracted text · page 10
Trigonometry
sin 2 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 2 2 2 sin n Solution : (i) sin + sin3 + sin5 + .... + sin(2n–1) = =
2 sin
sin
2
sin2º sin4º .... sin180º sin90º(sin91º) cos1º cot1º
(ii) = = = =
90 90sin1º 90sin1º 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
sin11 2sin11
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 2 2 E = a b sin cos
a 2 b 2 a 2 b 2
b a
Let = sin & = cos
a 2 b 2 a 2 b 2
2 2 b E = a b 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 + 3cos2x
1
Solution : (i) sinsin(90º – ) = sincos = × sin2
2
1
maximum value =
2
ADVTRI- 10
MAASTrigonometry
© MAAS. All rights reserved.Page 11 / 47
Extracted text · page 11
Trigonometry
2 2sinx 1 2 13 (ii) 1 + 2sinx + 3cos2x = – 3sin2x + 2sinx + 4=– 3 sin x +4=–3 sinx + 3 3 3
1 2 16 16 1 2 Now 0 sinx – – 3 sinx 0
3 9 9 3
1 2 13 13 – 1 – 3 sinx +
3 3 3
Self practice problems
(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
MAASTrigonometry
© MAAS. All rights reserved.Page 12 / 47
Extracted text · page 12
Trigonometry
General Solution of Some Standard Trigonometric Equations :
(i) If sin = sin = n + (1)n where , , n . 2 2
(ii) If cos = cos = 2 n ± 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
MAASTrigonometry
© MAAS. All rights reserved.Page 13 / 47
Extracted text · page 13
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
Self Practice Problems :
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.
3 x 3 x sin – cos
2 2 cosxExample # 15 :
2 sinx 3
3 x 3 x x x 2 x 2 x x x sin – cos sin – cos sin cos sin cos
2 2 cosx 2 2 2 2 2 2 cosxSolution : =
2 sinx 3 2 sinx 3
x x sin – cos 2 sinx 2 2 cosx x x 2 x 2 x = 3 sin – cos – 2 cos – sin 0 2 2 sinx 3 2 2 2 2
x x x x x x x
sin – cos 3 2sin 2cos 0 sin cos tan 1 2 2 2 2 2 2 2
x
n ,n x 2n ,n
2 4 2
Self Practice Problems :
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
MAASTrigonometry
© MAAS. All rights reserved.Page 14 / 47
Extracted text · page 14
Trigonometry
cosx 1
Example # 16 : Solve sin2x – =
4 4
cosx 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
Self Practice Problems :
(23) Solve 4sin2+ 3 1 – 3 = 0 2sin
(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
Self Practice Problems :
(25) Solve sin7= sin3 + sin (26) Solve 1 + cos3x = 2cos 2x
sin6x
(27) Solve 8cosx cos 2x cos4x =
sinx
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
MAASTrigonometry
© MAAS. All rights reserved.Page 15 / 47
Extracted text · page 15
Trigonometry
Example # 18 : Solve sec4 – sec2 = 2
1 1
Solution : – = 2
cos4 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)
2 x x 1 tan 2tan
2 2 cosx = & sinx =
2 x 2 x 1 tan 1 tan
2 2
equation (i) becomes
2 x x 1 tan 2tan 2 2 3 + 4 = 5 ........(ii)
x x 1 tan 2 1 tan 2
2 2
x
Let tan = t
2
1 t 2 2t
equation (ii) becomes 3 + 4 2 = 5 2 1 t 1 t
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
MAASTrigonometry
© MAAS. All rights reserved.Page 16 / 47
Extracted text · page 16
Trigonometry
Self Practice Problems :
(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– t2 in (i)
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 n Answers : (30) n + , n (31) n – or (–1) , 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
Self Practice Problems :
(33) Solve cos50x – sin50x = 1
(34) Solve 12 sin x + 5cosx = 2y2 – 8y + 21 for x & y
5
Answers : (33) n, n (34) x = 2nwhere = cos–1 , n y = 2 13
ADVTRI- 16
MAASTrigonometry
© MAAS. All rights reserved.Page 17 / 47
Extracted text · page 17
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
MAASTrigonometry
© MAAS. All rights reserved.Page 18 / 47
Extracted text · page 18
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.
y
1
0 x
–2 3 – 5 3 2 2 6 2 6 2 2
–1
y = sinx
From, the graph of y = sinx, it is obvious that, between 0 and 2sinx > 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)
nI
Self practice problems
(35) Solve the following inequations
(i) (sinx – 2) (2sinx–1) < 0 (ii) sinx + 3 cosx 1
5 Ans. (i) x 2n, 2n (ii) x [– +2n, 2n ] n 6 6 nI 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
MAASTrigonometry
© MAAS. All rights reserved.Page 19 / 47
Extracted text · page 19
Trigonometry
Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Measurement of Angles & Allied angles
A-1. Find the radian measures corresponding to the following degree measures
(i) 15° (ii) 240° (iii) 530°
A-2. Find the degree measures corresponding to the following radian measures
3 5 7
(i) (ii) – 4 (iii) (iv)
4 3 6
A-3. Prove that :
7 (i) 2 sin2 + cosec cos2 = 0
6 6 3
5 (ii) cot2 + cosec + 3 tan2 = 6
6 6 6
A-4. Find the value of :
(i) cos 210° (ii) sin 225° (iii) tan 330° (iv) cot (– 315°)
A-5. Prove that
cos( )cos( )
(i) = cot2 .
sin( )cos 2
(ii) cos + sin (270° + ) – sin (270° – ) + cos (180° + ) = 0.
3 3
(iii) cos cos (2 + ) cot cot (2 ) = 1.
2 2
A-6. If tan = – 5/12, is not in the second quadrant, then show that
sin(3600 ) tan(900 ) 181
=
sec(2700 ) cosec( ) 338
Section (B) : Graphs and Basic Identites (sin(A±B), cos(A±B), tan(A±B))
B-1. Sketch the following graphs :
(i) y = 3 sin 2x (ii) y = 2 tan x (iii) y = cosx
B-2. Find number of solutions of equation sinx = –4x + 1
2
B-3. If tan + sec = then sec is
3
B-4. Show that : (i) sin 20° . cos 40° + cos 20° . sin 40° = 3 /2
(ii) cos 100° . cos 40° + sin 100° . sin 40° = 1/2
9 5
B-5. Show that : cos 2 cos – cos 3 cos = sin 5 sin .
2 2 2
1 0
B-6 If A + B = 450, prove that (1 + tanA)(1 + tan B) = 2 and hence deduce that tan 22 = 2 – 1
2
B-7. Eliminate from the relations a sec = 1 – b tan , a2 sec2 = 5 + b2 tan2
ADVTRI- 19
MAASTrigonometry
© MAAS. All rights reserved.Page 20 / 47
Extracted text · page 20
Trigonometry
Section (C) : sin2A – sin2B, Multiple angles upto 3A, 2sinA cosB, sinC – sinD
C-1. Show that :
(i) sin2 750 – sin2 150 = 3 /2 (ii) sin2 450 – sin2 150 = 3 / 4
C-2. Find the value of
(i) 4 sin 18° cos 36° (ii) cos2 72° – sin2 54°
(iii) cos2 48º – sin2 12º
a 2 – b 2
C-3. If and are the solution of a cos + b sin = c, then show that cos( + ) =
a 2 b 2
A A 1
C-4. Show that : sin2 – sin2 = sin A
8 2 8 2 2
C-5. Show that : cos² + cos² (+ ) 2cos cos cos (+ ) = sin².
C-6. Prove that
sin2 A sin2 B
(i) = tan (A + B)
sinAcosA sinBcosB
4cos2A
(ii) cot (A + 15º) – tan (A – 15º) =
1 2sin2A
C-7. If 0 < < /4, then show that 2 2(1 cos4) = 2 cos
cos3 A cos3A sin3 A sin3A
C-8. Prove that + = 3
cosA sinA
C-9. Prove that
2
1 cot 4 9 (i) cos cot 4 sec = cosec 4. `
2 2 1 cot 2 4
1 1 (ii) cot 2 tan3 tan cot3 cot =
sec8A 1 tan8A (iii) =
sec 4A 1 tan2A
cosA sinA cosA sinA (iv) – = 2 tan 2A
cosA sinA cosA sinA
sin3
C-10. Prove that sin = and hence deduce the value of sin 15°.
1 2cos2
C-11. Prove that 4(cos3 20° + cos3 40°) = 3(cos 20° + cos 40°)
C-12. Prove that :
tan3x 2cos2x 1 2sinx tanx
(i) (ii) = 1
tanx 2cos2x 1 sin3x tan3x
C-13. Prove that :
tan tan (600 + ) tan(600 – ) = tan 3 and hence deduce that tan 200 tan 400 tan 600 tan 800 = 3.
ADVTRI- 20
MAASTrigonometry
© MAAS. All rights reserved.Page 21 / 47
Extracted text · page 21
Trigonometry
C-14. Prove that :
(i) (cosec – sin ) (sec – cos ) (tan + cot ) = 1
2sin tan (1 tan ) 2sin sec 2 2sin
(ii) =
(1 tan )2 (1 tan )
1 sinA
(iii) = ± (sec A – tan A)
1 sinA
cosAcosecA sinAsec A
(iv) = cosec A – sec A
cosA sinA
1 1 1 1
(v) – = –
sec tan cos cos sec tan
cos3 A sin3 A cos3 A sin3 A
(vi) + = 2
cosA sinA cosA sinA
Section (D) : Conditional Identities & Trigonometric Series
D-1. For all values of , prove that,
cos + cos + cos+ cos (+ + ) = 4 cos . cos . cos .
2 2 2
D-2. If x + y + z = show that, sin 2x + sin 2y + sin 2z = 4cosx cosy cosz.
2
D-3. If x + y = + z, then prove that sin2x + sin2y – sin2z = 2 sin x sin y cos z.
D-4. If A + B + C = 2S then prove that
A B C
cos (S – A) + cos(S – B) + cos (S – C) + cos S = 4 cos cos cos
2 2 2
D-5. If A + B + C = 0° then prove that sin 2A + sin 2B + sin 2C = – 4 sin A sin B sin C
D-6. If is the exterior angle of a regular polygon of n sides and is any constant, then prove that
sin + sin ( + ) + sin ( +2) + ............ up to n terms = 0
n sinn cos(n 1)
D-7. Prove that sin2 + sin2 2 + sin2 3 + ....... + sin2 n = –
2 2sin
D-8. Prove that :
2 4 6 1
(i) cos cos cos =
7 7 7 8
2 3 4 5 1
(ii) cos cos cos cos cos =
11 11 11 11 11 32
Section (E) : Range of Trigonometric Expressions
2 2 E-1. Find the extreme values of cos x cos x cos x 3 3
E-2. Find the maximum and minimum values of following trigonometric functions
(i) cos 2x + cos2x (ii) cos2 x + (sinx – cos x)2 4
E-3. Find the greatest and least value of y
(i) y = 10 cos²x 6 sin x cos x + 2 sin²x
+ 3 (ii) y = 3 cos + 5 cos 3
ADVTRI- 21
MAASTrigonometry
© MAAS. All rights reserved.Page 22 / 47
Extracted text · page 22
Trigonometry
Section (F) : Trigonometric Equations
F-1. What are the most general values of which satisfy the equations:
1
(i) sin = (ii) tan (x – 1) = 3
2
2
(iii) tan = – 1 (iv) cosec = .
3
(v) 2cot2 = cosec2
F-2. Solve
(i) sin9 = sin (ii) cot + tan = 2cosec
(iii) sin2 = cos3 (iv) cot = tan8
(v) cot – tan = 2. (vi) cosec = cot + 3
(vii) tan2 tan = 1
(viii) tan + tan2 + 3 tan tan2 = 3 .
F-3. Solve
(i) sin + sin3 + sin5 = 0.
(ii) cos + sin = cos 2 + sin 2.
(iii) cos2 x + cos2 2 x + cos2 3 x = 1 .
(iv) sin2n – sin2(n – 1) = sin2, where n is constant and n 0, 1
F-4. Solve
(i) tan2 – (1 + 3 ) tan + 3 = 0
(ii) 4 cos – 3 sec = 2 tan
2
(iii) tan x . tan x . tan x = 3 . 3 3
F-5. Solve
(i) 3 sin – cos = 2 (ii) 5 sin + 2 cos = 5
Section (G) : Trigonometric Inquations and Height & Distance
G-1. Solve tan2 x 1
G-2. Solve 2sin2x – sinx – 1 > 0
G-3. Solve 3 cot 1
G-4. Two pillars of equal height stand on either side of a roadway which is 60 m wide. At a point in the
roadway between the pillars, the angle of elevation of the top of pillars are 60º and 30º. Then find height
of pillars -
G-5. If the angles of elevation of the top of a tower from two points distance a and b from the base and in the
same straight line with it are complementary, then find the height of the tower :
G-6. From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of
depression of the foot of the tower. Then find height of the tower -
PART - II : ONLY ONE OPTION CORRECT TYPE
Section (A) : Measurement of Angles & Allied angles
A-1. cos (540° – ) – sin (630° – ) is equal to
(A) 0 (B) 2 cos (C) 2 sin (D) sin – cos
A-2. The value of tan 1° tan 2° tan 3° ... tan 89° is
1
(A) 1 (B) 0 (C) (D)
2
ADVTRI- 22
MAASTrigonometry
© MAAS. All rights reserved.Page 23 / 47
Extracted text · page 23
Trigonometry
2 4A-3. If x = y cos = z cos , then xy + yz + zx is equal to
3 3
(A) – 1 (B) 0 (C) 1 (D) 2
3
A-4. If 0° < x < 90° & cosx = , then the value of log10 sin x + log10 cos x + log10tan x is
10
(A) 0 (B) 1 (C) – 1 (D) 2
A-5. If tan + cot = a then the value of tan4 + cot4 =
(A) a4 4a2 + 2 (B) a4 4a2 + 2 (C) a4 4a2 2 (D) a4 2a2 + 2
Section (B) : Graphs and Basic Identites (sin(A±B), cos(A±B), tan(A±B))
B-1. STATEMENT-1 : sin 2 > sin 3
STATEMENT-2 : If x, y , , x < y, then sin x > sin y 2
(A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for
STATEMENT-1
(B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation
for STATEMENT-1
(C) STATEMENT-1 is true, STATEMENT-2 is false
(D) STATEMENT-1 is false, STATEMENT-2 is true
B-2. If cosec – cot = then cot is :
1 1 1 1 1 1 (A) (B) (C) (D) – 2 2
B-3. If a cos + b sin = 3 & a sin b cos = 4 then a2 + b2 has the value =
(A) 25 (B) 14 (C) 7 (D) 10
3 3 7
tan x . cos x sin x 2 2 2 B-4. when simplified reduces to:
3
cos x . tan x 2 2
(A) sin x cos x (B) sin2 x (C) sin x cos x (D) sin2x
4 3 4 6 6
B-5. The expression 3 sin sin (3 ) – 2 sin sin (5 ) is equal to
2 2
(A) 0 (B) 1 (C) 3 (D) sin 4 + sin 6
3 7 9 B-6. The value of the expression 1 cos 1 cos 1 cos 1 cos is 10 10 10 10
1 1 1
(A) (B) (C) (D) 0
8 16 4
sin24 cos6 sin6 sin66
B-7. The value of is
sin21 cos 39 cos51 sin69
(A) 1 (B) 1 (C) 2 (D) 0
ADVTRI- 23
MAASTrigonometry
© MAAS. All rights reserved.Page 24 / 47
Extracted text · page 24
Trigonometry
B-8. If tan A and tan B are the roots of the quadratic equation x2 ax + b = 0, then the value of sin2 (A + B).
a 2 a 2 a 2 a 2
(A) (B) (C) (D)
a 2 (1 b)2 a 2 b 2 (b c)2 b 2 (1 a)2
B-9. If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is equal to
1 1 1 1 1 1 1
(A) – (B) – (C) + (D)
y x x y x y x y
tan155 tan115
B-10. If tan 25° = x, then is equal to
1 tan155 tan115
1 x 2 1 x 2 1 x 2 1 x 2
(A) (B) (C) (D)
2x 2x 1 x 2 1 x 2
cot A cot B
B-11. If A + B = 225°, then the value of . is
1 cot A 1 cot B
1 1
(A) 2 (B) (C) 3 (D) –
2 2
B-12. The value of tan 203° + tan 22° + tan 203° tan 22° is
(A) – 1 (B) 0 (C) 1 (D) 2
Section (C) : sin2A – sin2B, Multiple angles upto 3A, 2sinA cosB, sinC – sinD
1 tan2 15
C-1. The value of is
1 tan2 15
3
(A) 1 (B) 3 (C) (D) 2
2
C-2. If A lies in the third quadrant and 3 tan A – 4 = 0, then 5 sin 2A + 3sinA + 4 cosA is equal to
24 24 48
(A) 0 (B) – (C) (D)
5 5 5
C-3. If cos A = 3/4, then the value of 16cos2 (A/2) – 32 sin (A/2) sin (5A/2) is
(A) – 4 (B) – 3 (C) 3 (D) 4
C-4. If tan2 = 2 tan2 + 1, then the value of cos 2 + sin2 is
(A) 1 (B) 2 (C) – 1 (D) Independent of
C-5. If , then the value of 1 sin 1 sin is equal to: 2
(A) 2 cos (B) 2 sin (C) 2 (D) none of these
2 2
1 1
C-6. The value of + is
cos 290 3 sin 250
2 3 4 3
(A) (B) (C) 3 (D) none
3 3
C-7. The value of tan 3A – tan 2A – tan A is equal to
(A) tan 3A tan 2A tan A (B) – tan 3A tan 2A tan A
(C) tan A tan 2A – tan 2A tan 3A – tan 3A tan A (D) none of these
ADVTRI- 24
MAASTrigonometry
© MAAS. All rights reserved.Page 25 / 47
Extracted text · page 25
Trigonometry
cos20 8sin70 sin50 sin10
C-8. is equal to:
sin2 80
(A) 1 (B) 2 (C) 3/4 (D) 0
C-9. The numerical value of sin 12° . sin48° . sin 54° is equal to
1 1 1 1
(A) (B) (C) (D)
2 4 16 8
C-10. If A = tan 6° tan 42° and B = cot 66° cot 78°, then
1
(A) A = 2B (B) A = B (C) A = B (D) 3A = 2B
3
Section (D) : Conditional Identities & Trigonometric Series
D-1. In a triangle tan A + tan B + tan C = 6 and tan A tan B = 2, then the values of tan A, tan B and tan C are
respectively
(A) 1, 2, 3 (B) 2, 3, 1 (C) 1, 2, 0 (D) none of these
D-2. tan + 2 tan 2 + 4 tan 4 + 8 cot 8 =
(A) tan (B) cot (C) cot 16 (D) 16 cot
2 3 4 5 6 D-3. The value of cos 0 + cos + cos + cos + cos + cos + cos is
7 7 7 7 7 7
(A) 1/2 (B) 1/2 (C) 0 (D) 1
2 4 8 16 D-4. The value of cos cos cos cos cos is :
10 10 10 10 10
10 2 5 cos /10 cos /10 10 2 5
(A) (B) – (C) (D) –
64 16 16 16
3 5 17 D-5. The value of cos + cos + cos +...... + cos is equal to :
19 19 19 19
(A) 1/2 (B) 0 (C) 1 (D) 2
Section (E) : Range of Trigonometric Expressions
E-1. If f() = sin4 + cos2 , then range of f() is
(A) 1, 1 (B) 1 , 3 (C) 3 , 1 (D) None of these 2 2 4 4
E-2. Range of function f(x) = cos2x + 4sec2x is
(A) [4, ) (B) [0, ) (C) [5, ) (D) (0, )
E-3. The difference between maximum and minimum value of the expression y = 1 + 2 sin x + 3 cos2 x is
16 13
(A) (B) (C) 7 (D) 8
3 3
E-4. The maximum value of 12 sin – 9 sin2 is -
(A) 3 (B) 4 (C) 5 (D) None of these
E-5. The greatest and least value of y = 10 cos²x 6 sin x cos x + 2 sin²x are respectively
(A) 11, 1 (B) 10, 2 (C) 12, – 4 (D) 11, – 1
ADVTRI- 25
MAASTrigonometry
© MAAS. All rights reserved.Page 26 / 47
Extracted text · page 26
Trigonometry
Section (F) : Trigonometric Equations
F-1. The solution set of the equation 4sin.cos – 2cos – 2 3 sin + 3 = 0 in the interval (0, 2) is
3 7 5 3 5 5 11
(A) , (B) , (C) , , , (D) , ,
4 4 3 3 4 3 3 6 6 6
F-2. All solutions of the equation 2 sin + tan = 0 are obtained by taking all integral values of m and n in:
2 2 (A) 2n + , n (B) n or 2m ± where n, m
3 3
(C) n or m ± where n, m (D) n or 2m ± where n, m
3 3
F-3. Total number of solutions of equation sinx . tan4x = cosx belonging to (–, 2) are :
(A) 4 (B) 7 (C) 8 (D) 15
F-4. If x 0 , , the number of solutions of the equation sin 7x + sin 4x + sin x = 0 is: 2
(A) 3 (B) 5 (C) 6 (D) 4
F-5. The general solution of equation sinx + sin5x = sin2x + sin4x is :
n n n 2 n
(A) ; n (B) ; n (C) ; n (D) ; n
2 5 3 3
F-6. The general solution of the equation 2cos2x = 3.2cos2x 4 is
(A) x = 2n, n (B) x = n, n (C) x = n, n (D) x = n, n
F-7. If 2 cos2 ( + x) + 3 sin ( + x) vanishes then the values of x lying in the interval from 0 to 2 are
(A) x = /6 or 5/6 (B) x = /3 or 5/3 (C) x = /4 or 5/4 (D) x = /2 or 5/2
cos 3 1
F-8. = if
2 cos 2 1 2
(A) = n + , n (B) = 2n , n (C) = 2n ± , n (D) = n + , n
3 3 6 6
F-9. If cos 2 + 3 cos = 0, then
17 3 17 3
(A) = 2n ± where = cos–1 (B) = 2n ± where = cos–1 4 4
17 3 17 3
(C) = 2n ± where = cos–1 (D) = 2n ± where = cos–1 4 4
F-10. If sin + 7 cos = 5, then tan (/2) is a root of the equation
(A) x2 6x + 1 = 0 (B) 6x2 x 1 = 0 (C) 6x2 + x + 1 = 0 (D) x2 x + 6 = 0
1
F-11. The most general solution of tan = – 1 and cos = is :
2
7 7 (A) n + , n (B) n + (– 1)n , n
4 4
7 3
(C) 2n + , n (D) 2n + , n
4 4
1
F-12. A triangle ABC is such that sin(2A + B) = . If A, B, C are in A.P. then the angle A, B, C are
2
respectively.
5 5 5 5 (A) , , (B) , , (C) , , (D) , ,
12 4 3 4 3 12 3 4 12 3 12 4
ADVTRI- 26
MAASTrigonometry
© MAAS. All rights reserved.Page 27 / 47
Extracted text · page 27
Trigonometry
Section (G) : Trigonometric Inquations and Height & Distance
G-1. The complete solution of inequality sec23x < 2 is
n n n n
(A) x – , , n (B) x – , , n
3 12 3 12 3 12 3 6
n n
(C) x n – ,n , n (D) x – , , n
12 12 3 6 3 6
G-2. The complete solution of inequality 2cos2x – 7 cosx + 3 < 0 is
(A) n – < x < + n (B) 2n – < x < + 2n
3 3 6 6
(C) 2n – < x < + 2n (D) n – < x < + n
3 3 6 6
G-3. The complete solution of inequality cos 2x cos x is
2 2 (A) x 2n – , 2n (B) x 2n – , 2n 3 3 3 3
2 2 (C) x 2n, 2n (D) x 2n – , 2n 3 3
G-4. Which of the following set of values of x satisfy the inequation tan2x – (1 + 3 ) tan x + 3 < 0
4n 1 3n 1 2n 1 2n 1
(A) , , (n Z) (B) , , (n Z)
4 3 4 3
4n 1 4n 1
, , (C) , (n Z) (D) x 3 4 2 4
G-5. A tree 12 m high, is broken by the wind in such a way that its top touches the ground and makes an
angle 60º with the ground. The height from the bottom of the tree from where it is broken by the wind is
appoximately
(A) 5.57 m (B) 5.21 (C) 5.36 (D) 5.9
G-6. AB is a vertical pole and C is the middle point. The end A is on the level ground and P is any point on
the level ground other than A. The portion CB subtends an angle at P. If AP : AB = 2 : 1, then is
equal to-
1 4 5 2
(A) tan–1 (B) tan–1 (C) tan–1 (D) tan–1 9 9 9 9
G-7. A round ballon of radius r subtends an angle at the eye of the observer, while the angle of elevation
of its centre is . The height of the centre of ballon is-
(A) r cosec sin (B) r sin cosec (C) r sin cosec (D) r cosec sin
2 2 2 2
G-8. If the angle of elevation of a cloud from a point 200 m above a lake is 30º and the angle of depression
of its reflection in the lake is 60º, then the height of the cloud above the lake, is
(A) 200 m (B) 500 m (C) 30 m (D) 400 m
G-9. A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards
it. If it takes 12 minutes for the angle of depression to change from 30º to 45º, then the car will reach
the tower in
(A) 17 minutes 23 seconds (B) 16 minutes 23 seconds
(C) 16 minutes 18 seconds (D) 18 minutes 22 seconds
ADVTRI- 27
MAASTrigonometry
© MAAS. All rights reserved.Page 28 / 47
Extracted text · page 28
Trigonometry
PART - III : MATCH THE COLUMN
1. Column - I Column - II
(A) tan 9° tan 27° tan 63° + tan 81° (p) 1
(B) cosec 10° – 3 sec 10° (q) 2
sec 5 cos 40
(C) 2 2 sin10 – 2 sin 35 (r) 3
2 sin 5
(D) 3 (cot 70º + 4 cos 70º) (s) 4
2. Column – Column –
1
(A) If for some real x, the equation x + = 2 cos holds, (p) 2
x
then cos is equal to
(B) If sin + cosec = 2, then sin2008 + cosec2008 is equal to (q) 1
(C) Maximum value of sin4 + cos4 is (r) 0
(D) Least value of 2 sin2 + 3 cos2 is (s) – 1
3. Column – Column –
(A) Number of solutions of sin2 + 3 cos = 3 (p) 0
in [– , ]
(B) Number of solutions of sin x . tan 4x = cos x (q) 1
in (0, )
(C) Number of solutions of equation (r) 4
tan2 (1 – tan ) (1 + tan ) sec2 + 2 = 0 where , 2 2
(D) If [sin x] + [ 2 cosx] = – 3, where x [0, 2] (s) 5
then [sin 2x] equals (Here [.] denotes G.I.F.)
Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
1. In a triangle ABC if tan A < 0 then:
(A) tan B. tan C > 1 (B) tan B. tan C < 1 (C) tan B. tan C = 1 (D) Data insufficient
2. If sin = 1/2 and cos = 1/3, then the values of + (if , are both acute) will lie in the interval
2 2 5 5 (A) , (B) , (C) , (D) , 3 2 2 3 3 6 6
ADVTRI- 28
MAASTrigonometry
© MAAS. All rights reserved.Page 29 / 47
Extracted text · page 29
Trigonometry
sinA 3 cosA 5
3. If = and = , 0 < A, B </2, then tan A + tan B is equal to
sinB 2 cosB 2
(A) 3 / 5 (B) 5 / 3 (C) 1 (D) ( 5 3)/ 5
4. In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the opposite
vertex. Then the other acute angles of the triangle are
3 3 (A) & (B) & (C) & (D) &
3 6 8 8 4 4 5 10
5. If 3 cos x + 2 cos 3x = cos y, 3 sin x + 2 sin 3x = sin y, then the value of cos 2x is
1 1 7
(A) – 1 (B) (C) – (D)
8 8 8
cos3
6. If cos + cos = a, sin + sin = b and – = 2, then cos =
(A) a2 + b2 – 2 (B) a2 + b2 – 3 (C) 3 – a2 – b2 (D) (a2 + b2) /4
3 17. If < < , then 2cot is equal to
4 sin2
(A) 1 + cot (B) – 1 – cot (C) 1 – cot (D) – 1 + cot
sin sin28. For << , lies in the interval 2 2 1 cos cos2
(A) () (B) (2, 2) (C) (0, ) (D) (1, 1)
9. The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos 2x + a3 sin2x = 0 for all x is
(A) 0 (B) 1 (C) 2 (D) infinite
3
10. If A + B + C = , then cos 2A + cos2B + cos2C is equal to
2
(A) 1 – 4cos A cosB cosC (B) 4 sin A sin B sin C
(C) 1 + 2 cos A cos B cos C (D) 1– 4 sin A sin B sin C
11. If A + B + C = & cosA = cosB. cosC then tanB. tanC has the value equal to:
(A) 1 (B) 1/2 (C) 2 (D) 3
12. The general solution of the equation tan2 +2 3 tan = 1 is given by:
n (A) = , n (B) = (2n + 1) , n
2 2
n (C) = (6n + 1) , n (D) = , n
12 12
2 13. The general solution of the equation tan x + tan x + tan x = 3 is 3 3
n n n n
(A) , n (B) , n (C) , n (D) , n
4 12 3 6 3 12 3 4
ADVTRI- 29
MAASTrigonometry
© MAAS. All rights reserved.Page 30 / 47
Extracted text · page 30
Trigonometry
sec 2 x
14. The complete solution set of the equation 1 + 2 cosec x = 2 is
2
(A) 2 n , n (B) n , n (C) 2 n , n (D) n , n
2 2 2 2
15. The principal solution set of the equation 2 cos x = 2 2sin2x is
13 13 13 13
(A) , (B) , (C) , (D) ,
8 8 4 8 4 10 8 10
16. The solution of |cosx| = cosx – 2sinx is
(A) x = n, n (B) x = n + , n
4
(C) x = n + (–1)n , n (D) x = (2n + 1) + , n
4 4
17. The solution of inequality 4tanx – 3.2tanx + 2 0 is
(A) x n, n ; n (B) x n, n – ; n 4 4
(C) x n, n ; n (D) x n, n – ; n 6 6
18. The solution of inequality 5 2sinx 6 sin x – 1 is
(A) [ (12n – 7)/6, (12n + 7)/6] (n Z) (B) [ (12n – 7)/6, (12n + 1)/6] (n Z)
(C) [ (2n – 7)/6, (2n + 1)/6] (n Z) (D) [ (12n – 7)/3, (12n + 1)/3] (n Z)
PART - II : NUMERICAL VALUE QUESTIONS
INSTRUCTION :
The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto two digit.
If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal
placed.
tan
2
1. If 19 sin = 29 sin, then find the value of .
tan
2
2. If , ( – 2nn ) are different values of satisfying the equation 5 cos – 12 sin = 11. then
absolute value of sin ( + ) is
3 x 4 2
3. If x , then 4 cos2 + 4sin x sin 2x is always equal to 2 4 2
4. If three angles A, B, C are such that
cos A + cos B + cos C = 0 and if
cos A cos B cos C = (cos 3A + cos 3B + cos 3C) then value of is :
5. Find sum of square of all possible integral values of for which equation 4cos x + 3 sin x = 2 + 1 has a
solution.
6. If a cos3 + 3a cos sin2 = m and a sin3 + 3a cos2 sin = n. if (m+n)2/3 + (m n)2/3 = paq, then find
value of p3 + q3
ADVTRI- 30
MAASTrigonometry
© MAAS. All rights reserved.Page 31 / 47
Extracted text · page 31
Trigonometry
7. If 2 cos x + sin x = 1, then find the sum of all possible values of 7 cos x + 6 sin x.
31 8. The number of roots of the equation cot x = + x in , is , 2 2
n 9. If 2tan2x – 5 secx – 1 = 0 has 7 different roots in 0, , n N, then find the greatest value of n. 2
10. Sum of all possible integral values of a for which the equation cos 2x + a sin x = 2a 7 possesses a
solution.
11. The number of solutions of the equation |sinx| = | cos3x| in [–2, 2] is
12. In any triangle ABC, which is not right angled cosA .cosecB.cosecC is equal to
13. If A + B + C = , then find value of tan B tan C + tan C tan A + tan A tan B – sec A sec B sec C.
14. If the arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos( + x) – 1 = 0 in the interval
[0, 315] is equal to k , then find the value of k
1
15. cos ( – ) = 1 and cos ( + ) = , where , [–, ]. Then number of ordered pairs (, ) which
e
satisfy both the equations.
16. Sum of all possible value of between 0° and 360° which satisfy the equation
sec2 .cosec2 + 2 cosec2 = 8
17. Find the number of all values of [0, 10.5] satisfying the equation
cos 6 + cos 4 + cos 2 + 1 = 0 .
18. In (0, 6), find the number of solutions of the equation tan+ tan 2 + tan 3 = tan .tan2.tan3
19. If 0 x 3 , 0 y 3and cos x . sin y = 1, then find the possible number of values of the ordered
pair (x, y)
20. Find the number of values of satisfying the equation sin3 = 4sin . sin 2. sin 4 in 0 2
21. Sum of all possible values of x which satisfy equation cos 6x + tan2 x + cos 6x . tan2 x = 1 in the interval
[0, 2].
3 7
22. Consider tan + sin = & tan² + cos² = , find maximum value of ( + ) if + (0, 2).
2 4
23. If range of values of n so that sinx(sinx + cosx) = n has at least one solution is [p,q] then values of
p3 + q3 is
24. Find the number of values of x in (0, 2) satisfying the equation cot x 2 sin 2x = 1 .
25. Find the number of solutions of sin + 2sin2 + 3sin3 + 4sin4 = 10 in(0, ).
26. Find the values of x satisfying the equation 2 sin x = 3 x2 + 2 x + 3.
27. Find the number of solution of sinx cosx – 3 cosx + 4 sinx – 13 > 0 in [0,2] .
ADVTRI- 31
MAASTrigonometry
© MAAS. All rights reserved.Page 32 / 47
Extracted text · page 32
Trigonometry
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
(cos11 sin11 )
1. The value of is
(cos11 sin11 )
(A) –tan 304° (B) tan 56° (C) cot 214° (D) cot 34°
1 t
2. If sin t + cos t = then tan can be
5 2
1 1
(A) 1 (B) – (C) 2 (D) –
3 6
sinx cosx
3. The value of =
cos3 x
(A) 1 + tan x + tan2x tan3x (B) 1 + tan x + tan2x + tan3x
(C) 1 tan x + tan2x + tan3x (D) (1 + tan x) sec2x
4. If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C) then
each side can be
(A) 1 (B) –1 (C) 0 (D) none
5. Which of the following is correct ?
(A) sin 1° > sin 1 (B) sin 1° < sin 1 (C) cos 1° > cos 1 (D) cos 1° < cos 1
6. If sin x + sin y = a & cos x + cos y = b, then which of the following may be true.
2 ab x y 4 a 2 b 2
(A) sin (x + y) = (B) tan =
a 2 b 2 2 a 2 b 2
x y 4 a 2 b 2 2ab
(C) tan = – (D) cos (x + y) =
2 a 2 b 2 a 2 b 2
3
7. If cos (A – B) = and tan A tan B = 2, then which of the following is/are correct
5
1 2
(A) cos A cos B = – (B) sin A sin B =
5 5
1 4
(C) cos (A + B) = – (D) sin A cos B =
5 5
8. If Pn = cosn + sinn and Qn = cosn – sinn, then which of the following is/are true.
(A) Pn – Pn – 2 = – sin2 cos2 Pn – 4 (B) Qn – Qn – 2 = – sin2 cos2 Qn – 4
(C) P4 = 1 – 2 sin2 cos2 (D) Q4 = cos2 – sin2
9. tan2 + 2tan. tan2 = tan2 + 2tan. tan2, if
(A) tan2 + 2tan. tan2 = 0 (B) tan + tan =0
(C) tan2 + 2tan. tan2= 1 (D) tan = tan
10. If the sides of a right angled triangle are {cos2cos2 + 2cos( + )} and
{sin2sin2 + 2sin( + )}, then the length of the hypotenuse is :
(A) 2[1+cos()] (B) 2[1 cos()] (C) 4 cos2 (D) 4sin2
2 2
11. For 0 < < /2, tan + tan 2 + tan 3 = 0 if
(A) tan = 0 (B) tan 2 = 0 (C) tan 3 = 0 (D) tan tan 2 = 2
ADVTRI- 32
MAASTrigonometry
© MAAS. All rights reserved.Page 33 / 47
Extracted text · page 33
Trigonometry
12. (a + 2) sin + (2a – 1) cos = (2a + 1) if tan =
3 4 2a 2a
(A) (B) (C) (D)
4 3 a 2 1 a 2 1
2b
13. If tan x = , (a c)
a c
y = a cos2x + 2b sin x cos x + c sin2x
z = a sin2x – 2b sin x cos x + c cos2x, then
(A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2 + 4b2
cosA cosB n sinA sinB n
14. The value of + is sinA sinB cosA cosB
A B A B
(A) 2 tann (B) 2 cotn : n is even
2 2
(C) 0 : n is odd (D) 0 : n is even
15. The equation sin6x + cos6x = a2 has real solution if
1 1 1 (A) a (–1, 1) (B) a 1, (C) a (D) a 1, 1 2 2 2 2
16. If sin(x y) = cos(x + y) = 1/2 then the values of x & y lying between 0 and are given by:
(A) x = /4, y = 3/4 (B) x =/4, y = /12
(C) x = 5/4, y = 5/12 (D) x = 11/12, y = 3/4
17. If 2 sec2 – sec4 – 2 cosec2 + cosec4 = 15/4, then tan can be
(A) 1/ 2 (B) 1/2 (C) 1/2 2 (D) –1/ 2
18. If 3 sin = sin (2 + ), then tan (+ ) – 2 tan is
(A) independent of (B) independent of
(C) dependent of both and (D) independent of but dependent of
19. If + + = 2, then
(A) tan + tan + tan = tan tan tan
2 2 2 2 2 2
(B) tan tan + tan tan + tan tan = 1
2 2 2 2 2 2
(C) tan + tan + tan = – tan tan tan
2 2 2 2 2 2
(D) tan tan + tan tan + tan tan = 1
4 4 4 4 4 4
20. If x + y = z, then cos2 x + cos2 y + cos2 z – 2 cos x cos y cos z is equal to
(A) cos2 z (B) sin2 z (C) cos (x + y – z) (D) 1
21. If tanA + tan B + tan C = tan A. tan B. tan C, then
(A) A, B, C may be angles of a triangle (B) A + B + C is an integral multiple of
(C) sum of any two of A, B, C is equal to third (D) none of these
1 2x 5x 2 22. Which of the following values of ' t ' may satisfy the condition 2 sin t = , t , . 2 3x 2x 1 2 2
3 3 (A) , (B) 0, (C) , (D) , 2 10 2 10 2 10 10
ADVTRI- 33
MAASTrigonometry
© MAAS. All rights reserved.Page 34 / 47
Extracted text · page 34
Trigonometry
23. sinx, sin2x, sin3x are in A.P if
(A) x = n/2, n (B) x = n, n (C) x = 2n, n (D) x = (2n +1), n
24. sin x + sin2x + sin 3x = 0 if
(A) sin x = 1/2 (B) sin 2x = 0 (C) sin 3x = 3 /2 (D) cos x = 1/2
25. cos4x cos8x cos5x cos9x = 0 if
(A) cos12x = cos 14 x (B) sin13 x = 0 (C) sinx = 0 (D) cosx = 0
26. sinx cos2x 1 assumes the least value for the set of values of x given by:
(A) x = n + (1)n+1 (/6) , n (B) x = n + (1)n (/6) , n
(C) x = n + (1)n (/3), n (D) x = n (1)n (/6) , n
27. Let 0 and x = X cos + Ysin , y = Xsin – Ycos such that x2 + 4xy + y2 = aX2 + bY2,
2
where a, b are constants then
(A) a = –1, b = 3 (B) = /4 (C) a = 3, b = – 1 (D) =
3
28. If the equation sin (x2) – sin(x2 + 2x) = 0 is solved for positive roots, then in the increasing sequence
of positive root
1 7 1 3
(A) first term is (B) first term is
2 2
1 11
(C) third term is 1 (D) third term is
2
29. The general solution of the equation cosx . cos6x = – 1, is :
(A) x = (2n + 1), n (B) x = 2n, n
(C) x = (2n – 1), n (D) none of these
30. Which of the following set of values of x satisfy the inequation sin 3x < sin x.
8n 1 8n 1 (8n 1)
(A) , 2n , n (B) , , n
4 4 4
8n 1 (8n 3) 8n 5 (C) , , n (D) 2n 1 , , n 4 4 4
x x
31. The equation 2sin . cos2x + sin2x = 2 sin . sin2x + cos2x has a root for which
2 2
1 1
(A) sin2x = 1 (B) sin2x = – 1 (C) cosx = (D) cos2x = –
2 2
32. cos 15 x = sin 5x if
n n (A) x = + ,n (B) x = + , n
20 5 40 10
3 n 3 n (C) x = + , n (D) x = + , n
20 5 40 10
33. 5 sin2 x + 3 sinx cosx + 6 cos2x = 5 if
(A) tan x = 1/ 3 (B) sin x = 0
(C) x = n + /2, n (D) x = n + /6, n
ADVTRI- 34
MAASTrigonometry
© MAAS. All rights reserved.Page 35 / 47
Extracted text · page 35
Trigonometry
34. sin2x + 2 sin x cos x 3cos2x = 0 if
(A) tan x = (B) tanx = 1
(C) x = n + /4, n (D) x = n + tan1 (3), n
35. Solution set of inequality sin3 x cos x > cos3 x sin x , where x (0, ), is
3 3 (A) , (B) , (C) 0 , (D) , 4 2 4 4 2 4
36. 4 sin4x + cos4x = 1 if
1 1
(A) x = n ; (n I) (B) x = n ± cos–1 ; (n I) 2 5
n (C) x = ; (n I) (D) x = – n ; (n I)
2
37. sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x if
1 n 2 (A) cos x = – (B) sin 2x = cos 2x (C) x = + (D) x = 2n ± , (n )
2 2 8 3
PART - IV : COMPREHENSION
Comprehenssion # 1
Let p be the product of the sines of the angles of a triangle ABC and q is the product of the cosines
of the angles.
1. In this triangle tan A + tan B + tan C is equal to
p
(A) p + q (B) p – q (C) (D) none of these
q
2. tan A tan B + tan B tan C + tan C tan A is equal to
1 q 1 p
(A) 1 + q (B) (C) 1 + p (D)
q p
3. The value of tan3 A + tan3 B + tan3 C is
p3 3pq2 q3 p 3 p 3 3pq
(A) (B) (C) (D)
q3 p 3 q3 q3
Comprehension # 2
Let a, b, c, d R. Then the cubic equation of the type ax3 + bx2 + cx + d = 0 has either one root real or
all three roots are real. But in case of trigonometric equations of the type a sin3 x + b sin2 x + c sinx + d
= 0 can possess several solutions depending upon the domain of x.
To solve an equation of the type a cos + b sin = c. The equation can be written as
cos ( – ) = c/ (a 2 b 2 ) .
The solution is = 2n + ± , where tan = b/a, cos = c/ (a 2 b 2 ) .
4. On the domain [–, ] the equation 4sin3 x + 2 sin2 x – 2sinx – 1 = 0 possess
(A) only one real root (B) three real roots
(C) four real roots (D) six real roots
10tanx
5. In the interval [–/4, /2], the equation, cos 4x + = 3 has
1 tan2 x
(A) no solution (B) one solution (C) two solutions (D) three solutions
1
6. |tan x| = tan x + (0 x 2) has
cosx
(A) no solution (B) one solution (C) two solutions (D) three solutions
ADVTRI- 35
MAASTrigonometry
© MAAS. All rights reserved.Page 36 / 47
Extracted text · page 36
Trigonometry
Comprehension # 3
To solve a trigonometric inequation of the type sin x a where |a| 1, we take a hill of length 2 in the
sine curve and write the solution within that hill. For the general solution, we add 2n. For instance, to
1 3 7 solve sinx – , we take the hill – , over which solution is – x . The general 2 2 2 6 6
7
solution is 2n– < x < 2n + , n is any integer. Again to solve an inequation of the type sin x a,
6 6
where |a| 1, we take a hollow of length 2 in the sine curve. (since on a hill, sinx a is satisfied over
two intervals). Similarly cos x a or cosx a, |a| 1 are solved.
7
7. Solution to the inequation sin6x + cos6x < must be
16
(A) n + < x < n + (B) 2n + < x < 2n +
3 2 3 2
n n
(C) x (D) none of these
2 6 2 3
8. Solution to inequality cos 2x + 5 cos x + 3 0 over [– , ] is
–5 5 –2 2 (A) [– , ] (B) , (C) [0, ] (D) , 6 6 3 3
9. Over [– , ], the solution of 2 sin2 x + 3 cos 2x 0 is 4
– 5 5 (A) [– , ] (B) , 6 6
–7 5 3 (C) [0, ] (D) – , – , , 12 4 12 4
Marked questions are recommended for Revision.
* Marked Questions may have more than one correct option.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1
1. The maximum value of the expression is
sin2 3sin cos 5cos 2
[IIT-JEE-2010, Paper-1, (3, 0)/84]
2. The positive integer value of n > 3 satisfying the equation
1 1 1
is [IIT-JEE 2011, Paper-1, (4, 0), 80]
2 3
sin sin sin n n n
n
3. The number of values of in the interval – , such that for n = 0, ±1, ± 2 and 2 2 5
tan = cot 5 as well as sin 2 = cos 4is [IIT-JEE-2010, Paper-1, (3, 0)/84]
4. Let P = { : sin – cos = 2 cos } and Q = { : sin + cos = 2 sin } be two sets. Then
(A) P Q and Q – P (B) Q P
(C) P Q (D) P = Q [IIT-JEE 2011, Paper-1, (3, –1), 80]
ADVTRI- 36
MAASTrigonometry
© MAAS. All rights reserved.Page 37 / 47
Extracted text · page 37
Trigonometry
5.* Let , [0, 2] be such that 2cos(1 – sin) = sin2 tan cot cos – 1, tan(2– ) > 0 and 2 2
3
–1 < sin < – . Then cannot satisfy [IIT-JEE 2012, Paper-1, (4, 0), 70]
2
4 4 3 3 (A) 0 < < (B) < < (C) < < (D) < < 2
2 2 3 3 2 2
6. For x (0, ), the equation sinx + 2 sin 2x – sin 3x = 3 has
[JEE (Advanced) 2014, Paper-2, (3, –1)/60]
(A) infinitely many solutions (B) three solutions
(C) one solution (D) no solution
57. The number of distinct solutions of the equation cos2 2x + cos4 x + sin4 x + cos6 x + sin6x = 2 in
4
the interval [0, 2] is [JEE (Advanced) 2015, P-1 (4, 0) /88]
13 1
8. The value of is equal to (k )1 k k 1 sin sin
4 6 4 6
[JEE (Advanced) 2016, Paper-2, (3, –1)/62]
(A) 3 3 (B) 2 3 3 (C) 2 3 1 (D) 2 2 3
9. Let S = x (– , :) x ,0 . The sum of all distinct solutions of the equation 2
3 sec x + cosec x + 2(tan x – cot x) = 0 in the set S is equal to
[JEE (Advanced) 2016, Paper-1, (3, –1)/62]
7 2 5
(A) – (B) – (C) 0 (D)
9 9 9
10. Let and be nonzero real numbers such that 2(cos – cos ) + cos cos = 1. Then which of the
following is/are true? [JEE(Advanced) 2017, Paper-2,(4, –2)/61]
(A) 3 tan – tan = 0 (B) tan – 3 tan = 0
2 2 2 2
(C) tan + 3 tan = 0 (D) 3 tan + tan = 0
2 2 2 2
11. Let a, b, c be three non-zero real numbers such that the equation 3 a cos x + 2b sinx = c,
b x , , has two distinct real roots and with + = . Then, the value of is ______. 2 2 3 a
[JEE(Advanced) 2018, Paper-1,(3, 0),60]
n k 1 k 2
sin sin k 0 n 2 n 2 12. For non-negative integer n, let f(n) = n
2 k 1 sin k 0 n 2
Assuming cos–1x takes values in [0, ] which of the following options is/are correct?
[JEE(Advanced) 2019, Paper-2 ,(4, –1)/62]
3
(A) f(4) = (B) If = tan (cos–1 f(6)), then 2 + 2 – 1 =0
2
1
(C) sin(7 cos–1 f(5)) = 0 (D) nlim f(n) = 2
ADVTRI- 37
MAASTrigonometry
© MAAS. All rights reserved.Page 38 / 47
Extracted text · page 38
Trigonometry
For Q. 13 & 14
Let f(x) = sin(cosx) and g(x) = cos(2sinx) be two functions defined for x > 0. Define the following sets
whose elements are written in increasing order
X = {x : f(x) = 0}, Y = {x : f (x) = 0}
Z = {x : g(x) = 0}, W = {x : g(x) = 0}
List-I contains sets X,Y,Z and W List-II contains some information regarding these set.
[JEE(Advanced) 2019, Paper-2 ,(4, –1)/62]
List - I List - II
3
() X (P) , , 4 , 7 2 2
() Y (Q) an arithmetic progression
() Z (R) NOT an arithmetic progression
7 13
(V) W (S) , , 6 6 6
2
(T) , , 3 3
3
(U) , 6 4
13. Which of the following is the only correct combination ? [JEE(Advanced) 2019, Paper-2 ,(4, –1)/62]
(1) IV – (Q), (T) (2) III – (R), (U) (3) III – (P), (Q), (U) (4) IV – (P), (R), (S)
14. Which is the following is only CORRECT combination ? [JEE(Advanced) 2019, Paper-2 ,(4, –1)/62]
(1) I – (Q), (U) (2) I – (P), (R) (3) II – (Q), (T) (4) II – (R), (S)
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
4 51. Let cos( + ) = and let sin( – ) = , where 0 , . Then tan 2 =
5 13 4
[AIEEE 2010 (4, –1), 144]
56 19 20 25
(1) (2) (3) (4)
33 12 7 16
2. If A = sin2 x + cos4 x, then for all real x : [AIEEE 2011 (4, –1), 120]
3 13 3 13
(1) A 1 (2) A 1 (3) 1 A 2 (4) A
4 16 4 16
3. In a PQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to :
[AIEEE-2012, (4, –1)/120]
5 3
(1) (2) (3) (4)
6 6 4 4
tanA cot A
4. The expression can be written as : [AIEEE - 2013, (4, –1),360]
1 cot A 1 tanA
(1) sinA cosA + 1 (2) secA cosecA + 1 (3) tanA + cotA (4) secA + cosecA
1
5. Let fk (x) = (sinkx + coskx) where x R and k 1. Then f4(x) – f6(x) equals k
[JEE(Main)2014,(4, – 1), 120]
1 1 1 1
(1) (2) (3) (4)
4 12 6 3
ADVTRI- 38
MAASTrigonometry
© MAAS. All rights reserved.Page 39 / 47
Extracted text · page 39
Trigonometry
6. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to
the foot of the tower, are 30º, 45º and 60º respectively, then the ratio, AB : BC , is
[JEE(Main)2015,(4, – 1), 120]
(1) 3 :1 (2) 3 : 2 (3) 1 : 3 (4) 2 : 3
7. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on
the path, he observes that the angle of elevation of the top of the pillar is 30°, After walking for 10
minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of
the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is :
[JEE(Main)2016,(4, – 1), 120]
(1) 10 (2) 20 (3) 5 (4) 6
8. If 0 x < 2, then the number of real values of x, which satisfy the equation
cosx + cos2x + cos3x + cos4x = 0, is [JEE(Main)2016,(4, – 1), 120]
(1) 5 (2) 7 (3) 9 (4) 3
9. If 5(tan2 x – cos2x) = 2cos2x + 9, then the value of cos4x is : [JEE(Main)2017,(4, – 1), 120]
3 1 2 7
(1) (2) (3) (4)
5 3 9 9
10. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a
point on the ground such that AP = 2AB. If BPC = , then tan is equal to
[JEE(Main)2017,(4, – 1), 120]
6 1 2 4
(1) (2) (3) (4)
7 4 9 9
1
11. If sum of all the solutions of the equation 8 cosx. cos x .cos x = 1 in [0, ] is k, then k
6 6 2
is equal to : [JEE(Main)2018,(4, – 1), 120]
8 20 2 13
(1) (2) (3) (4)
9 9 3 9
12. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the
angles of elevation of the top of the tower at P, Q and R are respectively 45º, 30º and 30º, then the
height of the tower (in m) is : [JEE(Main)2018,(4, – 1), 120]
(1) 100 3 (2) 50 2 (3) 100 (4) 50
3
13. The sum of all values of 0, satisfying sin2 2 + cos4 2 = is : 2 4
[JEE(Main) 2019, Online (10-01-19),P-1 (4, – 1), 120]
3 5
(1) (2) (3) (4)
2 8 4
14. The value of cos . cos . .... . cos . sin is :
2 2 2 3 210 210
[JEE(Main) 2019, Online (10-01-19),P-2 (4, – 1), 120]
1 1 1 1
(1) (2) (3) (4)
1024 2 512 256
15. If sin 4 4 cos 4 2 4 2 sin cos ; , [,0 ] , then cos( ) – cos( – ) is equal to
[JEE(Main) 2019, Online (12-01-19),P-2 (4, – 1), 120]
(1) – 2 (2) 0 (3) 2 (4) –1
ADVTRI- 39
MAASTrigonometry
© MAAS. All rights reserved.Page 40 / 47
Extracted text · page 40
Trigonometry
5 5 16. The number of solutions of the equation 1 + sin4x = cos2 3x, x , is 2 2
[JEE(Main) 2019, Online (12-04-19),P-1 (4, – 1), 120]
(1) 4 (2) 5 (3) 7 (4) 3
17. The value of sin10°sin30°sin50ºsin70° is : [JEE(Main) 2019, Online (09-04-19),P-2 (4, – 1), 120]
1 1 1 1
(1) (2) (3) (4)
18 32 16 36
sin2 x 2 sinx 5 118. All the pairs (x, y) that satisfy the inequality 2 . 1 also satisfy the equation :
sin2 y 4
[JEE(Main) 2019, Online (10-04-19),P-1 (4, – 1), 120]
(1) sin x = |siny| (2) 2|sin x| = 3 siny (3) sin x = 2 sin y (4) 2 sin x = 2 sin y
19. Two vertical poles of heights, 20 m and 80 m stand a part on a horizontal plane. The height (in meters)
of the point of intersection of the lines joining the top of each pole to the foot of the other, from this
horizontal plane is : [JEE(Main) 2019, Online (08-04-19),P-2 (4, – 1), 120]
(1) 18 (2) 16 (3) 15 (4) 12
20. Let and be two real roots of the (k+1) tan2 x – 2 . tanx = (1– k), where k(–1) and are real
numbers. If tan2( + ) = 50, then a value of is:
[JEE(Main) 2020, Online (07-01-20),P-1 (4, – 1), 120]
(1) 10 2 (2) 5 2 (3) 10 (4) 5
3 3 21. The value of cos3 . cos + sin3 . sin is :
8 8 8 8
[JEE(Main) 2020, Online (09-01-20),P-1 (4, –1), 120]
1 1 1 1
(1) (2) (3) (4)
2 2 2 4 2
22. The number of distinct solutions of the equation, log1/2|sinx| = 2 – log1/2 |cosx| in the interval [0, 2], is
[JEE(Main) 2020, Online (09-01-20),P-1 (4, 0), 120]
ADVTRI- 40
MAASTrigonometry
© MAAS. All rights reserved.Page 41 / 47
Extracted text · page 41
Trigonometry
EXERCISE - 1
PART - I
Section (A) :
4 53
A-1. (i) (ii) (iii)
12 3 18
A-2. (i) 135° (ii) – 720° (iii) 300° (iv) 210°
3 1 1
A-4. (i) (ii) – (iii) – (iv) 1 2 2 3
Section (B) :
B-1. (i)
(ii)
(iii)
1 2
13
B-2. 1 B-3. B-7. a2b2 + 4a2 = 9b2
12
Section (C)
5 1 3 1
C-2. (i) 1 (ii) – 5 / 4 (iii) C-10.
8 2 2
Section (E) :
1 1
E-1. – , E-2. (i) 2, –1 (ii) 3, 0
4 4
E-3. (i) ymax = 11; ymin = 1 (ii) ymax = 10; ymin = – 4
ADVTRI- 41
MAASTrigonometry
© MAAS. All rights reserved.Page 42 / 47
Extracted text · page 42
Trigonometry
Section (F) :
F-1. (i) n + (– 1)n , n (ii) n + + 1, n
4 3
(iii) n – , n (iv) n + (– 1)n , n (v) n ± , n
4 3 4
m (2m 1)F-2. (i) , m or , m
4 10
(ii) 2n ± ,n
3
1 (iii) 2n , n or 2n – , n 2 5 2
1 1 (iv) n , n (v) n , n 2 9 4 2
2
(vi) 2n + , n (vii) n ±
3 6
1 (viii) n , n 3 3
n 1 F-3. (i) , n or n , n 3 3
2n (ii) 2 n , n or + , n
3 6
(iii) x =(2n 1) , n or x =(2n 1) , n or x = n ± , n
4 2 6
m 1 (iv) m, m or , m or m , m n 1 2 n
F-4. (i) n + , n or n + , n
3 4
3 (ii) n + (– 1)n , n or n – (– 1)n , n
10 10
n (iii) x = , n
3 9
F-5. (i) n + + (– 1)n , n
6 4
3
(ii) 2n+ , n or 2n + 2 where = tan–1 ,n
2 7
ADVTRI- 42
MAASTrigonometry
© MAAS. All rights reserved.Page 43 / 47
Extracted text · page 43
Trigonometry
Section (G) :
7 11 G-1. x n ,n : n I G-2. 2n , 2n 4 4 6 6
G-3. (n/3, n + /2] G-4. 15 3 m
G-5. ab G-6. 50 m
PART - II
Section (A) :
A-1. (A) A-2. (A) A-3. (B) A-4. (C) A-5. (B)
Section (B) :
B-1. (A) B-2. (B) B-3. (A) B-4. (D) B-5. (B) B-6. (B) B-7. (A)
B-8. (A) B-9. (C) B-10. (A) B-11. (B) B-12. (C)
Section (C) :
C-1. (C) C-2. (A) C-3. (C) C-4. (D) C-5. (A) C-6. (B) C-7. (A)
C-8. (B) C-9. (D) C-10. (C)
Section (D) :
D-1. (A) D-2. (B) D-3. (D) D-4. (B) D-5. (A)
Section (E) :
E-1. (C) E-2. (C) E-3. (A) E-4. (B) E-5. (A)
Section (F) :
F-1. (D) F-2. (B) F-3. (D) F-4. (B) F-5. (C) F-6. (B) F-7. (A)
F-8. (B) F-9. (A) F-10. (B) F-11. (C) F-12. (B)
Section (G) :
G-1. (A) G-2. (C) G-3. (B) G-4. (A) G-5. (A) G-6. (D) G-7. (D)
G-8. (D) G-9. (B)
PART - III
1. (A) (s), (B) (s), (C) (s), (D) (r)
2. (A) (q, s), (B) (p), (C) (q), (D) (p)
3. (A) (q), (B) (s), (C) (r), (D) (p)
EXERCISE - 2
PART - I
1. (B) 2. (B) 3. (D) 4. (B) 5. (A) 6. (B) 7. (B)
8. (A) 9. (D) 10. (D) 11. (C) 12. (C) 13. (C) 14. (A)
15. (A) 16. (D) 17. (A) 18. (B)
ADVTRI- 43
MAASTrigonometry
© MAAS. All rights reserved.Page 44 / 47
Extracted text · page 44
Trigonometry
PART - II
1. 04.80 2. 00.71 3. 02.00 4. 00.08 5. 19.00 6. 08.29 or 08.30
7. 08.00 8. 17.00 9. 15.00 10. 20.00 11. 24.00 12. 02.00
13. 01.00 14. 50.00 15. 04.00 16. 25.13 17. 17.00 18. 17.00
19. 06.00 20. 15.00 21. 21.99 22. 04.45 23. 01.75 24. 06.00
25. 00.00 26. 00.00 27. 00.00
PART - III
1. (ABCD) 2. (BC 3. (BD) 4. (AB) 5. (BC) 6. (ABC)
7. (BC) 8. (ABCD) 9. (BCD) 10. (AC) 11. (CD) 12. (BD)
13. (BC) 14. (BC) 15. (BD) 16. (BD) 17. (AD) 18. (AB)
19. (AD) 20. (CD) 21. (AB) 22. (AC) 23. (ABCD) 24. (BD)
25. (ABC) 26. (AD) 27. (BC) 28. (BC) 29. (AC) 30. (ACD)
31. (ABCD) 32. (ABCD) 33. (AC) 34. (CD) 35. (AB) 36. (ABD)
37. (ABCD)
PART - IV
1. (C) 2. (B) 3. (D) 4. (D) 5. (C) 6. (B) 7. (C)
8. (D) 9. (D)
EXERCISE - 3
PART - I
1. 2 2. (n = 7) 3. 3 4. (D) 5. (ACD) 6. (D) 7. 8
8. (C) 9. (C) 10. Bonus 11. (0.5) 12. (ABC) 13. (4) 14. (3)
PART - II
1. (1) 2. (1) 3. (2) 4. (2) 5. (2) 6. (1) 7. (3)
8. (2) 9. (4) 10. (3) 11. (4) 12. (3) 13. (2) 14. (3)
15. (1) 16. (2) 17. (3) 18. (1) 19. (2) 20. (3) 21. (1)
22. 8
ADVTRI- 44
MAASTrigonometry
© MAAS. All rights reserved.Page 45 / 47
Extracted text · page 45
Trigonometry
1. Prove that :
cot 2 (sec 1) 1 sin
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1 (ii) = sec2 1 sin 1 sec .
2. Simplify the expression sin4 x 4cos2 x – cos 4 x 4sin2 x
3. Let a, b, c, d be numbers in the interval [0, ] such that sin a + 7 sin b = 4(sin c + 2sin d),
cos a + 7 cos b = 4(cos c + 2cos d). Prove that 2 cos (a – d ) = 7 cos (b – c ).
4. Prove that (4cos2 9º – 3) (4cos2 27º – 3) = tan 9º
4 55. If cos (+ ) = ; sin ( ) = & lie between 0 & , then find the value of tan 2 .
5 13 4
2ac
6. If & are two distinct roots of the equation a tan + b sec = c, then prove that tan (+ ) = .
a 2 c 2
p7. If tan = where = 6 being an acute angle, prove that;
q
1 2 2 (p cosec 2 q sec 2 ) = p q .
2
8. If sin () = a & sin () = b (0 <, ,</2) then find the value of
cos2 () 4 ab cos()
9. Show that:
1 1
(i) cot 7 or tan 82 = 3 2 2 1 or 2 3 4 6 2 2
1 (ii) tan 142 = 2 + 2 3 6 .
2
tan tan sin2 sin2 10. If tan = , prove that sin 2 = 1 tan . tan 1 sin2 .sin2 .
a 2 ac b 2
11. If & satisfy the equation a cos 2 + b sin 2 = c then prove that: cos² + cos² = .
a 2 b 2
12. Show that: 4 sin 27° = (5 5)1/ 2 (3 5)1/ 2
13. If xy + yz + xz = 1, then prove that
x y z 4xyz
+ + = .
1 x 2 1 y 2 1 z 2 (1 x 2 )(1 y 2 )(1 z 2 )
14. Let a =
7
(a) Show that sin2 3a – sin2 a = sin 2a sin 3a
(b) Show that cosec a = cosec 2a + cosec 4a
(c) Evaluate cos a – cos 2a + cos 3a
(d) Prove that cos a is a root of the equation 8x3 – 4x2 – 4x + 1 = 0
(e) Evaluate tan a tan 2a tan 3a
(f) Evaluate tan2 a + tan2 2a + tan2 3a
(g) Evaluate tan2 a tan2 2a + tan2 2a tan2 3a + tan2 3a tan2 a
(h) Evaluate cot2 a + cot2 2a + cot2 3a
ADVTRI- 45
MAASTrigonometry
© MAAS. All rights reserved.Page 46 / 47
Extracted text · page 46
Trigonometry
A B C A B C 15. In a ABC, prove that sin + sin + sin = 1 + 4 sin sin sin 2 2 2 4 4 4
16. Evaluate
2
cos a cos 2a cos 3a..........cos 999a, where a =
1999
17. Prove that the average of the numbers
2 sin 2º, 4 sin 4º, 6 sin 6º, .........180 sin 180º is cot 1º
2
18. Solve tan2 = tan .
19. Find the general values of x and y satisfying the equations
5 sin x cos y = 1, 4 tan x = tan y
3 x 3 x sin cos
2 2 cos x20. Solve = .
2 sin x 3
21. Solve the system of equations :
2 3 x + y = , sin x + sin y = and x, y 0, 3 2 2
22. Solve the following system of simultaneous equations for x and y:
4sinx + 31/cos y = 11
5.16sinx – 2.31/cos y = 2
23. Solve cos + sin = cos 2 + sin 2.
3 1
24. Solve 8 sin x = +
cosx sinx
25. Solve the equation sin3x cos 3x + cos3x sin 3x + 0.375 = 0.
3
26. Solve the equation sin x – cos x = cos2x.
2
3
27. Solve the equation sin4x + cos4x – 2 sin2x + sin22x = 0
4
28. Solve for x, the equation 13 18tanx = 6 tan x 3, where 2 < x < 2 .
29. Solve the equation 3 – 2cos – 4 sin – cos 2 + sin 2 = 0
30. Solve the equation sin2 4x + cos2x = 2 sin 4x . cos4x
31. Prove that : cos 5A = 16 cos5 A – 20 cos3 A + 5 cos A
1 1 1 k 1
32. If cos = a and cos 3 = a k then number of natural numbers 'k' less than 50 is 2 a 2 a
(given a R)
2 33. Consider the equation for 0 2 ; sin2 3 cos 2 5 = cos 2 . If greatest value of 6
k
is (k, p are coprime), then find the value of (k + p).
p
ADVTRI- 46
MAASTrigonometry
© MAAS. All rights reserved.Page 47 / 47
Extracted text · page 47
Trigonometry
562. cos2 x – sin2x = cos 2x 5. 8. 1 2a2 2b2
33
1
14. (c) (e) 7 (f) 21 (g) 35 (h) 5
2
1 n n2 2
16. 18. ± 1 , n
2999 4 16
20. x = (4 n + 1) , n 21. x = , y = or x = , y = .
2 2 6 6 2
2n22. x = n + (– 1)n , y = 2n ± 23. 2 n , n or + , n
6 3 3 6
n n 24. x = n + , n , x = – , n 25. x = + (–1)n+1 . ; n
6 2 12 4 24
1
, n 2 526. x = (2n + 1) : n , x = 2n + , n 27. x = n ± cos–1 3 2
2
28. 2 , , + , where tan = 29. = (4n + 1) /2, = 2n , n
3
30. x = (2 n + 1) , nI 32. 25 33. 31
2
ADVTRI- 47