TRIGONOMETRY, HEIGHTS AND DISTANCES
rigonometric functions their periodicity and graphs, addition and subtraction formulae, formulae involving
multiple and submultiple angles.
Definition of Trigonometric Functions In a right angled triangle ABC, CAB = A and BCA = 90° = /2. AC is the base, BC the altitude and AB is
the hypotenuse. We refer to the base as the adjacent side and to the altitude as the opposite side. There are
six trigonometric ratios, also called trigonometric functions or circular functions. With reference to angle A,
the six ratios are:
hypotenuse
side opposite
AB
BC
is called the sine of A, and written as sin A.
hypotenuse
side adjacent
AB
AC
is called the cosine of A, and written as cos A.
side adjacent
side opposite
AC
BC
is called the tangent of A, and written as tan A.
B
CA Fig. 2
Obviously, cosA
sinA= Atan . The reciprocals of sine, cosine and tangent are called the cosecant, secant and
cotangent of A respectively. We write these as cosec A, sec A, cot A respectively. Since the hypotenuse is
the greatest side in a right angle triangle, sin A and cos A can never be greater than unity and cosec A and
sec A can never be less than unity. Hence sin A 1, cos A 1,cosec A 1, sec A1, while tan A
and cot A may have any numerical value.
Notes:
(i) All the six trigonometric functions have got a very important property in common that is periodicity.
(ii) Remember that the trigonometrical ratios are real numbers and remain same as long as angle A is
real.
Basic Formulae
(i) cos2A + sin
2A = 1
cos2A = 1 – sin
2A or sin
2A = 1 – cos
2A
(ii) 1 + tan2A = sec
2A
sec2A – tan
2A = 1
(iii) cot2A + 1 = cosec
2A
cosec2A – cot
2A = 1
(iv) Asin
AcosAcotand
Acos
AsinAtan
Illustration 1: (i) Express tan in terms of cosec .
(ii) Express sin in terms of sec .
T
Solution: tan = 1
cot
1 + cot2 = cosec
2
cot2 = cosec
2 – 1
cot = 2cosec 1
cot = 2
1
cosec 1
sin = 21 cos
= 2
11
sec
=
2 2
2
sec 1 sec 1
secsec
Illustration 2: ABC has a right angle at A and AB = AC = 5cm., find sin B, cot B and tan C.
Solution: BC2 = 5
2 + 5
2 = 50
BC = 50 5 2
sin B = AC 5 1
BC 5 2 2
cot B = AB 5
1AC 5
tan C = AB 5
1AC 5
5
5
5 5
C A
B
Illustration 3: If tan =5
12, show that tan
2 – sin
2 = tan
2. sin
2
Solution: tan = 5
12
BC = 5k AB = 12k AC
2 = (5k)
2 + 12k)
2
= 25k2 + 144k
2
= 169k2
AC = 13k sin = 5k 5
13k 13
13k 5k
12k
A B
C
Now, tan2 – sin
2 =
2 25 5 25 25 25(169 144) 625
12 13 144 169 144 169 144 169
........ (i)
tan2 . sin
2 =
2 25 5 25 25 625
.12 13 144 169 144 169
......... (ii)
From (i) and (ii) tan2 – sin
2 = tan
2 . sin
2
Illustration4: Show that
1 + sinθ
1 sinθ = sec + tan
Solution: 1+ sinθ 1+ sinθ
21 sin θ
1+ sinθ
cosθ
1 sinθ
+cosθ cosθ
sec + tan
Trigonometric Ratios of any Angle
Consider the system of rectangular co-ordinate axes dividing the plane into four quadrants.
A line OP makes angle with the positive x-axis. The angle is said to be positive if measured in counter
clockwise direction from the positive x-axis and is negative if measured in clockwise direction. The positive
values of the trigonometric ratios in the various quadrants are shown, the signs of the other ratios may be
derived. Note that xoy = 2
, xox’ = , xoy’ =
2
3.
Oxx
y
y
P1
Q1
P4P3
P2
Q2
Q3 Q4
Oxx
y
y
quadrant I (A)
All ratio + ve
quadrant IV (C)
cos, sec + ve
quadrant II (S)
sin, cosec + ve
quadrant III (T)
tan, cot + ve
PiQi is positive if above the x-axis, negative if below the x-axis, OPi is always taken as positive. OQi is
positive if along the positive x-axis, negative if in the opposite direction.
i ii i
i
PQsin Q OP
OP , i i
i i
i
O Qcos Q OP
OP , i i
i i
i
PQtan Q OP
OQ ( i = 1, 2, 3 ).
Thus depending on signs of OQi and PiQi the various trigonometrical ratios will have different signs.
Table – I
equals sin cos tan cot sec cosec
– –sin cos –tan – cot sec –cosec
90° – cos sin cot tan cosec sec
90° + cos –sin –cot – tan –cosec sec
180°– sin –cos –tan – cot –sec cosec
180°+ –sin –cos tan cot –sec –cosec
360°– –sin cos – tan – cot sec –cosec
360°+ sin cos tan cot sec cosec
Note:
(i) Angle and 90° – are complementary angles, and 180° – are supplementary angles
(ii) sin(n + (– 1)n) = sin , n
(iii) cos(2n ± ) = cos , n
(iv) tan(n + ) = tan , n
i.e. sine of general angle of the form n + (–1)n will have same sign as that of sine of angle and so
on. The same is true for the respective reciprocal functions also.
Trigonometric Ratios of Compound Angles: An angle made up of the algebraic sum of two or more angles is called a compound angle. Some of the
formulae and results regarding compound angles are:
sin(A + B) = sin A cos B + cos A sin B
sin(A – B) = sin A cos B – cos A sin B
cos(A + B) = cos A cos B – sin A sin B
cos(A – B) = cos A cos B + sin A sin B
BtanAtan1
BtanAtanBAtan
BtanAtan1
BtanAtanBAtan
Illustration 5: Prove that 0 0
0
tan50 cot50
tan10
= 2.
Solution: tan 500 = tan(10
0 + 40
0) =
0 0
0 0
tan10 + tan40
1 tan10 .tan40
or, tan 500 (1– tan 10
0 tan 40
0) = tan 10
0 + tan 40
0
or, tan 500 = tan 50
0 tan 10
0 tan 40
0 + tan 10
0 + tan 40
0
= cot 400 tan 10
0 tan 40
0 + tan 10
0 + tan 40
0
= 2 tan 10
0 + tan 40
0
tan 500 = 2 tan 10
0 + cot 50
0
0 0
0
tan50 cot50
tan10
= 2.
Illustration 6: If A – B = 45, show that (1 + tan A) (1 – tan B) = 2.
Solution: tan(A – B) =tanA tanB
=11+ tanAtanB
tan A – tan B – tan A tan B + 1= 1 + 1
tan A(1 – tan B) + (1 – tan B) = 2
(1 + tan A) (1 – tan B) = 2.
Illustration 7: Find the value of
0 0
0 0
3sin5 2cos33+
cos85 sin57
Solution:
0 0
0
3sin5 2cos33
cos 90 5 cos33
0 0
0 0
3sin5 2cos33+
sin5 cos33
3 + 2 = 5
Exercise 1: Show that :
(i) sin (A + B) sin (A – B) = sin2 A – sin
2 B = cos
2 B – cos
2 A
(ii) cos (A + B) cos (A – B) = cos2 A – sin
2B = cos
2 B – sin
2 A
Trigonometric Ratios of Multiples of an Angle
sin 2A = 2sin A cos A = 2
2tanA
1+ tan A
cos 2A = cos2A – sin
2A = 1 – 2sin
2A = 2cos
2A – 1 =
2
2
1 tan A
1+ tan A
tan 2A = 2
2tanA
1 tan A
sin 3A = 3sin A – 4sin3A = 4sin(60° – A) sin A sin(60° + A)
cos 3A = 4cos3A – 3cosA = 4cos(60° – A) cos A cos(60° + A)
3
o o
2
3tanA tan Atan3A = = tan 60 A tanAtan 60 + A
1 3tan A
Illustration 8: Find the values of (i) cos 72°, (ii) tan15°.
Solution: (i) sin18
Let = 18° then 2 = 36° = 90° – 54° = 90° – 3.
Now sin 2 = 2sin cos and
sin(90° – 3) = cos 3 = 4cos3 – 3cos .
Hence we have 2sin cos = cos (4cos2 – 3) = cos (1 – 4sin
2).
Hence, 2 sin = 1 – 4sin2 (as cos 0)
4sin2 + 2sin – 1 = 0 sin =
2 ± 4 +16 1± 5=
2.4 4
But as sin > 0 we have sin = 5 1
4
i.e. sin 18° =
5 1
4
cos 72° = sin (90 – 72) = sin 18° = 5 1
4
(ii) tan 15°
Let = 15° 2 = 30°
2
2tanθ 1tan2θ = = tan30° =
1 tan θ 3
tan2 + 23 tan – 1 = 0 tan =
2 3 ± 12+ 4 2 3 ± 4= = 3 ± 2
2 2
Since tan > 0, ignore negative value
so that tan = 2 – 3 i.e. tan 15° = 2 – 3
Alternative solution:
tan 150 = tan (60
0 – 45
0)
=
2
3 13 1 4 2 3= = = 2 3
3 1 21+ 3
.
Sum of sines/cosines in Terms of Products
2
BAcos
2
BAsin2BsinAsin
2
BAcos
2
BAsin2BsinAsin
2
BAcos
2
BAcos2BcosAcos
2
B+Asin
2
ABsin2BcosAcos
(here notice (B – A)!)
tan A + tan B =
Bcos.Acos
BAsin
Conversely:
2sin A cos B = sin(A + B) + sin (A – B)
2cos A sin B = sin(A + B) – sin (A – B)
2cos A cos B = cos(A + B) + cos(A – B)
2sin A sin B = cos(A – B) – cos (A + B)
Illustration 9: If cos . sin = 1
2 then = ……..
Solution: Given cos sin =1
2
2 cos sin = 1
sin 2 = 1
sin 2 = 1 = sin 900
2 = 900
= 450
Illustration 10: Evaluate : cos4 – sin
4
Solution: cos4 – sin
4 = 1 (sin
2 + cos
2) (–sin
2 + cos
2)
1 (–sin2 + cos
2)
cos2 – sin
2 = cos2
Illustration 11: (i) Find the harmonic mean of sin2 and cos
2 .
(ii) The value of sin2 30
0, sin
2 45
0, sin
2 60
0 are in….. progression.
Solution: (i) H.M. =2 2
2 2
2.sin θ.cos θ
sin θ + cos θ
=2 22sin θcos θ
1
= 2 sin2 cos
2 =
2sin 2θ
2
(ii) sin230
0 =
21 1
2 4
sin245
0 =
21 1
=22
sin260
0 =
2
3 3=
2 4
1 1 3
, ,4 2 4
…. are in A.P.
Illustration 12: (i) Find the value of sin 4200….
(ii) Find the value of cos 2400?
(iii) Find the value of sin 11100
Solution: sin 4200 = sin (360
0 + 60
0)
= sin 600 =
3
2
cos 2400 = cos (180 + 60)
= –cos 600 =
1
2
sin 11100 = sin [360
0 3 + 30
0]
= sin 300 =
1
2
Exercise 2: Prove that:
(i) sin 3 = 3 sin – 4 sin3 (ii) sin
3 =
3sinθ sin3θ
4
(iii) cos 3 = 4 cos3 – 3 cos (iv) cos
3 =
3cosθ + cosθ
4
(v) 1 cosθ
sinθ
= tan
θ sinθ=
2 1 + cosθ
HEIGHTS AND DISTANCES
DEFINITIONS
Angle of elevation:
If ‘O’ be the observer’s eye and OX be the horizontal line
through O. If the object P is at a higher level than eye, then
angle POX = is called the angle of elevation
Line of sight
Horizontal Line
P
XO
Angle of Depression:
If the object P is at a lower level than O, then angle POX is called
the angle of depression.
Line of sight
Horizontal Line
P
X
O
Illustration 1: A ladder leaning against a vertical wall is inclined at an angle to the horizontal. On moving its
foot 2 m away from the wall, the ladder is now inclined at angle . Find the vertical distance
moved by the ladder.
Solution: From ABC
l = 2 sec …(1)
x + d = 2 tan …(2)
From CDE
x = l sin …(3)
From (1) and (3)
x = 2 sec sin Putting in (2)
2 sec sin + d = 2 tan
d = 2 tan – 2 sec sin
d = 2sin 2sin
cos cos
=
2
cos(sin – sin )
D
A
l
C
l
B
E
x
d
2 m
Illustration 2: A boat is being rowed away from a cliff 150 meter height. At the top of the cliff the angle of
depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat.
Solution : Let PQ be the cliff PQ = 150 meters.
Let the two position of the boat be at A and B
so that AP = h cot60 = 150/3
BP = 150 cot45° = 150
AB = 150 – 150 150
=3 3
(3 – 1)
PB A
Q
60°45°
The speed of the boat is2
1.150
3(3 – 1) 60 metres per hour.
= 4500
3(3 – 1) meters per hour.
Exercise 1:
(i) A person standing on the bank of a river observers that the angle subtended by a tree on the opposite bank is 60°. When he retires 40 feet from the bank, he finds the angle to be 30°. Find the height of the tree and the breadth of the river.
(ii) The angular depressions of the top and the foot of a chimney as seen from the top of a second
chimney which is 150 meters high and standing on the same level as the first are and
respectively. Find the distance between their tops, when tan = 4/3, tan = 5/2.
ASSIGNMENTS
SUBJECTIVE
LEVEL – I
1. (i) Express the sexagesimal measures as radii (circular) measure.
1. 150 2. 135
0 3. 270
0 4. 360
0
(ii) Express the circular measure in sexagesimal measure (1)
c
6
(2)
c3
4
(3)
c3
5
.
2. The radius of a circle is 14 cm. The angle subtended by an arc of the circle at the centre is
450. Find the length of the arc.
3. Given sin (A – B) =1
2; cos (A + B) =
1
2. Find A and B.
4. Given that: 8 tan = 15, find sin - cos for 00 < < 90
0
5. Show that: 1+ cosθ
1 cosθ = cos + cot
6. Prove that 1+ cosec + cot
1+ cosec cot
=
cosec + cot 1
cot cosec +1
.
7. Prove that (sin + cosec )2 + (cos + sec )
2 = tan
2 + cot
2 + 7
8. Prove that1 1 1 1
=cosecθ + cotθ sinθ sinθ cosecθ cotθ
.
9. Prove that cos (tan + 2) (2 tan + 1) = 2 sec + 5 sin .
10. From the top of a building 60 m high the angles of depression of the top and the bottom of a
tower are observed to 300 and 60
0. Find the height of the tower.
LEVEL – II
1. Prove that tan tan
sec 1 sec 1
= 2 cosec .
2. Prove that sin2A tan A + cos
2A cot A + 2 sin A cos A = tan A + cot A
3. Prove that tan cot
1 cot 1 tan
= sec cosec + 1.
4. Find the length of the side of a regular polygon inscribed in a circle of radius 1 m, if it has 6
sides.
5. The angles of elevation of the top of a hill at the city centres of two towns on either side of the hill are
observed to be 310 and 62
0. If the distance uphill from the first city centre is 9 km, find in kilometers,
the distance uphill from the other city centre correct upto two places of decimals.
OBJECTIVE
LEVEL – I
1. If p cot = 2 2q p , then the value of sin is
(A) p
q (B)
q
2p (C)
p
3q (D) none of these.
2. If p = r cos and q = r sin , then the value of 2 2
2
p + q
r is
(A) 0 (B) 1 (C) r2tan
2 (D) none of these.
3. If sec = 13
5, then the value of
2
2 3cotα
4 9 sec 1
is
(A) 2
13 (B)
1
21 (C)
15
352 (D) none of these
4. The value of cot230
0 – 2 cos
2 60
0
3
4 sec
2 45
0 – 4 sin
2 30
0 is
(A) 0 (B) 1 (C) –1 (D) none of these
5. tan sec 1
tan sec 1
= ……………..
(A) cos
1 sin
(B)
1 cos
sin
(C)
1 sin
cos
(D) none of these.
6. 2 2
2 2
tan A sec A
cos B cot B = ……….
(A) cot2 A – cot
2B (B) tan
2A = tan
2B (C)
2
2
tan A
sec A (D) none of these.
7. If tan + sin = m & tan – sin = n then m2 – n
2 = ………
(A) mn (B) m
n (C) 4 mn (D) none of these.
8. If tan A = n tan B and sin A = m sin B then 2
2
m 1
n 1
= ………….
(A) sin2A (B)
2
2
n 1
m 1
(C)
2
2
m 1
n 1
(D) cos
2A
9. If 7 sin2 + 3 cos
2 = 4 then sec + cosec = …………
(A) 2
23 (B)
22
3 (C)
2
3 (D) none of these.
10. A person walking 20 m towards a chimney in a horizontal line through its base observes that its
angle of elevation changes from 300 to 45
0. The height of chimney is ……
(A)20
3 1 (B)
20
3 1 (C) 20 3 1 (D) none of these
LEVEL – II
1. If sin (A + B) =3
2; cos B =
3
2 value of A is
(A) 450 (B) 60
0 (C) 30
0 (D) 90
0
2. 2
2
tan . 1 sin
1 cos
=
(A) sin (B) cos (C) sec (D) 1
3. If 3 sin + 5 cos = 5 then 5 sin – 3 cos =
(A) 3 (B) –3 (C) both A and B (D) none of these
4. If tan A = n tan B and sin A = m sin B then 2
2
m 1
n 1
= ……….
(A) sin2A (B)
2
2
n 1
m 1
(C)
2
2
m 1
n 1
(D) cos
2A
5. The value of sec 700 sin 20
0 + cos20
0 cosec 70
0 = ……….
(A) 1 (B) 0 (C) 2 (D) none of these
6. The value of tan 10 tan 2
0 tan 3
0…… tan 89
0 = ………
(A) 1 (B) 0 (C) –1 (D) none of these
7. A harbor lies in a direction 600 south of west, and at a distance 30 km from A, a ship sets out from
the harbor at noon and sails due east at 10 km per hour. The ship will be 70 km away from the fort at
(A) 7 P.M. (B) 8 P.M. (C) 5 P.M. (D) 10 P.M.
8. From the top of a lighthouse 60 meters high with its base at the sea level, the angle of depression of
about is 150. The distance of the boat from the foot of the lighthouse is
(A)3 1
3 1
. 60 meters (B)
3 1
3 1
. 60 meters (C)
3 1
3 1
meters (D) none of these
9. A pole of 50 meter high stands on a building and the pole subtend equal angles. The distance of the
observer from the top of the pole is
(A) 25 6 m (B) 50 m (C) 25 3 m (D) 25 m
10. The height of a house subtends a right angle at the opposite street light. The angle of elevation of
light from the base of the house is 60o. If the width of the road be 6 meters, then the height of the
house is
(A) 8 3 m (B) 8 m (C) 6 m (D) 6 3 m
ANSWERS
SUBJECTIVE
LEVEL – I
1. I (1) c
12
(2)
c3
4
(3)
c
(4) 2
c
II (1)1800 (2) 30
0 (3) 108
0
3. 50 4.
7
17
5. cosec + cot 10. h = 40 metres
LEVEL – II
4. 11 cms
5. 5.25 km(correct upto two places of decimals)
OBJECTIVE
LEVEL – I 1. A 2. B 3. C 4. A
5. C 6. B 7. C 8. D
9. B 10. B
LEVEL – II
1. C 2. D 3. C 4. C
5. C 6. A 7. B 8. B
9. A 10. A