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