Trigonometry 2

Chapter12

Trigonometry 12.1 Introduction

In Class IX, you have learnt some elementary concepts of trigonometry which included trigonometric ratios, finding out other trigonometric ratios if one trigonometric ratio is given, trigonometric ratios of some specific angles and use of trigonometric ratios in solving right triangles. In this chapter, we will discuss trigonometric ratios in more details, learn some fundamental identities and use these identities in proving other trigonometric identities. 12.2 Trigonometric Identities Consider any acute angle AOB (Fig.12.1). Let us take a point P on the ray OB and drop the perpendicular PQ on OA. Thus, we have a right triangle PQO. Let ∠ POQ be G. B

We recall that sin G =

PQ PQ , OQ , cos G = tan G = OQ , OP OP

P

OP 1 OP 1 cosec G = PQ = sin θ , sec G = OQ = cos θ , q

OQ 1 cot G = PQ = tan θ

and

O

PQ PQ OQ sin θ tan G = OQ = OP ÷ OP = cos θ .

Q

A Fig. 12.1

Further, in the right triangle PQO, we have OQ 2 + PQ 2 = OP 2

or

OQ 2 OP 2

+

PQ 2 OP 2

=

OP 2 OP 2

[Using Pythagoras Theorem] =1

[Dividing by OP2]

(1)

TRIGONOMETRY................................................................................................................................................ 211

 OQ 2  PQ 2   +  =1  OP   OP 

or or

(cos G)2 + (sin G)2 = 1

or

cos2 G + sin2 G = 1

(2)

Similarly, dividing both sides of (1) by OQ2, we have OQ OQ

2 2

+

PQ

2

OQ

2

=

OP

2

OQ

2

or

 PQ   OP  1+   =   OQ   OQ 

or

1+ (tan G)2 = (sec G)2

or

1 + tan2 G = sec2 G

2

2

(3)

Again dividing both sides of (1) by PQ2, we get 2

 OQ   OP    + 1 =   PQ    PQ 

2

(cot G)2 + 1 = (cosec G)2 cot2 G + 1 = cosec2 G

i.e.,

(4)

Relations (2), (3) and (4) found above are all identities. Each of these identities can be derived from the other e.g., identities (3) and (4) can be derived from identity (2) on dividing it by cos2 G and sin2 G respectively. (2), (3) and (4) are called fundamental identities. Sometimes, these are also referred as the Pythagorean identities. These might have been so named because of the first step, i.e., using Pythagoras theorem to obtain (1) above. We can use the fundamental identities to express each trigonometric ratio in terms of any other trigonometric ratio. For example, we can express sin G in terms of cos G, tan G in terms of sin G, for 0° < G < 90° and so on, as given below : cos2 G + sin2 G = 1, sin2 G = 1 – cos2 G sin G = ± 1 − cos 2 θ ∴

sin G = 1 − cos 2 θ

[sin G > 0 for acute angles]

212.................................................................................................................................................. MATHEMATICS

Similarly, sec2 G = 1 + tan2 G Therefore,

sec G = 1 + tan 2 θ

or

tan G =

sec θ − 1

cosec G =

1 + cot 2 θ

Also,

cot G =

or Further,

tan G =

2

cosec 2 θ − 1 sin θ = cos θ

sin θ

[For 0° < G < 90°, using Identity (2)]

1 − sin 2 θ

Similarly, we can find other relations expressing one trigonometric ratio in terms of other trigonometric ratios which are given in the following table : Table 12.1 sin G

cos G

sin G

sin G

1 − cos2 θ

cos G

1 − sin θ

cos G

sinθ

1 − cos 2 θ

2

tan G

tan G

1 − sin 2 θ

cos θ

cot G

1 − sin 2 θ sin θ

cos θ

sec G

cosec G

1 1 − sin 2 θ 1 sin θ

1 − cos θ 2

tan θ

cosec G

1

sec2 θ − 1 sec θ

1 cosec θ

1 sec θ

cosec 2 θ − 1

1 + cot θ

1

cot θ

1 + tan 2 θ

1 + cot 2 θ

2

tan G

1 cotθ

1 tan θ

cot G

1 + tan θ

1

1 + tan 2 θ

2

1 − cos θ

sec G

1 + tan θ 2

1 cos θ

2

cot G

tan θ

1 + cot 2 θ cot θ

1 + cot 2 θ

cosec θ 1

sec θ − 1 2

1 sec 2 θ − 1

sec G sec θ sec 2 θ − 1

cosec 2 θ − 1

cosec2 θ − 1 cosec θ cosec 2 θ − 1

cosec G

TRIGONOMETRY................................................................................................................................................ 213

Fundamental identities are also used to simplify expressions involving trigonometric ratios and proving other identities as illustrated in the following examples : Example 1: Simplify the expression sin G (cosec G – sin G). Solution : We have  1  sin G (cosec G – sin G) = sin G  sin θ − sin θ   

= 1 – sin2 G = cos2 G

[Using Identity (2)]

Example 2: Simplify : (sec G + tan G) (1 – sin G) Solution : We have  1

sin θ 

(sec G + tan G) (1 – sin G) =  cos θ + cos θ  (1 − sin θ )    1 + sin θ 

=  cos θ  (1 − sin θ)   =

1 − sin 2 θ cos θ

=

cos 2 θ = cos θ cos θ

[Using Identity (2)]

Example 3: Prove that 2

2

cos θ + tan θ − 1 2

sin θ

Solution : 2

LHS =

2

cos θ + tan θ − 1 2

sin θ tan θ 2

= =

sin 2θ

2

= tan θ .

sin θ

2

=

2

2

tan θ + cos θ − 1 2

sin θ

=

(

2

)

tan θ − 1 − cos θ 2

sin θ

2

−

sin 2θ

sin 2 θ 1 −1 . 2 cos θ sin 2 θ

[Using Identity (2)]

214.................................................................................................................................................. MATHEMATICS

= sec2 G – 1 = tan2 G = RHS

[Using Identity (3)]

Example 4: Prove that cot G + tan G = cosec G sec G. Solution : cos θ

sin θ

LHS = cot G + tan G = sin θ + cos θ cos 2 θ + sin 2 θ = sin θ cosθ 1

= sin θ cosθ 1

[Using Identity (2)]

1

= sin θ . cos θ = cosec G sec G = RHS Example 5 : Show that tan4A + tan2A = sec4A – sec2A. Solution : We have LHS = tan4A + tan2A = tan2 A (tan2A + 1) = (sec2A – 1) (sec2A) = sec4A – sec2A = RHS

[Using Identity (3)]

Example 6 : Prove the identity: tan 2 θ tan 2 θ − 1

+

cosec 2 θ sec 2 θ − cosec 2 θ

=

1 sin 2 θ − cos 2 θ

Solution : The right hand side has only sin G and cos G. This suggests us to change all the trigonometric ratios on the left hand side to sin G and cos G. We have sin 2 θ 2

tan θ 2

tan θ − 1

=

cos2 θ sin 2 θ cos2 θ

−1

=

sin 2 θ sin 2 θ − cos 2 θ

(1)

TRIGONOMETRY................................................................................................................................................ 215

1 cosec θ

2 sin θ

2

=

2 2 sec θ − cosec θ

1 cos2 θ

−

=

1 sin 2 θ

cos 2 θ

(2)

2 2 sin θ − cos θ

Adding (1) and (2), we get 2

=

2

tan θ

LHS =

2

tan θ − 1

+

2

2

2

2

2

cosec θ 2

2

sec θ − cosec θ

sin θ + cos θ sin θ − cos θ

=

1 sin θ − cos2 θ 2

2

sin θ

=

2

2

sin θ − cos θ

+

cos θ 2

2

sin θ − cos θ

= RHS

Example 7: Prove the identity: cosec A cosec A + = 2 sec 2 A cosec A +1 cosec A − 1

Solution : We have cosec A



cosec A

1

1



LHS = cosec A − 1 + cosec A + 1 = cosec A  cosec A − 1 + cosec A + 1    cosec A + 1 + cosec A − 1 

= cosecA  (cosec A − 1)(cosec A + 1)   =

cosec A . (2 cosec A ) cosec A − 1 2

= 2.

1 2

sin A

.

sin 2A 2

cos A

=

=

2 cosec 2 A cot 2 A

2 cos2A

= 2 sec 2 A

= RHS EXERCISE 12.1 1 + tan A 2

1.

Simplify :

2.

Prove that (cosec G + cot G) (1 – cos G) = sin G.

1 + cot A 2

[Using Identity (4)]

216.................................................................................................................................................. MATHEMATICS 2

tan θ

= sec θ − 1 .

3.

Prove that

4.

Using trigonometric identities, write the following expressions as an integer: (i) (ii) (iii) (iv)

5.

sec θ + 1

3 cot2 A – 3 cosec2 A 4 tan2 A – 4 sec2 A 5 sin2 G + 5 cos2 G 7 sec2 B – 7 tan2 B

Simplify the following expressions: 3

(i)

3

sin θ + cos θ sin θ + cos θ

2 − tan θ (ii) 2 cosec θ − sec θ

Prove the following identities : 6.

cos θ + sin θ = 1 + cot θ sin θ

7.

cos 2 θ − sin 2 θ = cot θ − tan θ sin θ cos θ

8.

9.

sin 4 θ − cos 4 θ sin 2 θ − cos 2 θ sec 2 θ − sin 2 θ tan θ 2

=1

= cosec θ − cos 2 θ 2

10.

1 1 + = 2sec 2 θ 1 − sin θ 1 + sin θ

11.

tan 3 θ − 1 = sec 2 θ + tan θ tan θ − 1

TRIGONOMETRY................................................................................................................................................ 217

12.

(1 + sin θ) (1 − sin θ) = sec12θ

13.

sin6 G + cos6 G = 1 – 3 sin2 G . cos2 G

14.

1 + cos2A 2

= 2 cosec 2 A − 1

sin A

15.

tan 2 φ − sin 2 φ = tan 2 φ sin 2 φ

16.

sec φ − 1 1 − cos φ = sec φ + 1 1 + cos φ

17.

1 − cos θ = (cosec θ − cot θ)2 1 + cos θ

18.

1 + sin A = sec A + tan A 1 − sin A

19.

1 − cos A = cosec A − cot A 1 + cos A

20.

1 + cos A sin A 2 + = sin A 1 + cos A sin A

21.

sin θ + cos θ sin θ − cos θ 2 2 + = = 2 sin θ − cos θ sin θ + cos θ 1 − 2 cos θ 2 sin 2 θ − 1

22.

(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1

23.

2 cos2 θ − 1 = cot θ − tan θ sin θ cos θ

218.................................................................................................................................................. MATHEMATICS

24.

(sec A – cosA) (cot A + tan A) = tan A sec A

sec θ + tan θ − 1

=

1 + sin θ

25.

tan θ − sec θ + 1

26.

cot θ cot θ − 1 = 1 + tan θ 2 − sec2 θ

27.

tan θ cot θ + = 1 + tan θ + cot θ 1 − cot θ 1 − tan θ

28.

tan2 φ + cot2 φ + 2 = sec2 φ cosec2 φ

29.

cos θ 1 + sin θ + = 2 sec θ cos θ 1 + sin θ

30.

tan A tan A + = 2cosec A sec A − 1 sec A + 1

12.3

cos θ

Trigonometric Ratios of Complementary Angles

Consider a right triangle PQO (Fig. 12.2). Let ∠ POQ = G. Then ∠ QPO = 90° – G. Clearly, angles POQ and QPO are complementary angles. We know that PQ OQ PQ  , cos θ = , tan θ = OP OP OQ   OP OP OQ  cosec θ = , sec θ = , cot θ = PQ OQ PQ 

P

sin θ =

(1)

Let us now consider the angle QPO, which is equal to 90° – G and its trigonometric ratios. Here, adjacent side is PQ and opposite side is OQ. So we have: OQ sin (90° – G) = = cos G OP

[Using (1) above]

Similarly, cos (90° – G) =

PQ = sin G OP

90°– θ

θ

Q

Fig. 12.2

O

TRIGONOMETRY................................................................................................................................................ 219

tan (90° – G) =

OQ PQ

= cot G

cot (90° – G) =

PQ = tan G OQ

sec (90° – G) =

OP = cosec G PQ

cosec (90° – G) =

OP = sec G OQ

Thus, sin (90° – θ) = cos θ , for 0°< θ < 90° cos (90° – θ) = sin θ

Also, sin 0° = 0 = cos 90° and sin 90° = 1 = cos 0° Therefore, sin (90° − θ ) = cos θ  , for 0° ≤ θ ≤ 90° cos (90° − θ) = sin θ

Similarly, tan (90° – G) = cot G, for 0° < G ≤ 90° cot (90° – G) = tan G, for 0° ≤ G < 90° sec (90° – G) = cosec G, for 0° < G ≤ 90° cosec (90° – G) = sec G, for 0° ≤ G < 90° You may note that tan 0° = 0 = cot 90°, sec 0° = 1 = cosec 90° and tan 90°, sec 90° etc. are not defined. Note : We have seen that the sine of any angle = the cosine of its complement, the tangent of any angle = the cotangent of its complement, the secant of any angle = the cosecant of its complement, and vice-versa. Perhaps, because of this, the word ‘co’ has been prefixed with sine, tangent and secant to obtain cosine, cotangent and cosecant.

220.................................................................................................................................................. MATHEMATICS

Example 8: Evaluate

sin 10° . cos 80°

[As cos G = sin (90°– G)]

Solution : cos 80° = sin (90°–80°) = sin 10° sin 10° sin 10° =1 cos 80° = sin 10°

Therefore,

Example 9 : Express sin 81° + tan 71° in terms of trigonometric ratios of angles between 0° and 45°. Solution : sin 81° + tan 71° = cos (90° – 81°) + cot (90° – 71°) = cos 9° + cot 19° Example 10 : Find G if sin (G + 36°) = cos G, where G + 36° is an acute angle. Solution : sin (G + 36°) = cos G (Given) or cos [90° – (G + 36°)] = cos G ∴ 90° – (G + 36°) = G or 2G = 54° or G = 27° Example 11 : If tan 2G = cot (G + 6°), where 2G and G + 6° are acute angles, find the value of G. Solution : We have: tan 2G = cot (G + 6°) or cot (90° – 2G) = cot (G + 6°) ∴ 90° – 2G = G + 6° or 3G = 84° or G = 28° EXERCISE 12.2 cos 59° . sin 31°

1.

Evaluate

2.

Find the value of : (i)

3.

tan 49° cot 41°

Show that

(ii)

cosec 39° sec 51°

= 1.

cot 50° tan 40°

TRIGONOMETRY................................................................................................................................................ 221

4.

Express cos 75° + cot 75° in terms of trigonometric ratios of angles between 0° and 45°.

5.

Evaluate : (i) sin2 20° + sin2 70°

6.

(ii)

cos 2 20° + cos 2 70° sin 2 59° + sin 2 31°

Evaluate :  sin 27° 2  cos 63° 2   +       cos 63°   sin 27° 

7.

Evaluate : cosec2 67° – tan2 23°

Prove that : 8.

cos (90° − θ ) sin θ + = 2, θ ≠ 0° sin θ cos (90° − θ )

9.

sec2 G – cot2 (90° – G) = cos2 (90° – G) + cos2 G

10.

cos (90° − θ ) cos θ + cos 2 (90° − θ) = 1 tan θ

11.

cos (81° + G) = sin (9° – G)

12.

cos 20° cos θ + =2 sin 70° sin (90° − θ)

13.

sin (90° – G) . cos (90° – G) =

14.

sin G cos (90° – G) + cos G sin(90° – G) = 1

15.

cos G cos (90° – G) – sin G sin(90° – G) = 0

16.

cot (90° – G) sin(90° – G) = sin G

tan θ 1 + tan 2 θ

222.................................................................................................................................................. MATHEMATICS

17.

tan 15° tan 20° tan 70° tan 75° = 1

18.

Evaluate : 3 sin 62° sec 42° − cos 28° cosec 48°

19.

If sin 3G = cos G – 6°), where 3G and G – 6°) are acute angles, find the value of G.

20.

If A and B are acute angles and sin A = cos B, prove that A + B = 90°.

21.

If A, B, C are the interior angles of a triangle ABC, show that B+C A = cos . 2 2 [Hint : Sum of the angles of a triangle is 180°.] sin