mathsvol1ch1to3ans

51


15 15
cos 2 cos 2

15 2 2 1 + cos 15
cot = = =
2 15 15 15 sin 15
sin 2 sin cos
2 2 2
√ √ √ √
1 3 1 1 3 + 1 6 + 2
cos 15 = cos(45 − 30) = cos 45 cos 30 + sin 45 sin 30 = √ + √ = √ =
2 2 2 2 √ 2 2 √ 4 √
√
1 3 1 1 3 − 1 6 − 2
sin 15 = sin(45 − 30) = sin 45 cos 30 − cos 45 sin 30 = √ − √ = √ =
2 2 2 2 2 2 4
√ √
!
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6 + 2
1 + √ √
4
15 1 + cos 15 4 + 6 + 2
cot = = √ √ ! = √ √
2 sin 15 6 − 2 6 − 2
4
√ √ √ √ √ √ √
√ √ √ √ √
15 4 + 6 + 2 6 + 2 4 6 + 6 + 2 3 + 4 2 + 2 3 + 2
cot = = = 6 + 4 + 3 + 2
2 4 4
n
n
10. Prove that (1 + sec 2θ)(1 + sec 4θ) . . . (1 + sec 2 θ) = tan 2 θ cot θ.
√ π π π π π
11. Prove that 32( 3) sin cos cos cos cos = 3.
48 48 24 12 6
Exercise 3.6:
1. Express each of the following as a sum or difference

◦
◦
(i) sin 35 cos 28 (ii) sin 4x cos 2x (iii) 2 sin 10θ cos 2θ (iv) cos 5θ cos 2θ (v) sin 5θ sin 4θ.
Solution:
1 1
◦
◦
◦
◦
◦
(i) sin 35 cos 28 ◦ = {sin(35 + 28) + sin(35 − 28) } = {sin(63) + sin(7) }
2 2
1 1
(ii) sin 4x cos 2x = {sin(4x + 2x) + sin(4x − 2x)} = {sin 6x + sin 2x}
2 2
(iii) 2 sin 10θ cos 2θ = sin(10θ + 2θ) + sin(10θ − 2θ) = sin 12θ + sin 8θ
1 1
(iv) cos 5θ cos 2θ = {cos(5θ + 2θ) + cos(5θ − 2θ)} = {cos 7θ + cos 3θ}
2 2
1 1
(v) sin 5θ sin 4θ = {cos(5θ − 4θ) + cos(5θ + 4θ)} = − {cos 9θ − cos θ}
2 2
2. Express each of the following as a product
◦
◦
◦
◦
◦
◦
(i) sin 75 − sin 35 ◦ (ii) cos 65 + cos 15 (iii) sin 50 + sin 40 ◦ (iv) cos 35 − cos 75 .
Solution:

75 − 35 75 + 35
◦
◦
(i) sin 75 − sin 35 ◦ = 2 sin cos = 2 sin 20 cos 55 ◦
2 2

65 + 15 65 − 15
◦
◦
◦
(ii) cos 65 + cos 15 = 2 cos cos = 2 cos 40 cos 25 ◦
2 2

50 + 40 50 − 40
◦
◦
(iii) sin 50 + sin 40 ◦ = 2 sin cos = 2 sin 45 cos 5 ◦
2 2

75 + 35 75 − 35
◦
◦
◦
(iv) cos 35 − cos 75 = 2 sin sin = 2 sin 55 sin 20 ◦
2 2

52

1
◦
◦
◦
3. Show that sin 12 sin 48 sin 54 = .
8
Solution:

1
sin 12 (sin 48 cos 36) = (cos 36 − cos 60) cos 36
2

1 1
2
= cos 36 − cos 36
2 4
√ √
   
! 2
1  5 + 1 1 5 + 1 1
=   −  =
4 4 4 8
2 
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π 2π 3π 4π 5π 6π 7π 1
4. Show that cos cos cos cos cos cos cos = .
15 15 15 15 15 15 15 128
Solution:
π
cos π cos 2π cos 3π cos 4π cos 5π cos 6π cos 7π = cos π cos 2π cos 3π cos 4π cos cos 6π cos 7π
15 15 15 15 15 15 15 15 15 15 15 3 15 15
= cos π cos 2π cos 3π cos 4π 1 cos 6π cos π − 8π
15 15 15 15 2 15 15
1
= − cos π cos 2π cos 4π cos 8π cos 3π cos 6π
2 15 15 15 15 15 15
= − 1 sin 2π sin 4π sin 8π sin 16π 3π cos 6π
15
15
8π cos
15
15
2 2 sin π 2 sin 2π 2 sin 4π 2 sin 15 15
15 15 15 15
= − 1 sin 16π π cos 3π cos 2 3π
15
4
2 2 sin 15 15
15
sin(π+ ) sin 6π sin 12π
π
1 15 15 15
= − 5 π 3π 6π
2 sin 2 sin 2 sin
15 15 15
sin(π− )
3π
= 1 15
2 7 sin 3π
15
= 1
128
sin 8x cos x − sin 6x cos 3x
5. Show that = tan 2x.
cos 2x cos x − sin 3x sin 4x
Solution:
sin 8x cos x − sin 6x cos 3x 2 sin 8x cos x − 2 sin 6x cos 3x
=
cos 2x cos x − sin 3x sin 4x 2 cos 2x cos x − 2 sin 3x sin 4x
sin 9x + sin 7x − sin 9x − sin 3x
=
cos 3x + cos x − cos x + cos 7x
sin 7x − sin 3x
=
cos 3x + cos 7x
2 sin 2x cos 5x
=
2 cos 5x cos 2x
= tan 2x
(cos θ − cos 3θ) (sin 8θ + sin 2θ)
6. Show that = 1.
(sin 5θ − sin θ) (cos 4θ − cos 6θ)
Solution:
(cos θ − cos 3θ) (sin 8θ + sin 2θ) (2 sin 2θ sin θ) (2 sin 5θ cos 3θ)
= = 1
(sin 5θ − sin θ) (cos 4θ − cos 6θ) (2 sin 2θ cos 3θ) (2 sin 5θ sin θ)
7. Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x).
Solution:

53

sin 2x + sin x + sin 3x = sin 2x + 2 sin 2x cos x = sin 2x (1 + 2 cos x)


sin 4x + sin 2x
8. Prove that = tan 3x.
cos 4x + cos 2x

Solution:
sin 4x + sin 2x 2 sin 3x cos x
= = tan 3x
cos 4x + cos 2x 2 cos 3x cos x

9. Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x.
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Solution:
2
1 + cos 6x + cos 2x + cos 4x = 2 cos 3x + 2 cos 3x cos x
= 2 cos 3x (cos 3x + cos x)

= 2 cos 3x (2 cos 2x cos x)

= 4 cos x cos 2x cos 3x

θ 7θ 3θ 11θ
10. Prove that sin sin + sin sin = sin 2θ sin 5θ.
2 2 2 2
Solution:

θ 7θ 3θ 11θ
2 sin sin + 2 sin sin = (cos 3θ − cos 4θ) + (cos 4θ − cos 7θ)
2 2 2 2
= cos 3θ − cos 7θ

= 2 sin 5θ sin 2θ

θ 7θ 3θ 11θ
Hence sin sin + sin sin = sin 2θ sin 5θ
2 2 2 2
1
◦
◦
◦
◦
11. Prove that cos (30 − A) cos (30 + A) + cos (45 − A) cos (45 + A) = cos 2A + .
4
Solution:
◦
◦
◦
◦
◦
◦
2 cos (30 − A) cos (30 + A) + 2 cos (45 − A) cos (45 + A) = (cos 60 + cos 2A) + (cos 90 + cos 2A)
1
= + 2 cos 2A
2
1
◦
◦
◦
◦
Hence cos (30 − A) cos (30 + A) + cos (45 − A) cos (45 + A) = + cos 2A
4
sin x + sin 3x + sin 5x + sin 7x
12. Prove that = tan 4x.
cos x + cos 3x + cos 5x + cos 7x
Solution:
sin x + sin 7x + sin 3x + sin 5x 2 sin 4x cos 3x + 2 sin 4x cos x
=
cos x + cos 7x + cos 3x + cos 5x 2 cos 4x cos 3x + 2 cos 4x cos x
2 sin 4x (cos 3x + cos x)
=
2 cos 4x (cos 3x + cos x)
= tan 4x

54

sin (4A − 2B) + sin (4B − 2A)
13. Prove that = tan (A + B).
cos (4A − 2B) + cos (4B − 2A)
Solution:

sin (4A − 2B) + sin (4B − 2A) 2 sin(A + B) cos(3A − 3B)
= = tan (A + B)
cos (4A − 2B) + cos (4B − 2A) 2 cos(A + B) cos(3A − 3B)
4 cos 2A
◦
◦
14. Show that cot (A + 15 ) − tan (A − 15 ) = .
1 + 2 sin 2A
Solution:
◦
◦
cos (A + 15 ) sin (A − 15 )
◦
◦
cot (A + 15 ) − tan (A − 15 ) = −
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◦
◦
sin (A + 15 ) cos (A − 15 )
◦
◦
◦
◦
cos (A + 15 ) cos (A − 15 ) − sin (A − 15 ) sin (A + 15 )
=
◦
◦
sin (A + 15 ) cos (A − 15 )
cos 2A
=
◦
◦
sin (A + 15 ) cos (A − 15 )
2 cos 2A
=
◦
◦
2 sin (A + 15 ) cos (A − 15 )
2 cos 2A
=
sin 2A + sin 30 ◦
4 cos 2A
=
1 + 2 sin 2A
Exercise 3.7:
◦
1. If A + B + C = 180 , prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Solution:
sin 2A + sin 2B + sin 2C = sin 2A + 2 sin(B + C) cos(B − C)
= 2 sin A cos A + 2 sin A cos(B − C)
= 2 sin A (cos A + cos(B − C))

A + B − C A − B + C
= 4 sin A cos cos
2 2

π − 2C π − 2B
= 4 sin A cos cos
2 2
= 4 sin A sin C sin B
A B C
(ii) cos A + cos B − cos C = −1 + 4 cos cos sin
2 2 2
Solution:

55


A + B A − B
cos A + cos B − cos C = 2 cos cos − cos C
2 2

C A − B C
= 2 sin cos − 1 + 2 sin 2
2 2 2

C A − B C
= 2 sin cos + sin − 1
2 2 2

C A − B A + B
= 2 sin cos + cos − 1
2 2 2

C A B
= 2 sin 2 cos cos − 1
2 2 2
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2
2
2
(iii) sin A + sin B + sin C = 2 + 2 cos A cos B cos C
Solution:
2
2
2
2 sin A + 2 sin B + 2 sin C = 3 − cos 2A − cos 2B − cos 2C
= 3 − cos 2A − (2 cos(B + C) cos(B − C))
2
= 3 − 2 cos A + 1 − 2 cos A cos(B − C)
= 4 − 2 cos A (cos A + cos(B − C))

= 4 − 2 cos A (cos(B + C) + cos(B − C))

= 4 + 2 cos A (2 cos B cos C)



2
2
2
(iv) sin A + sin B − sin C = 2 sin A sin B cos C
Solution:
2
2
2
2 sin A + 2 sin B − 2 sin C = 1 − cos 2A − cos 2B + cos 2C
= 1 − cos 2A + (2 sin(B + C) sin(B − C))
2
= 1 + 2 sin A − 1 + 2 sin A sin(B − C)

= 2 sin A (sin A + sin(B − C))

= 2 sin A (sin(B + C) + sin(B − C))

= 2 sin A (2 sin B cos C)



A B B C C A
(v) tan tan + tan tan + tan tan = 1
2 2 2 2 2 2
Solution:
A + B + C = π

A B C π
+ + =
2 2 2 2
A B π C
+ = −
2 2 2 2

56


A B π C
tan + = tan −
2 2 2 2
A B
tan + tan
2 2 = cot C
A B 2
1 − tan tan
2 2
C A C B A B
tan tan + tan tan = 1 − tan tan
2 2 2 2 2 2
A B B C C A
tan tan + tan tan + tan tan = 1
2 2 2 2 2 2
A B C
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(vi) sin A + sin B + sin C = 4 cos cos cos
2 2 2
Solution:

B + C B − C
sin A + sin B + sin C = sin A + 2 sin cos
2 2
A A A B − C
= 2 sin cos + 2 cos cos
2 2 2 2

A A B − C
= 2 cos sin + cos
2 2 2

A B + C B − C
= 2 cos cos + cos
2 2 2

A B C
= 2 cos 2 cos cos
2 2 2
(vii) sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4 sin A sin B sin C.

Solution:
sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = sin(π − 2A) + sin(π − 2B) + sin(π − 2C)

= sin 2A + sin 2B + sin 2C

= 2 sin A cos A + 2 sin(B + C) cos(B − C)
= 2 sin A cos A + 2 sin(A) cos(B − C)

= 2 sin A (cos A + cos(B − C))

= 2 sin A (− cos(B + C) + cos(B − C))

= 2 sin A (2 sin B sin C)

= 4 sin A sin B sin C

2. If A + B + C = 2s, then prove that sin(s − A) sin(s − B) + sin s sin(s − C) = sin A sin B.

Solution:

2 sin(s − A) sin(s − B) + 2 sin s sin(s − C)

= (cos(B − A) − cos(C)) + (cos(C) − cos(A + B))
= cos(B − A) − cos(B + A)

= 2 sin B sin A

57

2x 2y 2z 2x 2y 2z
3. If x + y + z = xyz, then prove that + + = .
2
2
1 − x 2 1 − y 2 1 − z 2 1 − x 1 − y 1 − z 2
Solution: Let x = tan A, y = tan B, z = tan C.
tan A + tan B + tan C = tan A tan B tan C

tan A + tan B = − tan C{1 − tan A tan B}
tan A + tan B
= − tan C
1 − tan A tan B
tan (A + B) = tan (−C)
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A + B + C = π

2A + 2B + 2C = 2π
tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C


2 tan A 2 tan B 2 tan C 2 tan A 2 tan B 2 tan C
+ + =
2
2
2
2
2
2
1 − tan A 1 − tan B 1 − tan C 1 − tan A 1 − tan B 1 − tan C
2x 2y 2z 2x 2y 2z
+ + =
2
2
1 − x 2 1 − y 2 1 − z 2 1 − x 1 − y 1 − z 2
π
4. If A + B + C = , prove the following
2
(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
(ii) cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B cos C.
Solution:
sin 2A + sin 2B + sin 2C = sin 2A + 2 sin(B + C) cos(B − C)
π
= 2 sin A cos A + 2 sin( − A) cos(B − C)
2
= 2 sin A cos A + 2 cos A cos(B − C)
= 2 cos A (sin A + cos(B − C))

= 2 cos A (cos(B + C) + cos(B − C))

= 2 cos A (2 cos B cos C)
= 4 cos A cos B cos C


cos 2A + cos 2B + cos 2C = cos 2A + 2 cos(B + C) cos(B − C)

2
= 1 − 2 sin A + 2 sin A cos(B − C)
= 1 − 2 sin A (sin A − cos(B − C))

= 1 − 2 sin A (cos(B + C) − cos(B − C))
= 1 + 2 sin A (2 sin B sin C)

= 1 + 4 sin A sin B sin C


π
5. If 4ABC is a right triangle and if ∠A = , then prove that
2
2
2
(i) cos B + cos C = 1

58

Solution:


2
2
(ii) sin B + sin C = 1
√ B C
(iii) cos B − cos C = −1 + 2 2 cos sin .
2 2
Exercise 3.8:
1. Find the principal solution and general solutions of the following:
1 √ 1
(i) sin θ = −√ (ii) cot θ = 3 (iii) tan θ = −√ .
2 3
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Solution:
1
(i) sin θ = −√ < 0 principal value lies in the IV quadrant
2
1 π π
sin θ = −√ = − sin θ = nπ + (−1) n − . n ∈ Z
2 4 4
√ 1
(ii) cot θ = 3 ⇒ tan θ = √ > 0 principal value lies in the I quadrant
3
1 π π
tan θ = √ = tan θ = nπ + n ∈ Z
3 6 6
1
(iii) tan θ = −√ < 0 principal value lies in the IV quadrant
3
1 π π
tan θ = −√ = − tan θ = nπ − n ∈ Z
3 6 6
◦
2. Solve the following equations for which solutions lies in the interval 0 ≤ θ < 360 ◦
4
2

2
2
2
4
(i) sin x = sin x sin x − sin x = 0 ⇒ sin x sin x − 1 = 0
sin (x) = 1, sin (x) = −1, sin (x) = 0
π 3π
sin (x) = 1 ⇒ x = : sin (x) = −1 ⇒ x = : sin (x) = 0 ⇒ x = 0 x = π
2 2 1
2
2
(ii) 2 cos x + 1 = −3 cos x 2 cos x + 3 cos x + 1 = 0 ⇒ cos (x) = − , cos (x) = −1
2
1 2π 4π
cos (x) = − ⇒ x = , : cos (x) = −1 ⇒ x = π
2 3 3
1
2
2
(iii) 2 sin x + 1 = 3 sin x 2 sin x − 3 sin x + 1 = 0 ⇒ sin (x) = 1, sin (x) =
2
π 1 π 5π
sin (x) = 1 ⇒ x = : sin (x) = ⇒ x = :
2 2 6 6
(iv) cos 2x = 1 − 3 sin x.

3
2
1 − 2 sin x = 1 − 3 sin x ⇒ sin x sin x − ⇒ sin x = 0 ⇒ x = 0, π
2
3. Solve the following equations:
(i) sin 5x − sin x = cos 3x
2 sin 2x cos 3x = cos 3x ⇒ cos 3x(2 sin 2x − 1) = 0
π π
When cos 3x = 0, 3x = (2n + 1) and hence x = (2n + 1)
2 6

59

1 π π π
when sin 2x = , 2x = (−1) n + nπ and hence x = (−1) n + n
2 6 12 2

2
(ii) 2 cos θ + 3 sin θ − 3 = 0
2
2(1 − sin θ) + 3 sin θ − 3 = 0
2
2 sin θ − 3 sin θ + 1 = 0
sin (θ) = 1, sin (θ) = 1
2
π

When sin (θ) = 1 ⇒ θ = nπ + (−1) n
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2
1
π
When sin (θ) = ⇒ θ = nπ + (−1) n
2 6
(iii) cos θ + cos 3θ = 2 cos 2θ
2
3
cos θ + 4 cos θ − 3 cos θ = 4 cos θ − 2
2
3
4 cos θ − 4 cos θ − 2 cos θ + 2 = 0
3
2
2 cos θ − 2 cos θ − cos θ + 1 = 0
2
2(cos θ)(cos θ − 1) − 1(cos θ − 1) = 0
2
(2 cos θ − 1)(cos θ − 1) = 0
1
cos θ = ±√ or cos θ = 1
2
π
Hence θ = 2nπ ± or θ = 2nπ
4
(iv) sin θ + sin 3θ + sin 5θ = 0

sin θ + sin 5θ + sin 3θ = 0

2 sin 3θ cos 2θ + sin 3θ = 0

sin 3θ(2 cos 2θ + 1) = 0

1
Hence sin 3θ = 0 or cos 2θ = −
2
π π
Solution is θ = nπ or 2θ = 2nπ ± ⇒ θ = nπ ± .
6 3

(v) sin 2θ − cos 2θ − sin θ + cos θ = 0

We have (sin 2θ − sin θ) − (cos 2θ − cos θ) = 0

3θ θ θ 3θ
Hence 2 sin cos + 2 sin sin = 0
2 2 2 2

3θ θ θ
sin cos + sin = 0
2 2 2

θ θ
cos + sin = 0
2 2

60


θ θ
Divide both sides by cos , cos 6= 0
2 2

θ θ
cos + sin
2 2 0
=

θ θ
cos cos
2 2

θ
tan + 1 = 0
2

θ
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tan + 1 − 1 = 0 − 1
2

θ
tan = −1
2
θ 3π
= + πn
2 4
3π
θ = + 2nπ
2
π

cos θ = 0 ⇒ θ = (2n + 1)
2
√
(vi) sin θ + cos θ = 2

1
Multiplying both sides by √
2
1 1
√ sin θ + √ cos θ = 1
2 2
π π
cos sin θ + sin cos θ = 1
4 4
π π

sin + θ = sin
4 2
π
θ =
4
π
Hence the solution is θ = nπ + (−1) n
√ 4
(vii) sin θ + 3 cos θ = 1

1
Multiplying both sides by
2
1 1 √ 1
sin θ + 3 cos θ =
2 2 2
π π π
cos sin θ + sin cos θ = sin
3 3 6
π π

sin + θ = sin
3 6
π
θ = −
3
π
Hence the solution is θ = nπ + (−1) n
3

61
√
(viii) cot θ + cosecθ = 3

Multiplying both sides by sin θ, We get
√
cos θ + 1 = 3 sin θ
√ 1
cos θ − 3 sin θ = −1 Multiplying both sides by
2
1 1 √ 1
cos θ − 3 sin θ = −
2 2 2
π π 2π
cos cos θ − sin sin θ = cos
3 3 3
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π 2π

cos θ + = cos
3 3
π
θ =
3
π
Hence the solution is θ = 2nπ ±
√

3
(ix) tan θ + tan θ + π + tan θ + 2π = 3
3 3

π 2π

tan θ + tan θ + + tan θ + = 3
3 3
π  2π 
 
tan θ + tan tan θ + tan
 = 3
tan θ +  3 π  +  3 

2π
1 − tan θ tan 1 − tan θ tan
√ 3 √ 3
! !
tan θ + 3 tan θ − 3
tan θ + √ + √ = 3
1 − 3 tan θ 1 + 3 tan θ
√ √ √ √
!
(1 + 3 tan θ)(tan θ + 3) + (1 − 3 tan θ)(tan θ − 3)
tan θ + √ √ = 3
(1 − 3 tan θ)(1 + 3 tan θ)
√ √ √ √
!
2
2
tan θ + 3 + 3 tan θ + 3 + tan θ − 3 − 3 tan θ + 3
tan θ + = 3
2
1 − 3 tan θ

8 tan θ
tan θ + = 3
2
1 − 3 tan θ
2
3
tan θ − 3 tan θ + 8 tan θ = 3 − 9 tan θ
3
2
3 tan θ − 9 tan θ − 9 tan θ + 3 = 0
3
2
tan θ − 3 tan θ − 3 tan θ + 1 = 0
2
3
tan θ + 1 − 3 tan θ − 3 tan θ = 0
2
(tan θ + 1)(tan θ − tan θ + 1) − 3 tan θ(tan θ + 1) = 0
2
(tan θ + 1)(tan θ − 4 tan θ + 1) = 0
√ √
tan θ = −1 tan θ = 2 + 3 tan θ = 2 − 3
π √ √
−1
−1
θ = nπ − θ = nπ − tan (2 + 3) θ = nπ − tan (2 − 3)
4

62
√
5 + 1
(x) cos 2θ =
4
◦
cos 2θ = cos 36 .
π
◦
θ = 18 =
5
π
θ = 2nπ ±
5
2
(xi) 2 cos x − 7 cos x + 3 = 0 (2 cos (x) − 1) (cos (x) − 3) = 0
1
cos(x) = or cos(x) = 3
2
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π
Since cos(x) = 3 has no solution,the only solution is x = + 2πn
3
Exercise 3.9:
sin A sin(A − B)
2
2
2
1. In a 4ABC, if = , prove that a , b , c are in Arithmetic Progression.
sin C sin(B − C)
Solution:
sin A sin(B − C) = sin C sin(A − B)
sin(B + C) sin(B − C) = sin(A + B) sin(A − B)

2
2
2
2
sin B − sin C = sin A − sin B
2
2
2
2
2
2
R (b − c ) = R (a − b )
2
2b 2 = a + c 2
2
2
2
Hence a , b , c are in Arithmetic Progression.
√ √
2. The angles of a triangle ABC, are in Arithmetic Progression and if b : c = 3 : 2, find ∠A.
Solution:
Since ∠A, ∠B, ∠C are in A.P, we have 2B = A + C.
π π 2π
But A + B + C = π ⇒ 2B + B = π. That is, B = . Hence A + C = π − =
3 3 3
sin B sin C
From Sine Rule, = .
b c

2π
sin − A
sin B 3
We have =
b c

63


2π c
sin − A = sin B
3 b
√
!
2
π
= √ sin
3 3
√ √
!
2 3
= √
3 2
1
= √
2
π

= sin
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4
2π π
− A =
3 4
8π − 5π π
A = =
12 4
sin A
3. In a 4ABC, if cos C = , show that the triangle is isosceles.
2 sin B
Solution:

2 sin B cos C = sin A
sin(B + C) + sin(B − C) = sin A

sin A + sin(B − C) = sin A

sin(B − C) = 0

B − C = 0

B = C
sin B c − a cos B
4. In a 4ABC, prove that = .
sin C b − a cos C
Solution:
c − a cos B 2R sin C − 2R sin A cos B
=
b − a cos C 2R sin B − 2R sin A cos C
2 sin C − sin(A + B) − sin(A − B)
=
2 sin B − sin(A + C) − sin(A − C)
2 sin C − sin C − sin(A − B)
=
2 sin B − sin B − sin(A − C)
sin C − sin(A − B)
=
sin B − sin(A − C)
sin(A + B) − sin(A − B)
=
sin(A + C) − sin(A − C)
2 sin B cos A
=
2 sin C cos A
sin B
=
sin C
5. In a 4ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.

64

Solution:
a cos A + b cos B + c cos C = 2R sin A cos A + 2R sin B cos B + 2R sin C cos C Using Sine Rule

= R (sin 2A + sin 2B + sin 2C)

= R (2 sin(A + B) cos(A − B) + sin 2C)

= R (2 sin C cos(A − B) + 2 sin C cos C)

= 2R sin C (cos(A − B) − cos(A + B))
= 2R sin C (2 sin A sin B)
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= 2R sin A sin C sin B

= a sin C sin B

B − C
◦
6. In a 4ABC, ∠A = 60 . Prove that b + c = 2a cos .
2
Solution:
b + c (sin B + sin C)
= R
2a 2R sin(A)

B + C B − C
2 sin cos
2 2
=
A A
4 sin cos
2
2
A B − C
cos cos
2 2
=
A A
2 sin cos
2 2
B − C
cos
= 2
2 sin(30) ◦
B − C
b + c = 2a cos
2
7. In a 4ABC, prove the following

A A
(i) a sin + B = (b + c) sin
2 2
Solution:
B + C B − C A B − C
2 sin cos 2 cos cos
b + c k sin B + k sin C sin B + sin C 2 2 2 2
= = = =
a k sin A sin A A A A A
2 sin cos 2 sin cos
2 2 2 2

A B − C B + C A π A
(b + c) sin = a cos = a cos B − = a cos B + − = a sin B +
2 2 2 2 2 2
A
(ii) a(cos B + cos C) = 2(b + c) sin 2
2
Solution:

65

a k sin A
=
b + c k sin B + k sin C
A A
2 sin cos
= 2 2
B + C B − C
2 sin cos
2 2
A A
2 sin cos
= 2 2
A B − C
2 cos cos
2 2
A
sin
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= 2
B − C
cos
2
A B + C
sin cos
= 2 2
B + C B − C
cos cos
2 2
A B + C
2 sin cos
= 2 2
B + C B − C
2 cos cos
2 2
A B + C
2 sin cos
= 2 2
cos B + cos C
A
2 sin 2
= 2
cos B + cos C

A
Hence a(cos B + cos C) = 2(b + c) sin 2
2
2
a − c 2 sin(A − C)
(iii) =
b 2 sin(A + C)
Solution:
2
a − c 2 (a + c)(a − c)
=
b sin(A − C) b sin(A − C)
2
2
R 2 sin A − sin C
=
R sin(A + C) sin(A − C)
2
2
sin A − sin C
= R
2
2
sin A − sin C
= R
2
a − c 2 sin(A − C)
Hence = Rb sin(A − C) =
b 2 sin(A + C)
a sin(B − C) b sin(C − A) c sin(A − B)
(iv) = =
2
2
2
b − c 2 c − a 2 a − b 2
Solution:

66

a sin(B − C) R sin A sin(B − C)
=
2
2
2
b − c 2 R 2 sin B − sin C
sin(B + C) sin(B − C)
=
R sin(B + C) sin(B − C)
1
=
R
b sin(C − A)
a sin(B−C)
Because of symmetricity, we can easily show that = =
2
b −c 2 c − a 2
2
c sin(A − B) 1
= .
2
a − b 2 R
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a + b A + B A − B
(v) = tan cot .
a − b 2 2
Solution:
a + b R (sin A + sin B)
=
a − b R (sin A − sin B)
A + B A − B
2 sin cos
= 2 2
A − B A + B
2 sin cos
2 2
A + B A − B
   
sin cos
2 2
   
A + B A − B
=    
cos sin
2
2
A + B A − B
= tan cot
2 2
2
2
2
2
2
2
8. In a 4ABC, prove that (a − b + c ) tan B = (a + b − c ) tan C.
Solution:
tan B sin B cos C 2 sin B cos C
= =
tan C cos B sin C 2 cos B sin C
2
2
a + b − c 2
sin B
= 2ab
2
2
a + c − b 2
sin C
2ac
2
2
2
c sin B (a + b − c )
=
2
2
2
b sin C (a + c − b )
2
2
a + b − c 2
=
2
2
a + c − b 2
9. An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park
to be developed must be of maximum area. Find out the dimensions of the park.
Solution:
For a fixed perimeter, the equilateral triangle has the maximum area and the maximum area
s 2
is given by ∆ = √ sq.units. Here s = 120 m. Hence the area of the park is given by
3 3
√
14400 1600 3
∆ = √ = Clearly the sides of the park are equal and given by 40 m.
3 3 .

67



10. A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find
the dimensions of the triangle so formed.

Solution:

Considering the length of rope as perimeter,the largest area of the triangle formed by this rope is
s 2
an equilateral triangle. The largest area formed is given by ∆ = √ sq.units. Here s = 12 m.
3 3
144 √
Hence the area is given by ∆ = √ = 16 3. Clearly the sides are equal and given by 4 m.
3 3
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11. Derive Projection formula from (i) Law of sines, (ii) Law of cosines.

Solution:
Projection Formula gives the relation between angles and sides of a triangle. We can find the length
of a side of the triangle if other two sides and corresponding angles are given using projection
formula. If a, b and c be the length of sides of a triangle and A, B and C are angles opposite to the
sides respectively, then projection formula is given below:
a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A

Let us prove the first one using law of cosines.
2
2
2
a = b + c + 2bc cos(A)
2
2
= b + c + 2bc cos(B + C)
2
2
= b + c + 2bc (cos B cos C − sin B sin C) − (b sin C − c sin B) 2
2
2
2
2
2
2
= b + c + 2bc cos B cos C − 2bc sin B sin C − b sin C − c sin B + 2bc sin B sin C
2
2
2
2
2
2
= b + c + 2bc cos B cos C − b sin C − c sin B

2
2

= b 2 1 − sin C + c 2 1 − sin B + 2bc cos C cos B
2
2
2
2
= b cos C + c cos B + 2bc cos C cos B
= (b cos C + c cos B) 2
Taking square root on both sides, the result follows. Let us prove using Law of Sines
a cos B + b cos A = R sin A cos B + R sin B cos A
= R sin(A + B)
= R sin C
= c
Exercise 3.10:

68

1. Determine whether the following measurements produce one triangle, two triangles or no triangle:
◦
∠B = 88 , a = 23, b = 2. Solve if solution exists.
Solution:
◦
Let h = a sin B = 23 sin 88 = 23(0.99939082701) = 22.9859890214..
Since b < h triangle is not possible.
2. If the sides of a 4ABC are a = 4, b = 6 and c = 8, then show that 4 cos B + 3 cos C = 2.

Solution:
2
2
a + c − b 2 16 + 64 − 36
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cos B = = = 0.6875
2ac 64
2
2
a + b − c 2 16 + 36 − 64
cos C = = = −0.25
2ab 48
Now,4 cos B + 3 cos C = 4(0.6875) − 3(0.25) = 2
√ √
◦
3. In a 4ABC, if a = 3 − 1, b = 3 + 1 and C = 60 , find the other side and other two angles.
Solution:
√ 2 √ 2 √ √
1
2
2
2
c = a + b − 2ab cos C = 3 − 1 + 3 + 1 − 2 3 − 1 3 + 1
√ √ 2
= 3 + 1 − 2 3 + 3 + 1 + 2 3 − (3 − 1) = 6
. Hence c = 2.
√ √ √
a sin C 3 3 − 1 3 − 1
◦
sin A = = √ = √ = sin 15 .
c 2 6 2 2
◦
◦
B = 180 − (A + C) = 105 .
2
2
b + c − a 2
4. In any 4ABC, prove that the area 4 = .
4 cos A
Solution:
1
Area of the triangle = bc sin A
2
2bc sin A
=
4
2
2
b + c − a 2
= sin A
4 cos A
◦
5. In a 4ABC, if a = 12 cm, b = 8 cm and C = 30 , then show that its area is 24 sq.cm.
Solution:
1 1
◦
Area of the Triangle = ab sin C = (12)(8) sin 30 = 24 sq.cm.
2 2

6. In a 4ABC, if a = 18 cm, b = 24 cm and c = 30 cm, then show that its area is 216 sq.cm.

Solution:
p p
Area of the triangle = s(s − a)(s − b)(s − c) = 36(18)(12)(6) = 216 sq.cm.
7. Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at
the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the