Compound and Multiple Angles Class 11 OP Malhotra Exe-5B ISC Maths Solutions Ch-5 Solutions. In this article you would learn about Converting Products into Sums or Differences and Vice Versa. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Compound and Multiple Angles Class 11 OP Malhotra Exe-5B ISC Maths Solutions Ch-5
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-5 | Compound and Multiple Angles |
| Writer | OP Malhotra |
| Exe-5(B) | The Product Formulae. |
Exercise- 5B
Compound and Multiple Angles Class 11 OP Malhotra Exe-5B Solution.
Que-1: Convert the following products into sum or difference. If angles are given in degrees, evaluate from tables.
(i) 2 sin 48° cos 12°
(ii) 2 sin 54° sin 66°
(iii) 2 cos 5θ cos 3θ
(iv) 2 cos 72° sin 56°
(v) cos (α + ß) cos (α – ß)
(vi) sin {(A+B)/2} cos {(A-B)/2}
Sol: (i) 2 sin 48° cos 12°
= sin (48° + 12°) + sin (48° – 12°) [∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= sin 60° + sin 36°
= (√3/2) + sin36°
= 1.45382
(ii) 2 sin 54° sin 66°
= cos (66° – 54°) – cos (66° + 54°)
= cos 12° – cos 120°
= cos 12° – cos (180° – 60°)
= cos 12° + cos 60° = 0.9781 + 0.5
= 1.4781 [∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
(iii) 2 cos 5θ cos 3θ
= cos (5θ + 3θ) + cos (5θ – 3θ)
= cos 8θ + cos 2θ [∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
(iv) 2 cos 72° sin 56°
= sin (72° + 56°) – sin (72° – 56°) [∵ 2 cos A sin B = sin (A + B) – sin (A – B)]
= sin 128° – sin 16°
= 0.7880 – 0.2756
= 0.5124
(v) cos (α + ß) cos (α – ß)
= (1/2) [2 cos (α + ß) cos (α – ß)]
= (1/2) [cos (α + ß + α – ß) + cos (α + ß – α + ß)]
= (1/2) [cos 2α + cos 2ß]
(vi) sin {(A+B)/2} cos {(A-B)/2}
= (1/2) [2 sin {(A+B)/2} cos{(A-B)/2}]
= (1/2) [sin{(A+B)2}+{(A-B)/2}] + [sin{(A+B)/2} – {(A-B)/2}]
= (1/2) (sin A + sin B)
Que-2: Convert the following sums or differences into products :
(i) sin 12A + sin 4A
(ii) sin 37° + sin 21°
(iii) sin 12A – sin 4A
(iv) cos 79° + cos 11°
(v) cos 12α + cos 8α
(vi) cos 25° – cos 37°
(vii) sin 61° – cos 39°
(viii) sin 4x + cos 2x
Sol: (i) sin 12A + sin 4A
= 2 sin {(12A+4A)/2} cos {(12A-4A)/2}
= 2 sin 8A cos 4A
(ii) sin 37° + sin 21°
= 2 sin {(37°+21°)/2} cos {(37°-21°)/2}
= 2 sin 29° cos 8°
(iii) sin 12A – sin 4A
= 2 cos {(12A+4A)/2} sin {(12A-4A)/2}
= 2 cos 8A sin 4A
(iv) cos 79° + cos 11°
= 2 cos {(79°+11°)/2} cos {(79°-11°)/2}
= 2 cos 45° cos 34°
= 2 (1/√2) cos 34°
= √2 cos 34°
(v) cos 12α + cos 8α
= 2 cos {(12α+8α)/2} cos {(12α-4α)/2}
= 2 cos 10α cos 4α
(vi) cos 25° – cos 37°
= 2 sin {(25°+37°)/2} sin {(25°-37°)/2}
= 2 sin 31° sin 6°
(vii) sin 61° – cos 39°
= sin 61° – cos (90°-51)
= sin 61° – sin 51°
= 2 cos {(61°+51°)/2} sin {(61°-51°)/2}
= 2 cos 56° sin 5°
(viii) sin 4x + cos 2x
= sin 4x + sin (90°-2x)
= 2 sin {(4x+90°-2x)/2} cos {(4x-90°+2x)/2}
= 2 sin (x+45°) cos (3x-45°)
Prove that
Que-3: (sinA+sinB)/(cosA+cosB) = tan{(A+B)/2}
Sol: LHS = (sinA + sinB)/(cosA + cosB)
= [2 sin {(A+B)/2} cos {(A-B)/2}] / [2 cos {(A+B)/2} cos {(A-B)/2}]
= tan {(A+B)/2}
Que-4: (sin75°−sin15°)/(cos75°+cos15°) = 1/√3
Sol: LHS = (sin75°−sin15°)/(cos75°+cos15°)
= [2 cos {(75°+15°)/2} sin {(75°-15°)/2}] / [2 cos {(75°+15°)/2} cos {(75°-15°)/2}]
= {2 cos 45° sin 30°}/{2 cos 45° cos 30°}
= tan 30°
= 1/√3.
Que-5: (sin7x+sin3x)/(cos7x+cos3x) = tan 5x.
Sol: LHS = (sin7x+sin3x)/(cos7x+cos3x)
= [2 sin {(7x+3x)/2} cos {(7x-3x)/2}] / [2 cos {(7x+3x)/2} cos {(7x-3x)/2}]
= {2 sin 5x cos 2x} / {2 cos 5x cos 2x}
= tan 5x = RHS.
Que-6: (cos2B−cos2A)/(sin2A+sin2B) = tan(A – B).
Sol: LHS = (cos2B−cos2A)/(sin2A+sin2B)
= [2 sin {(2B-2A)/2} sin {(2B-2A)/2}] / [2 sin {(2B-2A)/2} cos {(2B-2A)/2}]
= tan {(2A-2B)/2}
= tan (A-B) = RHS.
Que-7: {sin(4A−2B)+sin(4B−2A)}/{cos(4A−2B)+cos(4B−2A)} = tan(A + B).
Sol: {sin(4A-2B)+sin(4B-2A)}/{cos(4A-2B)+cos(4B-2A)}
= [sin{(4A-2B+4B-2A)/2} cos{(4A-2B-4B+2A)/2}]/[cos{(4A-2B+4B-2A)/2} cos{(4A-2B-4B+2A)/2}]
= [sin{(2A+2B)2} ⋅ cos{(6A-6B)/2}] / [cos{(2A+2B)/2} ⋅ cos{(6A-6B)/2}]
= {sin(A+B)}/{cos(A+B)}
= tan(A + B) = RHS.
Que-8: {cosα+2cos3α+cos5α}/{cos3α+2cos5α+cos7α} = cos 3α sec 5α
Sol: LHS = {cosα+2cos3α+cos5α}/{cos3α+2cos5α+cos7α}
= {(cos5α+cosα)+2cos3α}/{(cos7α+cos3α)+2cos5α}
= {(2cos3αcos2α) + 2cos3α}/{(2cos5αcos2α) + 2cos5α}
= {2cos3α (cos2α+1)}/{2cos5α (cos2α+1)}
= cos 3α sec 5α = RHS.
Que-9: {sinA+sin3A+sin5A+sin7}/{AcosA+cos3A+cos5A+cos7A} = tan 4A
Sol: {sinA+sin3A+sin5A+sin7A} / {cosA+cos3A+cos5A+cos7A}
= {(sinA+sin7A)+(sin3A+sin5A)} / {(cosA+cos7A)+(cos3A+cos5A)}
= {(2sin4Acos3A+2sin4AcosA)} / {(2cos4Acos3A+2cos4AcosA)}
= {2sin4A (cos3A+cosA)} / {2cos4A (cos3A+cosA)}
= sin4A / cos4A
= tan4A (Proved)
Que-10: cos 20° + cos 100° + cos 140° = 0
Sol: L.H.S. = cos 20° + cos 100° + cos 140°
= 2 cos {(100°+20°)/2} cos {(100°−20°)/2} + cos (180° – 40°)
= 2 cos 60° cos 40° – cos 40°
= 2 x (1/2) cos 40° – cos 40°
= 0 = R.H.S [∵ cos (180° – θ) = – cos θ]
Que-11: sin 10° + sin 20%+ sin 40° + sin 50° = sin 70° + sin 80°.
Sol: L.H.S = sin 10° + sin 20° + sin 40° + sin 50°
= (sin 50° + sin 10°) + (sin 40° + sin 20°)
= 2sin {(50°+10°)2} cos {(50°−10°)/2} + 2 sin {(30°+20°)/2} cos {(40°−20°)/2} [using C-D formulae]
= 2 sin 30° cos 20° + 2 sin 30° cos 10°
= 2 x (1/2) cos 20° + 2 x (1/2) cos 10°
= cos 20° + cos 10°
= cos (90° – 70°) + cos (90° – 80°)
= sin 70° + sin 80° = R.H.S.
Que-12: (i) cos 15° – sin 15° = 1/√2
(ii) sin 36° + cos 36° = √2 cos 9°
Sol: (i) cos 15° – sin 15°
= cos (45° – 30°) – sin (45° – 80°)
= cos 45° cos 30° + sin 45° sin 30° – sin 45° cos 30° + cos 45° sin 30°
= (1√2) × (√3/2) + (1√2) × (1/2) − (1/√2) × (√3/2) + (1/√2) × (1/2)
= 2/2√2
= 1√2 = R.H.S
(ii) L.H.S. = cos 36° + sin 36°
= cos 36° + sin (90° – 54°)
= cos 36° + cos 54°
= 2 cos {(36°+54°)/2} cos {(36°−54°)/2}
= 2 cos 45° cos (- 9°)
= 2 x (1√2)
= cos 9°
= √2 cos 9° = R.H.S
Que-13: cos 20° cos 40° cos 80° = 1/8
Sol: cos 20° cos 40° cos 80°
= cos 20° [cos(60° – 20°) cos(60° + 20°)]
= cos 20° [cos2 60° – sin2 20°]
= cos 20° [1/4 – (1 – cos2 20°)]
= 1/4 × cos20° – cos20° + cos320°
= cos320° – 3/4 × cos20°
= 1/4(4cos320° – 3cos20°)
= 1/4 × cos (3 × 20°)
= 1/4 × 1/2
= 1/8
Que-14: sin 10° sin 50° cos 70° = 1/8
Sol: LHS = sin10° sin50° sin70°.
= sin10° cos(90∘−50°) cos(90∘−70°).
= 2sin10° cos10° cos40° cos20° / 2cos10°
= 2sin20° cos20°cos40° / 4 cos10°
= 2sin40° cos40° / 8 cos10°
= sin80° / 8 cos10°
= cos10° / 8 cos10°
=1/8.
Que-15: 4 cos 12° cos 48° cos 72° = cos 36°
Sol: 4 cos 12° cos 48° cos 12° = 2 cos 12° [2 cos 12° cos 48°]
= 2 cos 12° [cos (12° + 48°) + cos (12° – 48°)] [∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= 2 cos 12° [cos 120° + cos 24°] = 2 cos 12° [cos (180°- 60°) + cos 24°]
= 2 cos 12° [- (1/2)
Que-16: tan 20° tan 40° tan 80° = tan 60°
Sol: cos 20° cos 40° cos 80°
= (1/2) cos 20° (2 cos 80° cos 40°)
= (1/2) cos 20° [cos (80° + 40°) + cos (80° – 40°)] [∵ 2 cosA cosB = cos (A + B) + cos (A – B)]
= (1/2) cos 20° [cos 120° + cos 40°] = (1/2) cos 20° [cos (180° – 60°) + cos 40°]
= – (1/4) cos 20° [- cos 60° + cos 40°] = (1/2) cos 20°[- (1/2) + cos 40°]
= – (1/4) cos 20° + (1/2) cos 40° cos 20°
= – (1/4) cos 20° + (1/4) (2 cos 40° cos 20°)
= – (1/4) cos 20° + (1/4) (2 cos 40° cos 20°)
= – (1/4) cos 20° + (1/4) x (1/2) + (1/4) cos 20° = 1/8
Now sin 20° sin 40° sin 80° = (sin20°)/2 (2 sin 80° sin 40°)
= (1/2) sin 20° [cos (80°-40°) – cos (80°+40°)]
= (1/2) sin 20° [cos 40° – cos 120°]
= (1/2) sin 20° [cos 40° – cos (180°-60°)]
= (1/2) sin 20° [cos 40° + (1/2)]
= (1/4) (2 cos 40° sin 20°) + (1/4) sin 20°
= (1/4) (sin 60° – sin 20°) + (1/4) sin 20°
= (1/4) sin 60° – (1/4) sin 20° + (1/4) sin 20°
= (1/2) × (√3/2) = √3/8
tan 20° tan 40° tan 80° = (sin 20° sin 40° sin 80°) / (cos 20° cos 40° cos 80°)
= (√3/8)/(1/8)
= √3 = tan 60°
Que-17: 2cos (π/13) cos(9π/13) + cos(3π/13) + cos(5π/13) = 0.
Sol: L.H.S. = 2 cos (π/13) cos (9π/13) + cos (3π/13) + cos (5π/13)
= cos {(π/13)+(9π/13)} + cos{(π/13)-(9π/13)} + cos (3π/13) + cos (5π/13)
= cos (10π/13) + cos (8π/13) + cos (3π/13) + cos (5π/13) [2cosx cosy = cos(x+y) + cos (x-y)]
= cos{(π-(13π/13)} + cos {(π- (8π/13)} + cos (3π/13) + cos (5π/13)
= -cos (3π/13) – cos (5π/13) + cos (3π/13) + cos (5π/13) = 0 [∵ cos (π – θ) = -cosθ]
= 0 = R.H.S.
Que-18: tan (A + 30°) + cot (A – 30°) = 1/sin2A−sin60°
Sol: LHS = tan (A + 30°) + cot (A – 30°)
= {sin(a+30)/cos(a+30)} + {cos(a-30)/sin(a-30)}
= [{sin(a+30)sin(a-30)} + {cos(a+30)cos(a-30)}]/{cos(a+30)sin(a-30)}
using cos A cos B+sin A sin B = cos(A-B) (in numerator)
= {cos(a+30) – (a-30)} / {cos(a+30) sin(a-30)}
= cos(60) / {cos(a+30) sin(a-30)}
putting cos 60 = 1/2
= (1/2) cos(a+30) sin(a-30)
but 2 cos A sin B = sin 2 A – sin 2 B
= 1/sin 2 a – sin 2(30)
= 1/sin 2a – sin60
Que-19: cos A + cos(120° – A) + cos(120° + A) = 0
Sol: L.H.S. = cosA + cos (120° – A) + cos(120° + A)
= cos A + [cos (1200A) + cos (1200 + A)]
= cos A + [{2cos(120°−A+120°+A)/2} {cos(120°−A−120°−A)/2}]
= cos A + [2 cos 120° cos (- A)]
= cos A + 2 (- 12) cos A
= cos A – cos A
= 0 = R.H.S.
Que-20: cos (A + B) + sin (A – B) = 2 sin (45° + A) cos (45° + B).
Sol: R.H.S = 2 sin (45° + A) cos (45° + B)
= sin (45° + A + 45° + B) + sin (45° + A – 45° – B) [∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= sin (90° + A + B) + sin (A – B)
= cos (A + B) + sin (A – B)
= L.H.S. [∵ sin (90° + θ) = cos θ]
Que-21: If cos A + cos B = 1/3 and sin A + sin B = 1/4 prove that tan (1/2) (A + B) = 3/4.
Sol: cos A + cos B = 1/3
= [2cos {(A+B)/2} cos{(A-B)/2}] = 1/3 …….. (i)
Also, sin A + sin B = 1/4
= [2sin {(A+B)/2} cos {(A-B)/2}] = 1/4 …… (ii)
On dividing eqn (ii) by (i) we get,
[2sin {(A+B)/2} cos {(A-B)/2}] / [2cos {(A+B)/2} cos{(A-B)/2}] = (1/4)/(1/3)
= tan {(A+B)/2} = 3/4.
Que-22: What is the value of (cos10°+sin20°)/(cos20°−sin10°)?
Sol: (cos10° + sin20°) / (cos20° – sin10°)
= tan60° * (cos60°/sin60°) * [(cos10° + sin20°) / (cos20° – sin10°)]
= tan60° * [(2cos60° cos10° + 2cos60° sin20°) / (2sin60°cos20° – 2sin60° sin10°)]
= √3 * [(cos70° + cos50° + sin80° – sin40°) / (sin80° + sin40° – cos50° + cos70°)]
= √3 * [(cos70° + cos50° + sin80° – cos50°) / (sin80° + cos50° – cos50° + cos70°)]
[because sin40° = cos50°]
= √3 * [(cos70° + sin80°) / (sin80° + cos70°)]
= √3
–: End Compound and Multiple Angles Class 11 OP Malhotra Exe-5B ISC Math Ch-5 Solution :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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