The availability of step-by-step S Chand Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(b) can make challenging problems more manageable.
S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(b)
Question 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 \(\frac{A+B}{2}\) cos \(\frac{A-B}{2}\)
Solution:
(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° = \(\frac{\sqrt{3}}{2}\) + 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) = \(\frac { 1 }{ 2 }\) [2 cos (α + ß) cos (α – ß)]
= \(\frac { 1 }{ 2 }\) [cos (α + ß + α – ß) + cos (α + ß – α + ß)] = \(\frac { 1 }{ 2 }\) [cos 2α + cos 2ß]
Question 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
Solution:
Prove that
Question 3.
\(\frac{\sin A+\sin B}{\cos A+\cos B}=\tan \left(\frac{A+B}{2}\right)\)
Solution:
Question 4.
\(\frac{\sin 75^{\circ}-\sin 15^{\circ}}{\cos 75^{\circ}+\cos 15^{\circ}}=\frac{1}{\sqrt{3}}\)
Solution:
Question 5.
\(\frac{\sin 7 x+\sin 3 x}{\cos 7 x+\cos 3 x}\) = tan 5x.
Solution:
Question 6.
\(\frac{\cos 2 B-\cos 2 A}{\sin 2 A+\sin 2 B}\) = tan(A – B).
Solution:
Question 7.
\(\frac{\sin (4 A-2 B)+\sin (4 B-2 A)}{\cos (4 A-2 B)+\cos (4 B-2 A)}\) = tan(A + B).
Solution:
Question 8.
\(\frac{\cos \alpha+2 \cos 3 \alpha+\cos 5 \alpha}{\cos 3 \alpha+2 \cos 5 \alpha+\cos 7 \alpha}\) = cos 3α sec 5α.
Solution:
Question 9.
\(\frac{\sin A+\sin 3 A+\sin 5 A+\sin 7 A}{\cos A+\cos 3 A+\cos 5 A+\cos 7 A}\) = tan 4A
Solution:
Question 10.
cos 20° + cos 100° + cos 140° = 0
Solution:
L.H.S. = cos 20° + cos 100° + cos 140°
= 2 cos \(\left(\frac{100^{\circ}+20^{\circ}}{2}\right)\) cos \(\left(\frac{100^{\circ}-20^{\circ}}{2}\right)\) + cos (180° – 40°)
= 2 cos 60° cos 40° – cos 40° = 2 x \(\frac { 1 }{ 2 }\) cos 40° – cos 40° = 0 = R.H.S [∵ cos (180° – θ) = – cos θ]
Question 11.
sin 10° + sin 20%+ sin 40° + sin 50° = sin 70° + sin 80°.
Solution:
L.H.S = sin 10° + sin 20° + sin 40° + sin 50° = (sin 50° + sin 10°) + (sin 40° + sin 20°)
= 2sin \(\left(\frac{50^{\circ}+10^{\circ}}{2}\right)\) cos \(\left(\frac{50^{\circ}-10^{\circ}}{2}\right)\) + 2 sin \(\left(\frac{30^{\circ}+20^{\circ}}{2}\right)\) cos \(\left(\frac{40^{\circ}-20^{\circ}}{2}\right)\) [using C-D formulae]
= 2 sin 30° cos 20° + 2 sin 30° cos 10° = 2 x \(\frac { 1 }{ 2 }\) cos 20° + 2 x \(\frac { 1 }{ 2 }\) cos 10°
= cos 20° + cos 10° = cos (90° – 70°) + cos (90° – 80°) = sin 70° + sin 80° = R.H.S.
Question 12.
(i) cos 15° – sin 15° = \(\frac{1}{\sqrt{2}}\)
(ii) sin 36° + cos 36° = \(\sqrt{2}\) cos 9°.
Solution:
(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°
= \(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\) = R.H.S
(ii) L.H.S. = cos 36° + sin 36° = cos 36° + sin (90° – 54°) = cos 36° + cos 54°
= 2 cos \(\frac{36^{\circ}+54^{\circ}}{2}\) cos \(\frac{36^{\circ}-54^{\circ}}{2}\) = 2 cos 45° cos (- 9°) = 2 x \(\frac{1}{\sqrt{2}}\) = cos 9° = \(\sqrt{2}\) cos 9°
= R.H.S
Question 13.
cos 20° cos 40° cos 80° = \(\frac { 1 }{ 8 }\).
Solution:
Question 14.
sin 10° sin 50° cos 70° = \(\frac { 1 }{ 8 }\).
Solution:
Question 15.
4 cos 12° cos 48° cos 72° = cos 36°
Solution:
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° [- \(\frac { 1 }{ 2 }\) + cos 24°] = – cos 12°+ 2 cos 24° cos 12°
= – cos 12° + cos 36° + cos 12° = cos 36°
Question 16.
tan 20° tan 40° tan 80° = tan 60°
Solution:
cos 20° cos 40° cos 80° = \(\frac { 1 }{ 2 }\) cos 20° (2 cos 80° cos 40°)
= \(\frac { 1 }{ 2 }\) cos 20° [cos (80° + 40°) + cos (80° – 40°)] [∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac { 1 }{ 2 }\) cos 20° [cos 120° + cos 40°] = \(\frac { 1 }{ 2 }\) cos 20° [cos (180° – 60°) + cos 40°]
= – \(\frac { 1 }{ 4 }\) cos 20° [- cos 60° + cos 40°] = \(\frac { 1 }{ 2 }\) cos 20°[- \(\frac { 1 }{ 2 }\) + cos 40°]
= – \(\frac { 1 }{ 4 }\) cos 20° + \(\frac { 1 }{ 2 }\) cos 40° cos 20° = – \(\frac { 1 }{ 4 }\) cos 20° + \(\frac { 1 }{ 4 }\) (2 cos 40° cos 20°)
= – \(\frac { 1 }{ 4 }\) cos 20° + \(\frac { 1 }{ 4 }\) (2 cos 40° cos 20°) = – \(\frac { 1 }{ 4 }\) cos 20° + \(\frac { 1 }{ 4 }\) x \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 4 }\) cos 20° = \(\frac { 1 }{ 8 }\)
Now sin 20° sin 40° sin 80° = \(\frac { sin 20° }{ 2 }\) (2 sin 80° sin 40°)
Question 17.
\(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}\) = 0.
Solution:
Question 18.
tan (A + 30°) + cot (A – 30°) = \(\frac{1}{\sin 2 A-\sin 60^{\circ}}\).
Solution:
Question 19.
cos A + cos(120° – A) + cos(120° + A) = 0
Solution:
L.H.S. = cosA + cos (120° – A) + cos(120° + A)
= cos A + [cos (1200A) + cos (1200 + A)]
= cos A + \(\left[2 \cos \left(\frac{120^{\circ}-\mathrm{A}+120^{\circ}+\mathrm{A}}{2}\right) \cos \left(\frac{120^{\circ}-\mathrm{A}-120^{\circ}-\mathrm{A}}{2}\right)\right]\)
= cos A + [2 cos 120° cos (- A)] = cos A + 2 (- \(\frac { 1 }{ 2 }\)) cos A
= cos A – cos A = 0 = R.H.S.
Question 20.
cos (A + B) + sin (A – B) = 2 sin (45° + A) cos (45° + B).
Solution:
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]
Question 21.
If cos A + cos B = \(\frac { 1 }{ 3 }\) and sin A + sin B = \(\frac { 1 }{ 4 }\) prove that tan \(\frac { 1 }{ 2 }\) (A + B) = \(\frac { 3 }{ 4 }\).
Solution:
Question 22.
What is the value of \(\frac{\cos 10^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}-\sin 10^{\circ}}\)?
Solution:
