{"id":170647,"date":"2024-05-31T16:31:48","date_gmt":"2024-05-31T11:01:48","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170647"},"modified":"2024-05-31T16:32:40","modified_gmt":"2024-05-31T11:02:40","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-ex-3-6","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-ex-3-6\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6"},"content":{"rendered":"

Utilizing ISC Mathematics Class 11 Solutions<\/a> Chapter 3 Trigonometry Ex 3.6 as a study aid can enhance exam preparation.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6<\/h2>\n

Very short answer\/objective questions (1 to 3) :<\/span><\/p>\n

Question 1.
\nConvert the following products into sums or differences:
\n(i) 2 sin 3x cos 2x
\n(ii) 2 cos 3x sin 2x
\n(iii) 2 sin 4x sin 2x
\n(iv) 2 cos 7x cos 3x.
\nSolution:
\n(i) 2 sin 3x cos 2x
\n= sin (3x + 2x) + sin (3x – 2x)
\n= sin 5x + sin x
\n[\u2235 2 sin A cos B = sin (A + B) + sin (A – B)]<\/p>\n

(ii) 2 cos 3x sin 2x
\n= sin (3x + 2x) – sin (3x – 2x)
\n= sin 5x – sin x
\n[\u2235 2 cos A sin B = sin(A + B) – sin (A – B)]<\/p>\n

(iii) 2 sin 4x sin 2x
\n= cos (4x – 2x) – cos (4x + 2x)
\n= cos 2x – cos 6x
\n[\u2235 2 sin A sin B = cos (A – B) – cos (A + B)]<\/p>\n

(iv) 2 cos 7x cos 3x
\n= cos (7x + 3x) + cos (7x – 3x)
\n= cos 10x + cos 4x
\n[\u2235 2 cos A cos B = cos (A + B) + cos(A – B)]<\/p>\n

Question 2.
\nExpress each of the following as the product of sines and cosines :
\n(i) sin 10x + sin 6x
\n(ii) sin 10x – sin 6x
\n(ill) cos 10x + cos 6x
\n(iv) cos 10x – cos 6x.
\nSolution:
\n(i) sin 10x + sin 6x
\n= 2 sin \\(\\left(\\frac{10 x+6 x}{2}\\right)\\) + cos \\(\\left(\\frac{10 x-6 x}{2}\\right)\\)
\n= 2 sin 8x cos 2x
\n[\u2235 sin C + sin D = 2 sin \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)]<\/p>\n

(ii) sin 10x – sin 6x
\n= 2 cos \\(\\left(\\frac{10 x+6 x}{2}\\right)\\) sin \\(\\left(\\frac{10 x-6 x}{2}\\right)\\)
\n= 2 cos 8x sin 2x
\n[\u2235 sin C + sin D = 2 sin \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)]<\/p>\n

(iii) cos 10x + cos 6x
\n= – 2 sin \\(\\left(\\frac{10 x+6 x}{2}\\right)\\) sin \\(\\left(\\frac{10 x-6 x}{2}\\right)\\)
\n= – 2 sin 8x sin 2x
\n[\u2235 cos C – cos D = – 2 sin \\(\\frac{C+D}{2}\\) sin \\(\\frac{C-D}{2}\\)]<\/p>\n

(iv) cos 10x – cos 6x
\n= – 2 sin \\(\\left(\\frac{10 x+6 x}{2}\\right)\\) sin \\(\\left(\\frac{10 x-6 x}{2}\\right)\\)
\n= – 2 sin 8x sin 2x
\n[\u2235 cos C – cos D = – 2 sin \\(\\frac{C+D}{2}\\) sin \\(\\frac{C-D}{2}\\)]<\/p>\n

\"ML<\/p>\n

Question 3.
\nProve that :
\n(i) 2 cos 45\u00b0 cos 15\u00b0 = \\(\\frac{\\sqrt{3}+1}{2}\\)
\n(ii) 2 sin 75\u00b0 sin 15\u00b0 = \\(\\frac{1}{2}\\)
\nSolution:
\n(i) 2 cos 45\u00b0 cos 15\u00b0 = cos (45\u00b0 + 15\u00b0) + cos (45\u00b0 – 15\u00b0)
\n[\u2235 2 cos A cos B = cos (A + B) + cos (A – B)]
\n= cos 60\u00b0 + cos 30\u00b0
\n= \\(\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\)
\n= \\(\\frac{1+\\sqrt{3}}{2}\\)<\/p>\n

(ii) 2 sin 75\u00b0 sin 15\u00b0
\n= cos (75\u00b0 – 15\u00b0) – cos (75\u00b0 + 15\u00b0)
\n[\u2235 2 sin A sin B cos (A – B) – cos (A + B)]
\n= cos 60\u00b0 – cos 90\u00b0
\n= \\(\\frac{1}{2}\\)<\/p>\n

Short answer questions (4 to 7) :<\/p>\n

Question 4.
\nProve that :
\n(i) sin 80\u00b0 – cos 70\u00b0 = cos 50\u00b0
\n(ii) cos 5\u00b0 – sin 25\u00b0 = sin 35\u00b0
\n(iii) sin 36\u00b0 + cos 36\u00b0 = cos 9\u00b0
\n(iv) cos 15\u00b0-sin 15\u00b0 =7
\nSolution:
\n(i) L.H.S. = sin 80\u00b0 – cos 70\u00b0
\n= sin 80\u00b0 – cos (90\u00b0 – 20\u00b0)
\n= sin 80\u00b0 – sin 20\u00b0
\n= 2 cos \\(\\left(\\frac{80^{\\circ}+20^{\\circ}}{2}\\right)\\) sin \\(\\left(\\frac{80^{\\circ}-20^{\\circ}}{2}\\right)\\)
\n[\u2235 sin C – sin D = 2 cos \\(\\frac{C+D}{2}\\) sin \\(\\frac{C-D}{2}\\)]
\n= 2 cos 50\u00b0 sin 30\u00b0
\n= 2 cos 50\u00b0 \u00d7 \\(\\frac{1}{2}\\)
\n= cos 50\u00b0
\n= R.H.S.<\/p>\n

(ii) cos 5\u00b0 – sin 25\u00b0
\n= cos 5\u00b0 – sin (90\u00b0 – 65\u00b0)
\n= cos 5\u00b0 – cos 65\u00b0
\n= 2 sin \\(\\left(\\frac{5^{\\circ}+65^{\\circ}}{2}\\right)\\) sin \\(\\left(\\frac{65^{\\circ}-5^{\\circ}}{2}\\right)\\)
\n[\u2235 cos C – cos D = 2 sin \\(\\frac{C+D}{2}\\) sin \\(\\frac{C-D}{2}\\)]
\n= 2 sin 35\u00b0 \u00d7 \\(\\frac{1}{2}\\)
\n= sin 35\u00b0<\/p>\n

(iii) sin 36\u00b0 + cos 36\u00b0
\n= sin (90\u00b0 – 54\u00b0) + cos 36\u00b0
\n= cos 54\u00b0 + cos 36\u00b0
\n= 2 cos \\(\\left(\\frac{54^{\\circ}+36^{\\circ}}{2}\\right)\\) cos \\(\\left(\\frac{54^{\\circ}-36^{\\circ}}{2}\\right)\\)
\n[\u2235 cos C + cos D = 2 cos \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)]
\n= 2 cos 45\u00b0 cos 9\u00b0
\n= \\(\\frac{2}{\\sqrt{2}}\\) cos 9\u00b0
\n= \u221a2 cos 9\u00b0<\/p>\n

(iv) cos 15\u00b0 – sin 15\u00b0
\n= cos 15\u00b0 – sin (90\u00b0 – 75\u00b0)
\n= cos 15\u00b0 – cos 75\u00b0
\n= 2 sin \\(\\left(\\frac{15^{\\circ}+75^{\\circ}}{2}\\right)\\) sin \\(\\left(\\frac{75^{\\circ}-15^{\\circ}}{2}\\right)\\)
\n= 2 sin 45\u00b0 sin 30\u00b0
\n= 2 \u00d7 \\(\\frac{1}{\\sqrt{2}} \\times \\frac{1}{2}\\)
\n= \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

\"ML<\/p>\n

Question 5.
\nProve that :
\n(i) \\(\\frac{\\sin 5 x+\\sin 3 x}{\\cos 5 x+\\cos 3 x}\\) = tan 4x
\n(ii) \\(\\frac{\\cos 7 x+\\cos 5 x}{\\sin 7 x-\\sin 5 x}\\) = cot x
\n(iii) \\(\\frac{\\sin x+\\sin y}{\\cos x+\\cos y}=\\tan \\frac{x+y}{2}\\)
\n(iv) \\(\\frac{\\sin x-\\sin y}{\\cos x+\\cos y}=\\tan \\frac{x-y}{2}\\)
\nSolution:
\n(i) L.H.S. = \\(\\frac{\\sin 5 x+\\sin 3 x}{\\cos 5 x+\\cos 3 x}\\)
\n= \\(\\frac{2 \\sin \\left(\\frac{5 x+3 x}{2}\\right) \\cos \\left(\\frac{5 x-3 x}{2}\\right)}{2 \\cos \\left(\\frac{5 x+3 x}{2}\\right) \\cos \\left(\\frac{5 x-3 x}{2}\\right)}\\)
\n[using C – D formulae]
\n= \\(\\frac{2 \\sin 4 x \\cos x}{2 \\cos 4 x \\cos x}\\)
\n= tan 4x
\n= R.H.S.<\/p>\n

(ii) \\(\\frac{\\cos 7 x+\\cos 5 x}{\\sin 7 x-\\sin 5 x}\\)
\n= \\(\\frac{2 \\cos \\left(\\frac{7 x+5 x}{2}\\right) \\cos \\left(\\frac{7 x-5 x}{2}\\right)}{2 \\cos \\left(\\frac{7 x+5 x}{2}\\right) \\sin \\left(\\frac{7 x-5 x}{2}\\right)}\\)
\n[using C – D formulae]
\n= \\(\\frac{2 \\cos 6 x \\cos x}{2 \\cos 6 x \\sin x}\\)
\n= cot x
\n= R.H.S.<\/p>\n

(iii) \\(\\frac{\\sin x+\\sin y}{\\cos x+\\cos y}\\)
\n= \\(\\frac{2 \\sin \\left(\\frac{x+y}{2}\\right) \\cos \\left(\\frac{x-y}{2}\\right)}{2 \\cos \\left(\\frac{x+y}{2}\\right) \\cos \\left(\\frac{x-y}{2}\\right)}\\)
\n[\u2235 sin C + sin D = 2 sin \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)
\ncos C + cos D = 2 cos \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)]
\n= tan \\(\\left(\\frac{x+y}{2}\\right)\\)
\n= R.H.S.<\/p>\n

(iv) L.H.S. = \\(\\frac{\\sin x-\\sin y}{\\cos x+\\cos y}\\)
\n= \\(\\frac{2 \\cos \\left(\\frac{x+y}{2}\\right) \\sin \\left(\\frac{x-y}{2}\\right)}{2 \\cos \\left(\\frac{x+y}{2}\\right) \\cos \\left(\\frac{x-y}{2}\\right)}\\)
\n= tan \\(\\left(\\frac{x-y}{2}\\right)\\)
\n[using C – D formulae]
\n= R.H.S.<\/p>\n

Question 6.
\n(i) \\(\\frac{\\cos 20^{\\circ}-\\cos 70^{\\circ}}{\\sin 70^{\\circ}-\\sin 20^{\\circ}}\\) = 1
\n(ii) \\(\\frac{\\sin 75^{\\circ}-\\sin 15^{\\circ}}{\\cos 75^{\\circ}+\\cos 15^{\\circ}}=\\frac{1}{\\sqrt{3}}\\)
\nSolution:
\n(i) L.H.S. = \\(\\frac{\\cos 20^{\\circ}-\\cos 70^{\\circ}}{\\sin 70^{\\circ}-\\sin 20^{\\circ}}\\)
\n= \\(\\frac{2 \\sin \\left(\\frac{20^{\\circ}+70^{\\circ}}{2}\\right) \\sin \\left(\\frac{70^{\\circ}-20^{\\circ}}{2}\\right)}{2 \\cos \\left(\\frac{70^{\\circ}+20^{\\circ}}{2}\\right) \\sin \\left(\\frac{70^{\\circ}-20^{\\circ}}{2}\\right)}\\)
\n= tan 45\u00b0
\n= 1
\n= R.H.S.<\/p>\n

(ii) \\(\\frac{\\sin 75^{\\circ}-\\sin 15^{\\circ}}{\\cos 75^{\\circ}+\\cos 15^{\\circ}}\\)
\n= \\(\\frac{2 \\cos \\left(\\frac{75^{\\circ}+15^{\\circ}}{2}\\right) \\sin \\left(\\frac{75^{\\circ}-15^{\\circ}}{2}\\right)}{2 \\cos \\left(\\frac{75^{\\circ}+15^{\\circ}}{2}\\right) \\cos \\left(\\frac{75^{\\circ}-15^{\\circ}}{2}\\right)}\\)
\n= tan 30\u00b0
\n= \\(\\frac{1}{\\sqrt{3}}\\)
\n= R.H.S.<\/p>\n

\"ML<\/p>\n

Question 7.
\nProve that :
\n(i) sin (\\(\\frac{\\pi}{4}\\) sin (\\(\\frac{\\pi}{2}\\) – x) = \\(\\frac{1}{2}\\) cos 2x
\n(ii) sec (\\(\\frac{\\pi}{4}\\) + x) sec (\\(\\frac{\\pi}{4}\\) – x) = 2 sec 2x
\n(iii) \\(\\sin \\left(\\frac{5 \\pi}{6}+x\\right)+\\sin \\left(\\frac{5 \\pi}{6}-x\\right)\\) = cos x
\nSolution:
\n(i) L.H.S. = sin (\\(\\frac{\\pi}{4}\\) sin (\\(\\frac{\\pi}{2}\\) – x)
\n= \\(\\frac{1}{2}\\left[2 \\sin \\left(\\frac{\\pi}{4}+x\\right) \\sin \\left(\\frac{\\pi}{4}-x\\right)\\right]\\)
\n= \\(\\frac{1}{2}\\left[\\cos \\left(\\frac{\\pi}{4}+x-\\frac{\\pi}{4}+x\\right)\\right.\\) – \\(\\left.\\cos \\left(\\frac{\\pi}{4}+x+\\frac{\\pi}{4}-x\\right)\\right]\\)
\n[\u2235 2 sin A sin B = cos (A – B) – cos (A + B)]
\n= \\(\\frac{1}{2}\\) [cos 2x – cos \\(\\frac{\\pi}{2}\\)]
\n= \\(\\frac{1}{2}\\) cos 2x
\n= R.H.S.<\/p>\n

(ii) L.H.S. = sec (\\(\\frac{\\pi}{4}\\) + x) sec (\\(\\frac{\\pi}{4}\\) – x)<\/p>\n

\"ML<\/p>\n

(iii) L.H.S. = \\(\\sin \\left(\\frac{5 \\pi}{6}+x\\right)+\\sin \\left(\\frac{5 \\pi}{6}-x\\right)\\)<\/p>\n

\"ML<\/p>\n

Question 8.
\nProve that :
\n(i) cos 7x + cos 5x + cos 3x + cos x = 4 cos x cos 2x cos 4x
\n(ii) sin x + sin 2x + sin 4x + sin 5x = 4cos \\(\\frac{x}{2}\\) cos \\(\\frac{3 x}{2}\\) sin 3x
\n(iii) cos 3x + cos 5x + cos 7x + cos 15x = 4 cos 4x cos 5x cos 6x
\n(iv) cos x cos \\(\\frac{x}{2}\\) – cos 3x cos \\(\\frac{9 x}{2}\\) = sin 4x sin \\(\\frac{7 x}{2}\\)
\nSolution:
\n(i) L.H.S. = cos 7x + cos 5x + cos 3x + cos x
\n= (cos 7x + cos x) + (cos 5x + cos 3x)
\n= 2 \\(\\cos \\left(\\frac{7 x+x}{2}\\right) \\cos \\left(\\frac{7 x-x}{2}\\right)\\) + 2 \\(\\cos \\left(\\frac{5 x+3 x}{2}\\right) \\cos \\left(\\frac{5 x-3 x}{2}\\right)\\)
\n= 2 cos 4x cos 3x + 2 cos 4x cos x
\n= 2 cos 4x (cos 3x + cos x)
\n= 2 cos 4x \\(\\left[2 \\cos \\left(\\frac{3 x+x}{2}\\right) \\cos \\left(\\frac{3 x-x}{2}\\right)\\right]\\)
\n[\u2235 cos C + cos D = 2 cos \\(\\left(\\frac{\\mathrm{C}+\\mathrm{D}}{2}\\right)\\) cos \\(\\left(\\frac{\\mathrm{C}-\\mathrm{D}}{2}\\right)\\)]
\n= 2 cos 4x (2 cos 2x cos x)
\n= 4 cos x cos 2x cos 4x
\n= R.H.S.<\/p>\n

(ii) L.H.S. = sin x + sin 2x + sin 4x + sin 5x
\n= (sin 5x + sin x) + sin 4x + sin 2x)
\n= 2 \\(\\sin \\left(\\frac{5 x+x}{2}\\right) \\cos \\left(\\frac{5 x-x}{2}\\right)\\) + 2 \\(\\sin \\left(\\frac{4 x+2 x}{2}\\right) \\cos \\left(\\frac{4 x-2 x}{2}\\right)\\)
\n= 2 sin 3x cos 2x + 2 sin 3x cos x
\n= 2 sin 3x (cos 2x + cos x)
\n= 2 sin 3x \\(\\left[2 \\cos \\left(\\frac{2 x+x}{2}\\right) \\cos \\left(\\frac{2 x-x}{2}\\right)\\right]\\)
\n= 4 cos \\(\\frac{x}{2}\\) cos \\(\\frac{3 x}{2}\\) sin 3x
\n= R.H.S.<\/p>\n

(iii) L.H.S. = cos 3x + cos 5x + cos 7x + cos 15x
\n= (cos 15x + cos 3x) + (cos 7x + cos 5x)
\n= \\(2 \\cos \\left(\\frac{15 x+3 x}{2}\\right) \\cos \\left(\\frac{15 x-3 x}{2}\\right)+2 \\cos \\left(\\frac{7 x+5 x}{2}\\right) \\cos \\left(\\frac{7 x-5 x}{2}\\right)\\)
\n= 2 cos 9x cos 6x + 2 cos 6x cos x
\n= 2 cos 6x [cos 9x + cos x]
\n= 2 cos 6x \\(\\left[2 \\cos \\left(\\frac{9 x+x}{2}\\right) \\cos \\left(\\frac{9 x-x}{2}\\right)\\right]\\)
\n= 2 cos 6x (2 cos 5x cos 4x)
\n= 4 cos 4x cos 5x cos 6x
\n= R.H.S.<\/p>\n

(iv) L.H.S. = cos x cos \\(\\frac{x}{2}\\) – cos 3x cos \\(\\frac{9 x}{2}\\)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 9.
\nProve that :
\n(i) cos 52\u00b0 + cos 68\u00b0 + cos 172\u00b0 = 0
\n(ii) cos 20\u00b0 + cos 100\u00b0 + cos 140\u00b0 = 0
\n(iii) \\(2 \\sin \\frac{\\pi}{17} \\sin \\frac{11 \\pi}{17}-\\cos \\frac{5 \\pi}{17}+\\cos \\frac{7 \\pi}{17}\\) = 0
\nSolution:
\n(I) L.H.S. = cos 52\u00b0 + OS 68\u00b0 COS 1720
\n= 2 cos \\(\\left(\\frac{52^{\\circ}+68^{\\circ}}{2}\\right)\\) cos \\(\\left(\\frac{52^{\\circ}-68^{\\circ}}{2}\\right)\\) + cos (180\u00b0 – 8\u00b0)
\n[cos C + cos D = 2 cos \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)]
\n= 2 cos 60\u00b0 cos 8\u00b0 – cos 8\u00b0
\n= 2 \u00d7 \\(\\frac{1}{2}\\) cos 8\u00b0 – cos 8\u00b0
\n= 0
\n= R.H.S.<\/p>\n

(ii) LH.S. = cos 20\u00b0 + cos 100\u00b0 + cos 140\u00b0
\n= 2 cos \\(\\left(\\frac{20^{\\circ}+100^{\\circ}}{2}\\right)\\) cos \\(\\left(\\frac{20^{\\circ}-100^{\\circ}}{2}\\right)\\) + cos (180\u00b0 – 40\u00b0)
\n= 2 cos 60\u00b0 cos 40\u00b0 – cos 40\u00b0
\n= 2 \u00d7 \\(\\frac{1}{2}\\) cos 40\u00b0 – cos 40\u00b0
\n= 0
\n= R.H.S.<\/p>\n

(iii) L.H.S = \\(2 \\sin \\frac{\\pi}{17} \\sin \\frac{11 \\pi}{17}-\\cos \\frac{5 \\pi}{17}+\\cos \\frac{7 \\pi}{17}\\)<\/p>\n

\"ML<\/p>\n

Question 10.
\nProve that :
\n(i) sin 10\u00b0 sin 50\u00b0 sin 70\u00b0 = \\(\\frac{1}{8}\\)
\n(ii) sin 10\u00b0 sin 30\u00b0 sin 50\u00b0 sin 70\u00b0 = \\(\\frac{1}{16}\\)
\n(iii) sin 20\u00b0 sin 40\u00b0 sin 60\u00b0 sin 80\u00b0 = \\(\\frac{3}{16}\\)
\nSolution:
\n(i) L.H.S. = sin 10\u00b0 sin 50\u00b0 sin 70\u00b0
\n= \\(\\frac{1}{2}\\) sin 10\u00b0 [2 sin 70\u00b0 sin 50\u00b0]
\n= \\(\\frac{1}{2}\\) sin 10\u00b0 [cos (20\u00b0) – cos 120\u00b0]
\n[\u2235 2 sin A sin B = cos (A – B) – cos (A + B)]
\n= \\(\\frac{1}{2}\\) sin 10\u00b0 [cos 20\u00b0 – cos (180\u00b0 – 60\u00b0)]
\n= \\(\\frac{1}{4}\\) (2 cos 20\u00b0 sin 10\u00b0) – \\(\\frac{1}{2}\\) (- cos 60\u00b0) sin 10\u00b0
\n= \\(\\frac{1}{4}\\) [sin (20\u00b0 + 10\u00b0) – sin (20\u00b0 – 10\u00b0)] + \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{1}{2}\\) sin 10\u00b0
\n= \\(\\frac{1}{4}\\) [\\(\\frac{1}{2}\\) – sin 10\u00b0] + \\(\\frac{1}{4}\\) sin 10\u00b0
\n= R.H.S.<\/p>\n

(ii) L.H.S. = sin 10\u00b0 sin 30\u00b0 sin 50\u00b0 sin 70\u00b0
\n= \\(\\frac{1}{2}\\) [sin 10\u00b0 sin 50\u00b0 sin 70\u00b0]
\n= \\(\\frac{1}{2 \\times 2}\\) sin 10\u00b0 [2 sin 70\u00b0 sin 50\u00b0]
\n= \\(\\frac{1}{4}\\) sin 10\u00b0 [cos (70\u00b0 – 50\u00b0) – cos (70\u00b0 + 50\u00b0)]
\n= \\(\\frac{1}{4}\\) sin 10\u00b0 [cos 20\u00b0 – cos(180\u00b0 – 60\u00b0)]
\n= \\(\\frac{1}{4}\\) sin 10\u00b0 [cos 20\u00b0 + cos 60\u00b0]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 sin 10\u00b0 sin 10\u00b0 \u00d7 \\(\\frac{1}{2}\\)
\n= \\(\\frac{1}{2}\\) [2 cos 20\u00b0 sin 10\u00b0] + \\(\\frac{1}{8}\\) sin 10\u00b0
\n= \\(\\frac{1}{8}\\) [sin (20\u00b0 + 10\u00b0) – sin (20\u00b0 – 10\u00b0)] + \\(\\frac{1}{8}\\) sin 10\u00b0
\n= \\(\\frac{1}{8}\\left[\\frac{1}{2}-\\sin 10^{\\circ}\\right]\\)
\n= \\(\\frac{1}{16}\\)
\n= R.H.S.<\/p>\n

(iii) L.H.S. = sin 20\u00b0 sin 40\u00b0 sin 60\u00b0 sin 80\u00b0
\n= \\(\\frac{\\sqrt{3}}{2}\\) [sin 20\u00b0 sin 40\u00b0 sin 80\u00b0]
\n= \\(\\frac{\\sqrt{3}}{2}\\) sin 20 \u00d7 \\(\\frac{1}{2}\\) [2 sin 80\u00b0 sin 40\u00b0]
\n= \\(\\frac{\\sqrt{3}}{4}\\) sin 20\u00b0 [cos (80\u00b0 – 40\u00b0) – cos(80\u00b0 + 40\u00b0)]
\n[\u2235 2 sin A sin B = cos (A – B) – cos(A + B)]
\n= \\(\\frac{\\sqrt{3}}{4}\\) sin 20\u00b0 [cos 40\u00b0 – cos 120\u00b0]
\n= \\(\\frac{\\sqrt{3}}{4}\\) sin 20\u00b0 [cos 40\u00b0 – cos (180\u00b0 – 60\u00b0)]
\n= \\(\\frac{\\sqrt{3}}{4}\\) sin 20\u00b0 [cos 40\u00b0 + \\(\\frac{1}{2}\\)]
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 40\u00b0 sin 20\u00b0 + \\(\\frac{\\sqrt{3}}{8}\\) sin 20\u00b0
\n= \\(\\frac{\\sqrt{3}}{8}\\) (2 cos 40\u00b0 sin 20\u00b0) + \\(\\frac{\\sqrt{3}}{8}\\) sin 20\u00b0
\n= \\(\\frac{\\sqrt{3}}{8}\\) [sin (40\u00b0 + 20\u00b0) – sin (40\u00b0 – 20\u00b0)] + \\(\\frac{\\sqrt{3}}{8}\\) sin 20\u00b0
\n[\u22352 cos A sin B = sin (A + B) – sin (A – B)]
\n= \\(\\frac{\\sqrt{3}}{8}\\) [sin 60\u00b0 – sin 20\u00b0] + \\(\\frac{\\sqrt{3}}{8}\\) sin 20\u00b0
\n= \\(\\frac{\\sqrt{3}}{8} \\times \\frac{\\sqrt{3}}{2}-\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}\\)
\n= \\(\\frac{3}{16}\\)
\n= R.H.S.<\/p>\n

\"ML<\/p>\n

Question 11.
\nProve that :
\n(i) cos 20\u00b0 cos 40\u00b0 cos 60\u00b0 cos 80\u00b0 = \\(\\frac{1}{16}\\)
\n(ii) cos 10\u00b0 cos 30\u00b0 cos 50\u00b0 cos 70\u00b0 = \\(\\frac{3}{16}\\)
\nSolution:
\n(i) L.H.S. = cos 20\u00b0 cos 40\u00b0 cos 60\u00b0 cos 80\u00b0
\n= \\(\\frac{1}{2}\\) [cos 20\u00b0 cos 40\u00b0 cos 80\u00b0]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 [2 cos 80\u00b0 cos 40\u00b0]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 [cos (80\u00b0 + 40\u00b0) + cos (80\u00b0 – 40\u00b0)]
\n[\u2235 2 cos A cos B = cos (A + B) + cos (A – B)]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 [cos 120\u00b0 + cos 40\u00b0]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 [cos (180\u00b0 – 60\u00b0) + cos 40\u00b0]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 [- cos 60\u00b0 + cos 40\u00b0]
\n= \\(\\frac{1}{4}\\) cos 20\u00b0 [- \\(\\frac{1}{2}\\) + cos 40\u00b0]
\n= \\(\\frac{1}{8}\\) cos 20\u00b0+ \\(\\frac{1}{8}\\) (2 cos 40\u00b0 cos 20\u00b0)
\n= – \\(\\frac{1}{4}\\) cos 20\u00b0 + \\(\\frac{1}{8}\\) [cos (40\u00b0 + 20\u00b0) + cos (40\u00b0 – 20\u00b0)]
\n= – \\(\\frac{1}{8}\\) cos 20\u00b0+ \\(\\frac{1}{8}\\) [\\(\\frac{1}{2}\\) + cos 20\u00b0]
\n= – \\(\\frac{1}{4}\\) cos 20\u00b0 + \\(\\frac{1}{16}\\) + \\(\\frac{1}{8}\\) cos 20\u00b0
\n= \\(\\frac{1}{16}\\)
\n= R.HS.<\/p>\n

(ii) L.H.S. = cos 10\u00b0 cos 30\u00b0 cos 50\u00b0 cos 70\u00b0
\n= \\(\\frac{\\sqrt{3}}{2}\\) cos 10\u00b0 (cos 70\u00b0 cos 50\u00b0)
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 10\u00b0 (2 cos 70\u00b0 cos 50\u00b0)
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 10\u00b0 [cos (70\u00b0 + 50\u00b0) + cos (70\u00b0 – 50\u00b0)]
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 10\u00b0 [cos 120\u00b0 + cos 20\u00b0]
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 10\u00b0 [cos (180\u00b0 – 60\u00b0) + cos 20\u00b0]
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 10\u00b0 [- cos 60\u00b0 + cos 20\u00b0]
\n= \\(\\frac{\\sqrt{3}}{4}\\) cos 10\u00b0 [- \\(\\frac{1}{2}\\) + cos 20\u00b0]
\n= – \\(\\frac{\\sqrt{3}}{8}\\) cos 10\u00b0 + \\(\\frac{\\sqrt{3}}{8}\\) (2 cos 20\u00b0 cos 10\u00b0)
\n= – \\(\\frac{\\sqrt{3}}{8}\\) cos 10\u00b0 + \\(\\frac{\\sqrt{3}}{8}\\) [cos 30\u00b0 + cos 10\u00b0]
\n= – \\(\\frac{\\sqrt{3}}{8}\\) cos 10\u00b0 + \\(\\frac{\\sqrt{3}}{8}\\) [\\(\\frac{\\sqrt{3}}{2}\\) + cos 10\u00b0]
\n= \\(-\\frac{\\sqrt{3}}{8} \\cos 10^{\\circ}+\\frac{3}{16}+\\frac{\\sqrt{3}}{8} \\cos 10^{\\circ}=\\frac{3}{16}\\)
\n= R.H.S.<\/p>\n

Question 12.
\nProve that :
\ntan 10\u00b0 tan 50\u00b0 tan 70\u00b0 = tan 30\u00b0.
\nSolution:
\nL.H.S. = tan 10\u00b0 tan 50\u00b0 tan 70\u00b0
\n= \\(\\frac{\\sin 10^{\\circ} \\sin 50^{\\circ} \\sin 70^{\\circ}}{\\cos 10^{\\circ} \\cos 50^{\\circ} \\cos 70^{\\circ}}\\) …………………(1)
\nNow sin 10\u00b0 sin 50\u00b0 sin 70\u00b0 = \\(\\frac{1}{2}\\) sin 10\u00b0 (2 sin 70\u00b0 sin 50\u00b0)
\n= \\(\\frac{1}{2}\\) sin 10\u00b0 [cos 20\u00b0 – cos 120\u00b0]
\n[\u2235 2 sinA sin B = cos (A – B) – cos (A + B)]
\n= \\(\\frac{1}{2}\\) sin 10\u00b0 [cos 20\u00b0 – cos (180\u00b0 – 60\u00b0)]
\n= \\(\\frac{1}{2}\\) sin 10\u00b0 [cos 20\u00b0 + \\(\\frac{1}{2}\\)]
\n= \\(\\frac{1}{4}\\) (2 cos 20\u00b0 sin 10\u00b0) + \\(\\frac{1}{4}\\) sin 10\u00b0
\n= \\(\\frac{1}{4}\\) [sin (20\u00b0 + 10\u00b0) – sin (20\u00b0 – 10\u00b0)] + \\(\\frac{1}{4}\\) sin 10\u00b0
\n= \\(\\frac{1}{4}\\) [\\(\\frac{1}{2}\\) – sin 10\u00b0] + \\(\\frac{1}{4}\\) sin 10\u00b0
\n= \\(\\frac{1}{8}\\) …………………..(2)
\n= cos 10\u00b0 cos 50\u00b0 cos 70\u00b0
\n= \\(\\frac{1}{2}\\) cos 10\u00b0 (2 cos 70\u00b0 cos 50\u00b0)
\n= \\(\\frac{1}{2}\\) cos 10\u00b0 [cos 120\u00b0 + cos 20\u00b0]
\n= \\(\\frac{1}{2}\\) cos 10\u00b0 [cos (180\u00b0 – 60\u00b0) + cos 20\u00b0]
\n= \\(\\frac{1}{2}\\) cos 10\u00b0 [- \\(\\frac{1}{2}\\) + cos 20\u00b0]
\n= – \\(\\frac{1}{4}\\) cos 10\u00b0 + \\(\\frac{1}{4}\\) (2 cos 20\u00b0 cos 10\u00b0)
\n= – \\(\\frac{1}{4}\\) cos 10\u00b0 + \\(\\frac{1}{4}\\) [cos 30\u00b0 + cos 10\u00b0]
\n= \\(\\frac{1}{4}\\) cos 30\u00b0
\n= \\(\\frac{1}{4} \\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{3}}{8}\\) …………………..(3)
\nUsing eqn. (2), (3) in eqn. (1) ; we have
\nL.H.S. = \\(\\frac{\\frac{1}{8}}{\\frac{\\sqrt{3}}{8}}=\\frac{1}{\\sqrt{3}}\\)
\n= tan 30\u00b0
\n= R.H.S.<\/p>\n

\"ML<\/p>\n

Question 13.
\n(i) cos x + cos (\\(\\frac{2 \\pi}{3}\\) – x) + cos (\\(\\frac{2 \\pi}{3}\\) + x) = 0
\n(ii) \\(\\cos \\frac{\\pi}{8}+\\cos \\frac{3 \\pi}{8}+\\cos \\frac{5 \\pi}{8}+\\cos \\frac{7 \\pi}{8}\\) = 0.
\nSolution:
\n(i) L.H.S. = cos x + cos (\\(\\frac{2 \\pi}{3}\\) – x) + cos (\\(\\frac{2 \\pi}{3}\\) + x)
\n= cos x + 2 \\(\\cos \\left(\\frac{\\frac{2 \\pi}{3}-x+\\frac{2 \\pi}{3}+x}{2}\\right) \\cos \\left(\\frac{\\frac{2 \\pi}{3}-x-\\frac{2 \\pi}{3}-x}{2}\\right)\\)
\n[\u2235 cos C + cos D = 2 cos \\(\\frac{C+D}{2}\\) cos \\(\\frac{C-D}{2}\\)]
\n= cos x + 2 cos \\(\\frac{2 \\pi}{3}\\) cos (- x)
\n= cos x + 2 cos (\u03c0 – \\(\\frac{\\pi}{3}\\)) cos x
\n= cos x + 2 (- cos \\(\\frac{\\pi}{3}\\)) cos x
\n= cos x – 2 \u00d7 (\\(\\frac{1}{2}\\)) cos x
\n= cos x – cos x
\n= 0
\n= R.H.S.<\/p>\n

(ii) L.H.S. = \\(\\cos \\frac{\\pi}{8}+\\cos \\frac{3 \\pi}{8}+\\cos \\frac{5 \\pi}{8}+\\cos \\frac{7 \\pi}{8}\\)<\/p>\n

\"ML<\/p>\n

Question 14.
\nProve that :
\n(i) 4 sin x sin (\\(\\frac{\\pi}{3}\\) – x) sin (\\(\\frac{\\pi}{3}\\) + x) = sin 3x
\n(ii) 4 cos x cos (\\(\\frac{\\pi}{3}\\) – x) cos (\\(\\frac{\\pi}{3}\\) + x) = cos 3x
\nSolution:
\n(i) L.H.S. = 4 sin x sin (\\(\\frac{\\pi}{3}\\) – x) sin (\\(\\frac{\\pi}{3}\\) + x)<\/p>\n

\"ML<\/p>\n

(ii) L.H.S. = 4 cos x cos (\\(\\frac{\\pi}{3}\\) – x) cos (\\(\\frac{\\pi}{3}\\) + x)<\/p>\n

\"ML<\/p>\n

= – cos x + 2 cos 2x cos x
\n= – cos x + cos (2x + x) + cos (2x – x)
\n= – cos x + cos 3x + cos x
\n= cos 3x
\n= R.H.S.<\/p>\n

\"ML<\/p>\n

Question 15.
\nProve that :
\n(i) (cos x + cosy)2<\/sup> + (sin x + siny)2<\/sup> = 4 c0s2<\/sup>
\n(ii) (cos x + cos y)2<\/sup> + (sin x – sin y)2<\/sup> = 4 cos2<\/sup>
\n(iii) sin2<\/sup> x + sin2<\/sup> (x – y) – 2 sin x cos y sin (x – y) = sin2<\/sup> y.
\nSol.
\n(i) L.H.S. = (cos x + cos y)2<\/sup> – (sin x + sin y)2<\/sup><\/p>\n

\"ML<\/p>\n

(ii) L.H.S. = (cos x + cos y)2<\/sup> + (sin x – sin y)2<\/sup><\/p>\n

\"ML<\/p>\n

(iii) L.H.S. = sin2<\/sup> x + sin2<\/sup> (x – y) – 2 sin x cos y sin (x – y)
\n= sin2<\/sup> x + sin (x – y) [sin (x – y) – 2 sin x cos y]
\n= sin2<\/sup> x + sin (x – y) [sin (x – y) – sin (x + y) – sin (x – y)]
\n= sin2<\/sup> x + sin (x – y) {- sin (x + y)}
\n= sin2<\/sup> x – sin (x – y) sin (x + y)
\n= sin2<\/sup> x – {sin2<\/sup> x – sin2<\/sup> y}
\n= sin2<\/sup> x – sin2<\/sup> x + sin2<\/sup> y
\n= sin2<\/sup> y
\n= R.H.S.<\/p>\n

Question 16.
\nProve that :
\n(i) \\(\\frac{\\sin x+\\sin y}{\\sin x-\\sin y}=\\tan \\frac{x+y}{2} \\cot \\frac{x-y}{2}\\)
\n(ii) \\(\\frac{\\sin x+\\sin 3 x+\\sin 5 x}{\\cos x+\\cos 3 x+\\cos 5 x}\\) = tan 3x
\n(iii) \\(\\frac{\\sin 5 x+2 \\sin 8 x+\\sin 11 x}{\\sin 8 x+2 \\sin 11 x+\\sin 14 x}=\\frac{\\sin 8 x}{\\sin 11 x}\\)
\n(iv) \\(\\frac{\\sin 8 x \\cos x-\\sin 6 x \\cos 3 x}{\\cos 2 x \\cos x-\\sin 3 x \\sin 4 x}\\) = tan 2x
\n(v) \\(\\frac{\\cos x+\\cos 3 x+\\cos 5 x+\\cos 7 x}{\\sin x+\\sin 3 x+\\sin 5 x+\\sin 7 x}\\) = cot 4x
\nSolution:
\n(i) L.H.S. = \\(\\frac{\\sin x+\\sin y}{\\sin x-\\sin y}\\)
\n= \\(\\frac{2 \\sin \\left(\\frac{x+y}{2}\\right) \\cos \\left(\\frac{x-y}{2}\\right)}{2 \\cos \\left(\\frac{x+y}{2}\\right) \\sin \\left(\\frac{x-y}{2}\\right)}\\)
\n= \\(\\tan \\left(\\frac{x+y}{2}\\right) \\cot \\left(\\frac{x-y}{2}\\right)\\)
\n= R.H.S<\/p>\n

(ii) L.H.S. = \\(\\frac{\\sin x+\\sin 3 x+\\sin 5 x}{\\cos x+\\cos 3 x+\\cos 5 x}\\)<\/p>\n

\"ML<\/p>\n

(iii) L.H.S. = \\(\\frac{\\sin 5 x+2 \\sin 8 x+\\sin 11 x}{\\sin 8 x+2 \\sin 11 x+\\sin 14 x}\\)<\/p>\n

\"ML<\/p>\n

(iv) L.H.S. = \\(\\frac{\\sin 8 x \\cos x-\\sin 6 x \\cos 3 x}{\\cos 2 x \\cos x-\\sin 3 x \\sin 4 x}\\)<\/p>\n

\"ML<\/p>\n

(v) \\(\\frac{\\cos x+\\cos 3 x+\\cos 5 x+\\cos 7 x}{\\sin x+\\sin 3 x+\\sin 5 x+\\sin 7 x}\\)
\n= cot4<\/sup> x<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 17.
\nProve that :
\ncos \u03b1 + cos \u03b2 + cos \u03b3 + cos (\u03b1 + \u03b2 + \u03b3) = 4 \\(\\cos \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\beta+\\gamma}{2} \\cos \\frac{\\gamma+\\alpha}{2}\\)
\nSolution:
\nL.H.S. = cos \u03b1 + cos \u03b2) + cos \u03b3 + cos (\u03b1 + \u03b2 + \u03b3)
\n= (cos \u03b1 + cos \u03b2) + (cos (\u03b1 + \u03b2 + \u03b3) + cos \u03b3)<\/p>\n

\"ML<\/p>\n

Question 18.
\nIf cos x + cos y = \\(\\frac{1}{3}\\) and sin x + sin y = \\(\\frac{1}{4}\\), prove that \\(\\frac{x+y}{2}=\\frac{3}{4}\\).
\nSolution:
\nGiven cos x + cos y = \\(\\frac{1}{3}\\)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 19.
\n(i) \\(\\frac{\\sin (x+y)}{\\sin (x-y)}=\\frac{a+b}{a-b}\\) then show that \\(\\frac{\\tan x}{\\tan y}=\\frac{a}{b}\\).
\n(ii) If cos (x + 2y) = m cos x, prove that cot y = \\(\\frac{1+m}{1-m}\\) tan (x + y).
\nSolution:
\nGiven \\(\\frac{\\sin (x+y)}{\\sin (x-y)}=\\frac{a+b}{a-b}\\)
\nApplying componendo and dividendo, we have<\/p>\n

\"ML<\/p>\n

(ii) Given, cos (x + 2y) = m cos x
\n\u21d2 \\(\\frac{\\cos (x+2 y)}{\\cos x}=\\frac{m}{1}\\)
\nApplying componendo and dividendo, we have<\/p>\n

\"\"<\/p>\n","protected":false},"excerpt":{"rendered":"

Utilizing ISC Mathematics Class 11 Solutions Chapter 3 Trigonometry Ex 3.6 as a study aid can enhance exam preparation. ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 Very short answer\/objective questions (1 to 3) : Question 1. Convert the following products into sums or differences: (i) 2 sin 3x cos …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170647"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170647"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170647\/revisions"}],"predecessor-version":[{"id":170668,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170647\/revisions\/170668"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170647"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170647"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170647"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}