Tiêu đề TRIGONOMETRIC FUNCTIONS A-5.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :5 781 (a) We have, 2 sin θ = r 4 ― 2 r 2 + 3 ⇒2sin θ = (r 2 ― 1) 2 + 2 Clearly, LHS ≤ 2 and RHS ≥ 2 So, the equation is meaningful if each side is equal to 2 Clearly, RHS = 2 for r 2 = 1 For r 2 = 1, we have 2 sin θ = 2 ⇒ sin θ = 1⇒θ = π 2 , 5π 2 , 9π 2 [ ∵ 0 ≤ θ ≤ 5π] Also, r 2 = 1⇒r =± 1 Hence, the total number of pairs of the form (r, θ) is 2 × 3 = 6 783 (a) We have, 1 p 2 1 + 1 p 2 2 + 1 p 2 3 = a 2 + b 2 + c 2 4∆ 2 Also, cotA + cosB + cot C = 2R abc (b 2 + c 2 ― a 2 + c 2 + a 2 ― b 2 + a 2 + b 2 ― c 2 ) ⇒ cotA + cosB + cot C = R(a 2 + b 2 + c 2 ) abc = a 2 + b 2 + c 2 4∆ Hence, 1 p 2 1 + 1 p 2 2 + 1 p 2 3 = cotA + cotB + cot C ∆ 784 (c) We have, sin 2 θ + 2 = 4 sin θ + cos θ ⇒2 sin θ cos θ ― cos θ + 2 ― 4 sin θ = 0 ⇒ cos θ(2 sin θ ― 1 ) ― 2(2 sin θ ― 1) = 0 ⇒(2 sin θ ― 1)(cos θ ― 2) = 0 ⇒2sin θ ― 1 = 0 [ ∵ cos θ ― 2 ≠ 0] ⇒ sin θ = 1 2 Topic :-TRIGONOMETRIC FUNCTIONS Solutions
⇒θ = 2 π + π 6 ,2 π + 5 π 6 ,4 π + π 6 ,4 π + 5 π 6 Hence, the equation has 4 solutions ALITER The curves y = sin x and y = 1 2 intersect at 4 points in [π, 5 π]. So, the equation has 4 solutions 785 (c) For a triangle inscribed in a circle, we have a 2 sinA = b 2 sinB = c 2 sin C = R ∴ sin2A + sin2B + sin2 C = a 2 4R 2 + b 2 4R 2 + c 2 4R 2 (a 2 + b 2 + c 2 ) It is given that a 2 + b 2 + c 2 2 = 2(2R) 2⇒a 2 + b 2 + c 2 = 16R 2 ∴ sin2A + sin2B + sin2 C = 1 4R 2 (16R 2 ) = 4 786 (d) We have, tan 9° ― tan 27° ― tan 63° + tan 81° = (tan 9° + tan 81°) ― (tan 27° + tan 63°) = (tan 9° + cot 9°) ― (tan 27° + cot 27°) = 1 sin 9° cos 9° ― 1 sin 27° cos 27° = 2 sin 18° ― 2 sin 54° = 2 sin 54° ― sin 18° sin 54° sin 18° = 2 cos 36° ― sin 18° sin 18° cos 36° = 4 787 (c) Given, sin 4A + sin 2A = cos 4A + cos 2A ⇒ 2 sin 3A cosA = 2 cos 3A cosA ∴ tan 3A = 1 and cosA = 0 ⇒ A = π 12 and A = π 2 ∉ (0, π 4 ) ∴ tan 4A = tan π 3 = 3 788 (c) We have, sinA + sinB = a + b c ⇒ sinA + sinB = sinA + sinB sin C ⇒ sin C = 1 789 (a)
We have, sin(α + β) = 1, sin(α ― β) = 1 2 ⇒α + β = π 2 and, α ― β = π 6 ⇒α = π 3 , β = π 6 ∴ tan(α + 2β) tan(2α + β) = tan ( 2π 3 ) tan 5π 6 = ( ― cot π 6 )( ― cot π 3 ) = 1 790 (c) Let a0 = cos θ. Then, a1 = 1 2 (1 + a0) = 1 2 (1 + cos θ) = cos θ 2 a2 = 1 2 (1 + a1) = 1 2 (1 + cos θ 2 ) = cos ( θ 2 2) and so on Now, 1 ― a 2 0 a1a2a3...to ∞ = sin θ cos θ 2 cos θ 2 2 cos θ 2 3 ...to ∞ = lim n→∞ sin θ cos θ 2 cos θ 2 2 cos θ 2 3 ...to ∞ = lim n→∞ {2 n sin (θ/2 n ) } sin θ sin (2 n × θ/2 n ) = lim n→∞ sin (θ/2 n ).θ (θ/2 n ) = θ = a0 791 (b) Le the angles of triangle ABC be A = θ,B = 2 θ and C = 7 θ. Then, A + B + C = 180°⇒10 θ = 180°⇒θ = 18° ∴ A = 18°,B = 36° and C = 126° Clearly, c is the greatest side and a is the smallest side. Now, a sinA = c sin C ⇒ c a = sin C sinA = sin 126° sin 18° = cos 36° sin 18° = 5 + 1 5 ― 1 792 (b) We have, A = 2 π 3 and ∆ = 9 3 2 cm2 ∴ ∆ = 1 2 bc sinA ⇒ 9 3 2 = 1 2 bc sin 2 π 3 ⇒bc = 18
Also, cosA = b 2 + c 2 ― a 2 2bc ⇒ cos 2 π 3 = (b ― c) 2 + 2bc ― a 2 2bc ⇒ ― 1 2 = 27 + 36 ― a 2 36 ⇒a 2 = 81⇒a = 9 cm 793 (b) We have, s = 8k and, ∆ = 8k × 3k × 2k × 3k = 12 k 2 ∴ r = ∆ s ⇒ 12 k 2 8k = 6⇒k = 4 794 (d) We have, 2 sin x + 2 cos x 2 ≥ 2 sin x2 cos x [ ∵ AM ≥ GM] ⇒2 sin x + 2 cos x ≥ 2 sin x+cos x ⇒2 sin x + 2 cos x ≥ 2 2 ― 2 [ ∵ ― 2 ≤ sin x + cos x ≤ 2] ⇒2 sin x + 2 cos x ≥ 2 1― 1 2 795 (a) Let A = 1 3 sin θ ― 4 cos θ + 7 Now, A will be minimum when 3sin θ ― 4 cos θ + 7 is maximum ∴ Maximum value of 3 sin θ ― 4 cos θ + 7 = 3 2 + 4 2 + 7 = 12 ∴ Minimum value of 1 3 sin θ ― 4 cos θ + 7 is 1 12 796 (b) We have, cosec θ = p + q p ― q Now, cos ( π 4 + θ 2 ) = 1 ― tan θ 2 1 + tan θ 2 = cos θ 2 ― sin θ 2 cos θ 2 + sin θ 2 ⇒ cos ( π 4 + θ 2 ) = ( cos θ 2 ― sin θ 2 cos θ 2 + sin θ 2 ) 2 = 1 ― sin θ 1 + sin θ