Tiêu đề TRIGONOMETRIC FUNCTIONS A-1.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :1 702 (c) We have, sin x + sin2 x = 1⇒ sin x = cos2 x Now, cos12 x + 3 cos10 x + 3 cos8 x + cos6 x ― 1 = cos6 x(cos6 x + 3 cos4 x + 3 cos2 x + 1) ― 1 = cos6 x(cos2 x + 1) 3 ― 1 = sin3 x(sin x + 1) 3 ― 1 = (sin2 x + sin x) 3 ― 1 = (sin2 x + cos2 x) 3 ― 1 [ ∵ sin x = cos2 x] = 1 ― 1 = 0 703 (c) Let a = 3 x + 4 y,b = 4 x + 3 y and c = 5 x + 5 y. Clearly, c is the largest side and thus the largest angle C is given by cos C a 2 + b 2 ― c 2 2 ab = ―2 xy 2(12 x 2 + 25 xy + 12 y 2 ) < 0 ⇒C is an obtuse angle 704 (a) Let a = x 2 +x + 1,b = x 2 ―1 and c = 2x + 1. Then, a ― b = x + 2 > 0 [ ∵ x > 1] a ― c = x 2 ― x > 0 [ ∵ x > 1] So, a is the largest side Hence, the largest angle is given by cos θ = b 2 + c 2 ― a 2 2bc ⇒ cos θ = (x 2 ― 1) 2 + (2x + 1) 2 ― (x 2 + x + 1) 2 2(x 2 ― 1)(2x + 1) = ― 1 2 ⇒θ = 2 π/3 = 120° 705 (c) Topic :-TRIGONOMETRIC FUNCTIONS Solutions
We have, 1 2 a p1 = ∆, 1 2 b p2 = ∆, 1 2 c p3 = ∆ ⇒p1 = 2 ∆ a ,p2 = 2 ∆ b ,p3 = 2 ∆ c ∴ 1 p 2 1 + 1 p 2 2 + 1 p 2 3 = a 2 + b 2 + c 2 4 ∆ 2 1 p1 + 1 p2 ― 1 p3 = a 2 ∆ + b 2 ∆ ― c 2 ∆ = a + b ― c 2 ∆ = 2(s ― c) 2 ∆ = s ― c ∆ 706 (c) We have, cos C = 63 65 ⇒ a 2 + b 2 ― c 2 2ab = 63 65 ⇒ 262 + 302 ― c 2 2 × 26 × 30 = 63 65 ⇒676 + 900 ― c 2 = 1260⇒c 2 = 64⇒c = 8 Thus, we have a = 26,b = 30 and c = 8 ∴ 2s = a + b + c⇒2s = 26 + 30 + 8 = 64⇒s = 32 Also, ∆ = s(s ― a)(s ― b)(s ― c) = 32 × 6 × 2 × 24 = 96 Hence, r2 = ∆ s ― b = 96 32 ― 30 = 48 707 (c) cos 1° cos 2° cos 3°... cos 90°... cos 100° = cos 1° cos 2° cos 3°...0... cos 100° = 0 708 (b) We have, sin π 2 + sin 2π 7 + sin 3π 7 = 1 2 sin π 7 {2 sin2 π 7 + 2 sin π 7 sin 2π 7 + 2 sin π 7 sin 3π 7 } = 1 2 sin ( π 7 ) {1 ― cos 2π 7 + cos π 7 ― cos 3π 7 + cos 2π 7 ― cos 4π 7 } = 1 2 sin π 7 {1 + cos π 7 } = 2 cos2 π 14 4 sin π 14 cos π 14 = 1 2 cot π 14 709 (a) Let f(x) = 3cos x + sin x
⇒f(x) = 2( 3 2 cos x + 1 2 sin x) = 2 sin (x + π 3 ) Since, ―1 ≤ sin (x + π 3 ) ≤ 1 Hence, f(x)is maximum, if x + π 3 = π 2 ⇒x = π 6 = 30° 710 (b) sin2 17.5° + sin2 72.5 ° = sin2 17.5° + cos2 17.5° [ ∵ sin(9θ ― θ) = cos θ ] = 1 = tan2 45° 711 (a) We have, a sinA = b sinB ⇒a ∙ ak = b ∙ bk⇒a = b⇒∆ABC is isosceles 712 (b) We know that sin2 θ ≥ 1 ⇒ 4xy (x + y) 2 ≥ 1 ⇒ 4xy ≥ (x + y) 2 ⇒ (x ― y) 2 ≤ 0 ⇒ x ― y = 0 ⇒ y = x And x ≠ 0, y ≠ 0 713 (b) Given that, cos θ = 1 2 (x + 1 x ) ⇒x + 1 x = 2cos θ ....(i) We know that, x 2 + 1 x 2 = (x + 1 x ) 2 ―2 = (2 cos θ) 2 ― 2 = 4 cos2 θ ― 2 = 2cos 2θ [from Eq.(i)] ∴ 1 2 (x 2 + 1 x 2) = 1 2 × 2 cos 2θ = cos 2θ 714 (d) sech―1 (sin θ) = cosh―1 (cosec θ)
= log [cosec θ + (cosec2 θ ― 1)] = log [ 1 sin θ + cos θ sin θ ] = log cot θ 2 715 (d) Consider the curves y = 2 cos x and y = |sin x|. Clearly, both the curves are symmetrical about y- axis as cos x and |sin x| are even functions Also, y = 2 cos x and y = |sin x| intersect at two points in [0, 2 π] Hence, there are four solutions of the given equation 716 (d) We have, cos(λ sin θ ) = sin(λ cos θ) ⇒ cos(λ sin θ ) = cos ( π 2 ― λ cos θ) ⇒λ sin θ = π 2 ― λ cos θ⇒ cos θ + sin θ = π 2 λ This equation will have a solution if | π 2 λ ≤ 2| [ ∵ |a cos θ + b sin θ| ≤ a 2 + b 2 ] ⇒ π 2 λ ≤ 2⇒λ ≥ π 2 2 [ ∵ λ > 0] 717 (c) We have, c1 + c2 = 2bcosA and c1c2 = b 2 ― a 2 ∴ c1 ― c2 = (c1 + c2 ) 2 ― 4c1c2 ⇒c1 ― c2 = 4b 2 cos2A ― 4(b 2 ― a 2 ) = 2 a 2 ― b 2 sin2A 718 (b) We have, tanα = (1 + 2 ―x ) ―1 = 2 x 2 x + 1 and tanβ = 1 2 x+1 + 1 ∴ tan(α + β) = tanα + tanβ 1 ― tanα tanβ ⇒ tan(α + β) = 2 x (2 x+1 + 1) + (2 x + 1) (2 x + 1)(2 x+1 + 1) ― 2 x ⇒ tan(α + β) = 2(2 x ) 2 + 2.2x + 1 2(2 x ) 2 + 2.2x + 1 = 1⇒α + β = π/4 719 (a) Given, f(x) = sin x(1 + cos x )