Tiêu đề TRIGONOMETRIC FUNCTIONS A-10.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :10 881 (b) Let f(x) = sin x cos x = 1 2 sin 2x We know that, ― 1 ≤ sin 2x ≤ 1 ⇒ ― 1 2 ≤ 1 2 sin 2x ≤ 1 2 Thus, the greatest and least value of f(x) are 1 2 and 1 2 respectively 882 (b) We have, x 2 + 4xy + y 2 = (X cos θ ― Y sin θ) 2 + 4(X cos θ ― Y sin θ)(X sin θ + Y cos θ) + (X sin θ + Y cos θ) 2 = (1 + 4 sin θ cos θ)X 2 + 4(cos2 θ ― sin2 θ)XY + (1 ― 4 sin θ cos θ)Y 2 ∴ x 2 + 4xy + y 2 = AX 2 + BY 2 ⇒(1 + 2 sin 2 θ)X 2 + 4 cos 2 θ XY + (1 ― 2 sin 2 θ)Y 2 = AX 2 + BY 2 ⇒ cos 2 θ = 0, A = 1 + 2 sin 2 θ, B = 1 ― 2 sin 2 θ ⇒θ = π 4 and A = 1 + 2 = 3,B = 1 ― 2 = ―1 883 (d) Given, 3cos 2x ― 10 cos x + 7 = 0 ⇒ 6 cos2 x ― 10 cos x + 4 = 0 [ ∵ cos 2x = 2 cos2 x ― 1] ⇒ 2(3 cos x ― 2)(cos x ― 1) = 0 ⇒ cos x = 1 or cos x = 2 3 Since, cos x is positive in Ist and IIIrd quadrant. Hence, total number of solutions are 4 884 (a) cosα sin(β ― γ) + cosβ sin(γ ― α) + cos γ sin(α ― β) Topic :-TRIGONOMETRIC FUNCTIONS Solutions
=cosα [sinβ cos γ ― cosβ sin γ] + cosβ[ sin γ cosα ― cos γ sinα ] + cos γ[ sinα cosβ ― cosα sinβ ] = 0 885 (d) Given, A + B = 45° ⇒ cot(A + B) = 1 ⇒ cotA cotB ― 1 cotA + cotB = 1 ⇒ cotAcot B ― ( cotA + cotB ) = 1 ...(i) Now, (cotA ― 1)(cotB ― 1) = cotAcotB ― (cotA + cotB) + 1 = 1 + 1 = 2 [from Eq. (i)] 886 (c) Let I = [sin x + cos x] 1+sin 2x = [ 2sin ( π 4 + x)] 1+sin 2x At x = π 4 , I = [ 2sin ( π 4 + π 4 )] 1+sin 2π 4 = ( 2) 2 = 2 887 (d) The given expression can be written as cos6 x(cos6 x + 3 cos4 x + 3 cos2 x + 1) + 2 cos4 x + cos2 x ― 2 = sin3 x(cos2 x + 1) 3 + 2 cos4 x + cos2 x ― 2 = sin3 x(sin x + 1) 3 + 2 sin2 x + cos2 x ― 2 [ ∵ sin x + sin2 x = 1⇒ sin x = cos2 x] = (sin x + sin2 x) 3 + sin2 x + (sin2 x + cos2 x) ― 2 = 1 3 + sin2 x + 1 ― 2 = sin2 x 888 (b) sin4 π 8 + sin4 3π 8 + sin4 5π 8 + sin4 7π 8 = 1 4[(2 sin2 π 8 ) 2 + (2 sin2 3π 8 ) 2 ] + 1 4[(2 sin2 π 8 ) 2 + (2 sin2 3π 8 ) 2 ] = 1 4[(1 ― cos π 4 ) 2 + (1 ― cos 3π 4 ) 2 ] + 1 4[(1 ― cos π 4 ) 2 + (1 ― cos 3π 4 ) 2 ] = 1 4[(1 ― 1 2 ) 2 + (1 + 1 2 ) 2 ] + 1 4[(1 ― 1 2 ) 2 + (1 + 1 2 ) 2 ]
= 1 4 (3) + 1 4 (3) = 3 2 889 (c) Given, sin(x + 3α) sin(α ― x) = 3 Applying componendo and dividendo, we get sin(x + 3α) + sin(α ― x) sin(x + 3α) ― sin(α ― x) = 3 + 1 3 ― 1 ⇒ 2 sin 2α cos(α + x) 2 cos 2α sin(α + x) = 2 ⇒ tan 2α tan(α + x) = 2 ⇒ 2 tanα 1 ― tan2α × (1 ― tanα tan x ) ( tanα + tan x ) = 2 ⇒ tanα ― tan2α tan x = tanα + tan x ― tan3α ― tan2α tan x ⇒ tan x = tan3α 890 (a) We have, cosα sin(β ― γ) + cosβ sin(γ ― β) + cos γ sin(α ― β) = 1 2 {sin(α + β ― γ) + sin(β ― γ ― α) + sin(γ ― α + β) + sin(γ ― α ― β) + sin(α ― β + γ) + sin(α ― β ― γ)} = 1 2 {sin(α + β ― γ) ― sin(α ― β + γ) ― sin(α ― β ― γ) ― sin(α + β ― γ) + sin(α ― β + γ) + sin(α ― β ― γ)} = 1 2 × 0 = 0 891 (c) sinA + cosA = m [given] ⇒ sin3A + cos3 A + 3 cosA sinA ( sinA + cosA ) = m3 ⇒ n + 3msinA cosA = m3 ....(i) [ ∵ sin3A + cos3A = n ] Again, sinA + cosA = m ⇒ sin2A + cos2A + 2sin A cosA = m2 ⇒ sinA cosA = m2 ― 1 2 ...(ii) From Eqs. (i) and (ii), we get n + 3m (m2 ― 1) 2 = m3 ⇒ 2n + 3m3 ― 3m = 2m3 ⇒ m3 ― 3m + 2n = 0 892 (b)
We have, (sec θ ― 1) = ( 2 ― 1)tan θ ⇒1 ― cos θ = ( 2 ― 1) sin θ ⇒2 sin2 θ 2 = 2( 2 ― 1)sin θ 2 cos θ 2 ⇒ sin θ 2 = 0 or, tan θ 2 = 2 ― 1 = tan π 8 ⇒ θ 2 = n π or, θ 2 = n π + π 8 ,n ∈ Z ⇒θ = 2 n π, θ = 2 n π + π 4 ,n ∈ Z 893 (d) Given, cos θ + sin 2θ = 0 ⇒ cos θ + 2 sin θ cos θ = 0 ⇒ cos θ(1 + 2 sin θ) = 0 ⇒ cos θ = 0 or sin θ = ― 1 2 For θ ∈ [ ― π, π] θ = π 2 , ― π 2 Or θ = ― π 6 , ― 5π 6 894 (b) We have, tan ( π 2 sin θ) = cot ( π 2 cos θ) ⇒ tan ( π 2 sin θ) = tan ( π 2 ― π 2 cos θ) ⇒ π 2 sin θ = r π + π 2 ― π 2 cos θ, r ∈ Z ⇒ sin θ + cos θ = (2r + 1) ,r ∈ Z ⇒ 1 2 sin θ + 1 2 cos θ = 2r + 1 2 ,r ∈ Z ⇒ cos (θ ― π 4 ) = 2r + 1 2 ,r ∈ Z ⇒ cos (θ ― π 4 ) = 1 2 or ― 1 2 [For r = 0, ― 1] ⇒θ ― π 4 = 2 r π ± π 4 ,r ∈ Z ⇒θ = 2 r π ± π 4 + π 4 , r ∈ Z ⇒θ = 2 r π, 2 r π + π 2 ,r ∈ Z But, θ = 2 r π + π 2 ,r ∈ Z gives extraneous roots as it does not satisfy the given equation. Therefore, θ = 2 r π,r ∈ Z 895 (b)