Chapter 2 Inverse Trigonometric Functions
- cos-1(cos\(\frac { 13π }{ 6 }\)) cos⁻¹(cos 13π / 6 ) = cos⁻¹cos(2π + π / 6 = cos⁻¹cos π / 6 ) = π / 6 1
- tan-1(tan\(\frac { 7π }{ 6 }\)) tan⁻¹(tan 7π / 6 ) = tan⁻¹tan(π + π / 6 = tan⁻¹tan π / 6 ) = π / 6 . 1
- tan-1\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos-1\(\frac { 1-x }{ 1+x }\), x ∈ [0, 1] Put x = tan²θ. ∴ θ = tan⁻¹ √(x) R.H.S. = 1 / 2 cos⁻¹ 1-x / 1+x = 1 / 2 cos⁻¹ (frac1-tan ² θ1+tan ² θ) = 1 / 2 cos⁻¹(cos2θ) = 1 / 2 x 2θ = θ.... 1
- 2tan-1(cos x) = tan-1(cosec x) Now L.H.S = 2tan⁻¹(cos x) = tan⁻¹ (frac2 cos x1-cos ² x) = tan⁻¹ (frac2 cos xsin ² x) Putting this value in the given equation, we get tan⁻¹... 1
- sin tan-1x, |x| < 1 is equal to (A) \(\frac{x}{\sqrt{1-x^{2}}}\) (B) \(\frac{1}{\sqrt{1-x^{2}}}\) Let tan⁻¹x = α ∴ tan α = x. So, sin α = fracxsqrt1+x² or α = sin⁻¹ fracxsqrt1+x² Now sin tan⁻¹x = sin α = sin(sin⁻¹ fracxsqrt1+x² ) = fracxs... 1
- If sin-1(1-x)-2 sin-1x = \(\frac { π }{ 2 }\), then x is equal to (A) 0, \(\frac { 1 }{ 2 }\) sin⁻¹(1-x)-2 sin⁻¹x = π / 2 Putting π / 2 = sin⁻¹(1 – x) + cos⁻¹(1 – x), sin⁻¹(1 – x) – 2 sin⁻¹ = sin⁻¹(1 – x) + cos⁻¹(1 – x) ⇒ – 2 sin⁻¹x =... 1
- 3sin-1x = sin-1(3x – 4x³), x ∈ [- \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\) ] Let sin⁻¹ = θ ∴ sin θ = x Also, ∴ sin 3θ = 2 sinθ – 4sin³θ = 3x – 4x³ ∴ 3θ = sin⁻¹(3x – 4x³) or 3 sin⁻¹x = sin⁻¹(3x – 4x³), x ∈ [- 1 / 2 , 1... 1
- 3cos-1x = cos-1(4x³ – 3x), x ∈ [\(\frac { 1 }{ 2 }\), 1] Let cos⁻¹x = θ. cos θ = x. Now, cos 3θ = 4cos³θ – 3 cos θ = 4x³ – 3x. or 3θ = cos⁻¹(4x³ – 3x) or 3 cos⁻¹x = cos⁻¹(4x³ – 3x), x ∈ [- 1 / 2 ,... 1
- If sin(sin-1\(\frac { 1 }{ 5 }\) + cos-1x) = 1, then find the value of x. sin(sin⁻¹ 1 / 5 + cos⁻¹x) = sin π / 2 ⇒ sin⁻¹ 1 / 5 + cos⁻¹x = sin π / 2 or (sin⁻¹ 1 / 5 + cos⁻¹ 1 / 5 ) + (- cos⁻¹ 1 / 5 x) = π / 2 or π /... 1
- If tan-1\(\frac { x-1 }{ x+2 }\) + tan-1\(\frac { x+1 }{ x+2 }\) = \(\frac { π }{ 4 }\), then find the value of x. L.H.S. = tan⁻¹ x-1 / x+2 + tan⁻¹ x+1 / x+2 = tan⁻¹[ x-1 / x+2 + x+1 / x+2 /1-( x-1 / x+2 )( x+1 / x+2 )] = tan⁻¹[2x²-4/-3] = πover4=tan⁻¹(1)... 1
- If sin-1(sin\(\frac { 2π }{ 3}\)) sin⁻¹(sin 2π / 3 ) = sin⁻¹[ sin(π – π / 3 ) ] = sin⁻¹[sin π / 3 ] = π / 3 . Please note : sin⁻¹(sin 2π / 3 ) ≠ sin 2π / 3 , since the range... 1
- tan-1(tan\(\frac { 3π }{ 4 }\)) tan⁻¹(tan 3π / 4 ) = tan⁻¹ tan(π – π / 4 ) = tan⁻¹(- tan π / 4 ) = tan⁻¹[tan( -π / 4 ) ] = – π / 4 Again note that tan⁻¹ tan 3π / 4 ≠ 3π / 4... 1
- cos-1(cos \(\frac { 7π }{ 6 }\)) is equal to (A) \(\frac { 7π }{ 6 }\) (B) \(\frac { 5π }{ 6 }\) (B) 5π / 6 cos⁻¹(cos 7π / 6 ) = cos⁻¹cos ( π + π / 6 ) = cos⁻¹(- cos π / 6 ) = cos⁻¹[cos ( π – π / 6 ) = cos⁻¹cos 5π / 6 = 5π / 6 . Note tha... 1
- sin[ \(\frac { π }{ 3 }\)-sin-1(-\(\frac { 1 }{ 2 }\)) ] is equal to (A) \(\frac { 1 }{ 2 }\) Answer: (D) 1 The principal value branch of sin⁻¹ is [-π/2, π/2] , and sin(-π/6) = -1/2 . So sin⁻¹(-1/2) = -π/6. Then sin[π/3 + π/6] = sinπ/... 1
- tan-1\(\sqrt{3}\) – cot-1(-\(\sqrt{3}\)) is equal to (A) π (B) \(\frac { π }{ 2 }\) (C) 0 (D) 2\(\sqrt{3}\) (B) π / 2 tan⁻¹ √(3) = tan⁻¹(tan π / 3 ) = π / 3 and cot⁻¹ √(3) = cot⁻¹(-cot π / 6 ) = cot⁻¹cot ( π – π / 6 ) = cot⁻¹cot 5π / 6 = 5π / 6 . s... 1
- sin-1(-\(\frac { 1 }{ 2 }\)) Let sin⁻¹(- 1 / 2 ) = y ∴ sin y = 1 / 2 = – sin π / 6 = sin(- π / 6 ) The range of principal value branch of sin⁻¹ is [- π / 2 , π / 2 ]. He... 1
- cos-1\(\left(-\frac{\sqrt{3}}{2}\right)\) Answer: cos⁻¹(-√(3)/2) = 5π/6 The principal value branch of cos⁻¹ is [0, π] . We know cosπ/6 = √(3)/2 . Using cos(π - θ) = -cosθ : cos(π - π... 1
- cos-1(-\(\frac { 1 }{ 2 }\)) Let cos⁻¹(- 1 / 2 ) = y ∴ cos y = – 1 / 2 = – cos π / 3 = cos(π – π / 3 = cos 2π / 3 The range of principal value branch is (0, π) ∴ princip... 1
- Find the principal value of tan-1(-1) Answer: tan⁻¹(-1) = -π/4 The principal value branch of tan⁻¹ is (-π/2, π/2) . Since tanπ/4 = 1 and tan is an odd function, tan(-π/4) = -1. -... 1
- sec-1\(\left(-\frac{2}{\sqrt{3}}\right)\) Answer: sec⁻¹(-2/√(3)) = 5π/6 The principal value branch of sec⁻¹ is [0, π] - \π/2\ . We know secπ/6 = 2/√(3) . For a negative value we use... 1
- cot-1 (\(\sqrt{3}\)) Let cot⁻¹( √(3) ) = y ∴ cot y = √(3) = cot π / 6 The range of principal value branch of cot⁻¹ is [0, π] ∴ principal value of cot⁻¹ √(3) is π... 1
- cos-1\(\left(-\frac{1}{\sqrt{2}}\right)\) Let cos⁻¹ (-1/√(2)) = y ∴ cos y = (-1/√(2)) = cos( π / 4 ) = cos(π – π / 4 ) The range of principal value branch of cos⁻¹ is [0, π]. ∴ princ... 1
- cosec-1 (-\(\sqrt{2}\)) Let cosec-1(- √(2) ) = y ∴ cosec y = – √(2) = – cosec π / 4 = cosec (- π / 4 ). The range of principal value branch of cosec-1 is (- π / 2 ,... 1
- tan-1(1) + cos-1(-\(\frac { 1 }{ 2 }\)) + sin-1(\(\frac { -1 }{ 2 }\)) Answer: 3π/4 Use the principal value branches: tan⁻¹(1) = π/4 , since tanπ/4 = 1 and π/4in(-π/2, π/2) . cos⁻¹(-1/2) = π - cos⁻¹1/2 = π - π/3... 1