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Solve sin-1 (1 - x) - 2 sin-1 x = pi/2, then x is - Trigonometry

Misc 16 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Misc 16 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3
Misc 16 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4
Misc 16 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 5

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Misc 16 Solve sin−1(1 – x) – 2sin−1 x = π/2 , then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2 sin−1 (1 – x) – 2sin−1 x = π/2 –2sin−1 x = 𝝅/𝟐 – sin−1 (1 – x) − 2sin−1 x = cos−1 (1 – x) We know that sin−1 x + cos−1x = 𝝅/𝟐 Replace x by (1 − x) sin-1 (1 − x) + cos−1 (1 − x) = 𝜋/2 cos-1 (1 − x) = 𝜋/2 – sin−1 (1 − x) Let sin−1 x = a, Hence our equation becomes −2a = cos−1 (1 – x) cos (−2a) = 1 – x cos (2a) = (1 – x) 1 – 2 sin2 a = 1 – x We assumed that sin−1 x = a 1 – 2 [sin(sin−1 x)]2 = 1 – x 1 – 2x2 = 1 – x 1 – 2x2 – 1 + x = 0 1 – 1 – 2x2 + x = 0 –2x2 + x = 0 0 = 2x2 – x 2x2 – x = 0 x (2x – 1) = 0 So, x = 0 and x = 1/2 But x = 𝟏/𝟐 does not satisfy the equation Taking equation sin−1(1 – x) – 2sin−1 x = π/2 Putting x = 𝟏/𝟐 in L.H.S sin−1(1− 1/2) – 2 sin−1 (1/2) = sin−1(1/2) – 2 sin−1 (1/2) = 𝜋/6 – 2 × 𝜋/6 = (𝜋 − 2𝜋)/6 = (− 𝜋)/6 ≠ 𝝅/𝟐 Hence x = 1/2 not possible ∴ x = 0 is the only solution Option C is correct Answer

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.