Ex 2.2, 16 - Find sin-1 (sin⁡ 2pi/3) - Chapter 2 NCERT - Ex 2.2

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Ex 2.2, 16 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise

Transcript

Ex 2.2, 16 Find the values of sin-1(sin⁡〖2π/3〗 ) Let y = sin-1 (sin 2𝜋/3) sin y = sin 2𝜋/3 sin y = sin (120°) But, Range of sin-1 is [(−π)/2, π/2] i.e. [-90° ,90° ] Hence, y = 120° not possible Now, sin y = sin (120°) sin y = sin (180° – 60°) sin y = sin (60°) sin y = sin (60 × 𝜋/180) sin y = sin 𝜋/3 Hence, y = 𝝅/𝟑 Since this is in range of sin-1 i.e. [(−π)/2, π/2] Hence, sin-1(sin 𝟐𝛑/𝟑) = 𝝅/𝟑 (As sin (180 – θ) = sin θ)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.