Ex 2.2, 16 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Dec. 8, 2016 by Teachoo
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 principal value 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 Which is in the range of principal value of sin-1 i.e. [(−π)/2, π/2] Hence, sin-1(sin 2π/3) = y = 𝜋/3
Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.