Last updated at May 12, 2021 by Teachoo

Transcript

Ex 2.1, 1 (Method 1) Find the principal value of sin-1 (−1/2) Let y = sin-1 ((−1)/2) y = − sin-1 (1/2) y = − 𝛑/𝟔 Since Range of sin −1 is [(−𝜋)/2, ( 𝜋)/2] Hence, Principal Value is (−𝝅)/𝟔 We know that sin−1 (−x) = − sin −1 x Since sin 𝜋/6 = 1/2 𝜋/6 = sin−1 (𝟏/𝟐) Ex 2.1, 1 (Method 2) Find the principal value of sin-1 (−1/2) Let y = sin-1 (−1/2) sin y = (−1)/2 sin y = sin ((−𝝅)/𝟔) Since Range of of sin −1 is [(−𝜋)/2, ( 𝜋)/2] Hence, Principal Value is (−𝝅)/𝟔 Rough We know that sin 30° = 1/2 θ = 30° = 30 × 𝜋/180 = 𝜋/6 Since (−1)/2 is negative Principal value is −θ i.e. (−𝜋)/6

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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 10 years. He provides courses for Maths and Science at Teachoo.