Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg
Ex 2.2, 20 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Feb. 13, 2020 by Teachoo
Transcript
Ex 2.2, 20 Find the values of sin (π/3 −"sin−1" (−1/2)) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1 Solving sin-1 (−𝟏/𝟐) Let y = sin−1 (−1/2) sin y = (−1)/2 sin y = sin ((−π)/6) Therefore, y = (−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 Thus, sin−1 (−1/2) = (−π)/6 Solving sin (𝝅/𝟑 " – sin−1 " ((−𝟏)/𝟐)) = sin [𝜋/3 " − " ((−𝜋)/6)] = sin (𝜋/3 " + " 𝜋/6) = sin ((180° )/3 " + " (180° )/6) = sin (60° + 30°) = sin 90° = 1 Hence, D is correct answer
Ex 2.2
Ex 2.2,1
Ex 2.2, 2 Important
Ex 2.2, 3 Important
Ex 2.2, 4 Important
Ex 2.2, 5 Important
Ex 2.2, 6 Important
Ex 2.2, 7 Important
Ex 2.2, 8 Important
Ex 2.2, 9
Ex 2.2, 10 Important
Ex 2.2, 11
Ex 2.2, 12 Important
Ex 2.2, 13 Important
Ex 2.2, 14 Important
Ex 2.2, 15 Important
Ex 2.2, 16
Ex 2.2, 17
Ex 2.2, 18 Important
Ex 2.2, 19 Important
Ex 2.2, 20 Important You are here
Ex 2.2, 21 Important
Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.