1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
  3. Ex 2.2

Ex 2.2, 20 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at May 12, 2021 by

Ex 2.2, 20 - Find sin (pi/3 - sin-1 (-1/2)) - Class 12 NCERT

Ex 2.2, 20 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Ex 2.2, 20 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3

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

    3. Ex 2.2

    Examples →

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) y = − sin-1 (1/2) y = − 𝛑/𝟔 We know that sin−1 (−x) = − sin −1 x Since sin 𝜋/6 = 1/2 𝜋/6 = sin−1 (𝟏/𝟐) 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

Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.