1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
  3. Ex 2.2
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Ex 2.2, 14 (MCQ) - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at Dec. 16, 2024 by Teachoo

 

Transcript

Ex 2.2, 14 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
  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise

    3. Ex 2.2

    Examples →
Examples →
Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo