Ex 2.2, 20 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Jan. 7, 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) y = (−π)/6 ∴ 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
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Chapter 2 Class 12 Inverse Trigonometric Functions
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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.