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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
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Chapter 2 Class 12 Inverse Trigonometric Functions
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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.