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

Transcript

Ex 2.1, 12 Find the value of cosβˆ’1 (1/2) + 2 sinβˆ’1 (1/2) Solving cosβˆ’1 (𝟏/𝟐) Let y = cosβˆ’1 (1/2) cos y = (1/2) cos y = cos (Ο€/3) Range of principal value of cosβˆ’1 is [0 , πœ‹] Hence, the principal value is Ο€/3. Rough We know that cos 60Β° = 1/2 ΞΈ = 60Β° = 60Β° Γ— πœ‹/180 = πœ‹/3 Since 1/2 is positive Principal value is ΞΈ i.e. πœ‹/3 Solving sinβˆ’1 (𝟏/𝟐) Let y = sinβˆ’1 (1/2) sin y = 1/2 sin y = sin (Ο€/6) Range of principal value of sinβˆ’1 is [(βˆ’πœ‹)/2 " , " πœ‹/2] Hence, the principal value is Ο€/6 Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ— πœ‹/180 = πœ‹/6 Since 1/2 is positive Principal value is ΞΈ i.e. πœ‹/6 Now we have cosβˆ’1 1/2 = πœ‹/3 & sinβˆ’1 1/2 = πœ‹/6 Solving cosβˆ’1 𝟏/𝟐 + 2sinβˆ’1 𝟏/𝟐 = πœ‹/3 + 2 Γ— πœ‹/6 = πœ‹/3 + πœ‹/3 = (πœ‹ + πœ‹)/3 = πŸπ…/πŸ‘

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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.