Example 3 (i) - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at Dec. 16, 2024 by

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Example 3 Show that sinβˆ’1 (2x√(1βˆ’π‘₯2)) = 2 sin-1x Solving L.H.S. sinβˆ’1 ( 2x √(1βˆ’π‘₯2) ) Putting x = sin ΞΈ = sinβˆ’1 ("2 sin ΞΈ " √(πŸβˆ’π’”π’Šπ’πŸ" ΞΈ" )) = sinβˆ’1 ("2 sin ΞΈ " √(π’„π’π’”πŸ" ΞΈ" )) = sinβˆ’1 (2sin ΞΈ cos ΞΈ) = sinβˆ’1 (sin 2ΞΈ) We need to make 2x √(πŸβˆ’π’™πŸ) in terms of sin When we get √(1βˆ’π‘₯2) , we put x = cos ΞΈ or sin ΞΈ = 2ΞΈ = 2 Γ— sinβˆ’1 x = 2 sinβˆ’1 x = R.H.S. Since L.H.S. = R. H. S. Hence proved As x = sin ΞΈ sin-1 x = ΞΈ

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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