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Ex 2.2, 1 3sin-1 π‘₯ = sin-1 (3π‘₯ βˆ’ 4π‘₯^3), π‘₯∈ [βˆ’1/2,1/2] Solving R.H.S sin^(βˆ’1) (3π‘₯ βˆ’ 4π‘₯^3) Putting x = sin πœƒ = sin^(βˆ’1) (3 sin πœƒ βˆ’ 4 γ€–"sin" γ€—^3πœƒ) = sin^(βˆ’1) (sin 3πœƒ ) = 3πœƒ = 3 γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) x Now, x = sin πœƒ sinβˆ’1 x = πœƒ (sin 3x = 3 sin x βˆ’ 4 〖𝑠𝑖𝑛〗^3x) (〖𝑠𝑖𝑛〗^(βˆ’1) (sin x) = x ) = L.H.S Hence, proved.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.