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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.