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Ex 2.2, 2 - 3cos -1 x = cos-1 (4x3 - 3x) - Chapter 2 Class 12

Ex 2.2, 2 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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Ex 2.2, 2 3cosβˆ’1 π‘₯ = cosβˆ’1 (4π‘₯^3βˆ’ 3π‘₯ ), π‘₯∈ [1/2,1] Solving R.H.S cos^(βˆ’1) (4π‘₯^3βˆ’ 3π‘₯ ) Putting x = cos πœƒ = cos^(βˆ’1) (4 γ€–"cos" γ€—^πŸ‘πœƒ βˆ’ 3cos πœƒ) = cos^(βˆ’1) (cos 3πœƒ) = 3πœƒ = 3 cos^(βˆ’1) x (cos 3x = 4 cos^3x βˆ’ 3 cos x) Now, x = cos πœƒ ∴ cos^(βˆ’1) (x) = πœƒ = L.H.S Hence, proved.

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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.