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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


Transcript

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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