Ex 2.2,1 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at May 12, 2021 by Teachoo
Last updated at May 12, 2021 by Teachoo
Transcript
Ex 2.2, 1 3sin-1 π₯ = sin-1 (3π₯ β 4π₯^3), π₯β [β1/2,1/2] Solving R.H.S sin^(β1) (3π₯ β 4π₯^3) Putting x = sin π = sin^(β1) (3 sin π β 4 γ"sin" γ^3π) = sin^(β1) (sin 3π ) = 3π = 3 γπ ππγ^(β1) x (sin 3x = 3 sin x β 4 sin^3x) Now, x = sin π sinβ1 x = π = L.H.S Hence, proved.
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