1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Concept wise
  3. Formulae based

Ex 2.2,1 - Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Last updated at May 12, 2021 by

Ex 2.2, 1 - 3 sin-1 x = sin-1 (3x - 4x3) - Chapter 2 Inverse

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Ex 2.2,1 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Concept wise

    3. Formulae based

    Changing of trignometric variables and then applying formula →

Transcript

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 〖𝑠𝑖𝑛〗^(βˆ’1) x (sin 3x = 3 sin x βˆ’ 4 sin^3x) Now, x = sin πœƒ sinβˆ’1 x = πœƒ = L.H.S Hence, proved.

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
Concept wise

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