Ex 2.1, 12 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Ex 2.1, 12 Find the value of cos−1 (1/2) + 2 sin−1 (1/2) Solving cos−1 (𝟏/𝟐) Let y = cos−1 (1/2) cos y = (1/2) cos y = cos (π/3) Range of principal value of cos−1 is [0 , 𝜋] Hence, the principal value is π/3. Rough We know that cos 60° = 1/2 θ = 60° = 60° × 𝜋/180 = 𝜋/3 Since 1/2 is positive Principal value is θ i.e. 𝜋/3 Solving sin−1 (𝟏/𝟐) Let y = sin−1 (1/2) sin y = 1/2 sin y = sin (π/6) Range of principal value of sin−1 is [(−𝜋)/2 " , " 𝜋/2] Hence, the principal value is π/6 Rough We know that sin 30° = 1/2 θ = 30° = 30 × 𝜋/180 = 𝜋/6 Since 1/2 is positive Principal value is θ i.e. 𝜋/6 Now we have cos−1 1/2 = 𝜋/3 & sin−1 1/2 = 𝜋/6 Solving cos−1 𝟏/𝟐 + 2sin−1 𝟏/𝟐 = 𝜋/3 + 2 × 𝜋/6 = 𝜋/3 + 𝜋/3 = (𝜋 + 𝜋)/3 = 𝟐𝝅/𝟑