Last updated at May 12, 2021 by Teachoo
Transcript
Ex 2.1, 5 Find the principal value of cos−1 (−1/2) Let y = cos−1 ((−1)/2) cos y = (−1)/2 cos y = cos (𝟐𝝅/𝟑) Since Range of cos−1 is [0 , 𝜋] Hence, the principal value is 𝟐𝝅/𝟑 Rough We know that cos 60° = 1/2 θ = 60° = 60° × 𝜋/180 = 𝜋/3 Since (−1)/2 is negative Principal value is 𝝅 – θ i.e. π −𝜋/3 = 𝟐𝝅/𝟑 Ex 2.1, 5 (Method 1) Find the principal value of cos−1 (−1/2) Let y = cos−1 ((−1)/2) y = 𝜋 − cos−1 (1/2) y = 𝜋 − 𝝅/𝟑 y = 𝟐𝝅/𝟑 Since Range of cos−1 is [0, 𝜋] Hence, the principal value is 𝟐𝝅/𝟑 We know that cos−1 (−x) = 𝜋 − cos−1 x Since cos 𝜋/3 = 1/2 𝜋/3 = cos−1 (1/2) Ex 2.1, 5 (Method 2) Find the principal value of cos−1 (−1/2) Let y = cos−1 ((−1)/2) cos y = (−1)/2 cos y = cos (𝟐𝝅/𝟑) Since Range of cos−1 is [0, 𝜋] Hence, the principal value is 𝟐𝝅/𝟑 Rough We know that cos 60° = 1/2 θ = 60° = 60° × 𝜋/180 = 𝜋/3 Since (−1)/2 is negative Principal value is 𝝅 – θ i.e. π −𝜋/3 = 𝟐𝝅/𝟑
Chapter 2 Class 12 Inverse Trigonometric Functions
Ex 2.1, 8 Important
Ex 2.1, 12 Important
Ex 2.1, 14 Important
Example 5 Important Deleted for CBSE Board 2021 Exams only
Example 8 Deleted for CBSE Board 2021 Exams only
Ex 2.2, 12 Important Deleted for CBSE Board 2021 Exams only
Ex 2.2, 15 Important Deleted for CBSE Board 2021 Exams only
Ex 2.2, 19 Important Deleted for CBSE Board 2021 Exams only
Ex 2.2, 21 Important Deleted for CBSE Board 2021 Exams only
Example 10 Important Deleted for CBSE Board 2021 Exams only
Example 12 Important Deleted for CBSE Board 2021 Exams only
Example 13 Important Deleted for CBSE Board 2021 Exams only
Misc. 2 Important
Misc. 7 Important Deleted for CBSE Board 2021 Exams only
Misc. 10 Important Deleted for CBSE Board 2021 Exams only
Misc. 11 Important Deleted for CBSE Board 2021 Exams only
Misc 12 Important Deleted for CBSE Board 2021 Exams only
Misc. 17 Important Deleted for CBSE Board 2021 Exams only
Chapter 2 Class 12 Inverse Trigonometric Functions
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