
Get live Maths 1-on-1 Classs - Class 6 to 12
Miscellaneous
Misc. 2 Important
Misc. 3 Deleted for CBSE Board 2023 Exams
Misc. 4 Important Deleted for CBSE Board 2023 Exams
Misc. 5 Deleted for CBSE Board 2023 Exams
Misc. 6 Deleted for CBSE Board 2023 Exams
Misc. 7 Important Deleted for CBSE Board 2023 Exams
Misc. 8 Important Deleted for CBSE Board 2023 Exams
Misc. 9 Important Deleted for CBSE Board 2023 Exams
Misc. 10 Important Deleted for CBSE Board 2023 Exams
Misc. 11 Important Deleted for CBSE Board 2023 Exams
Misc 12 Important Deleted for CBSE Board 2023 Exams
Misc. 13 Important Deleted for CBSE Board 2023 Exams
Misc. 14 Deleted for CBSE Board 2023 Exams
Misc 15 (MCQ) Important Deleted for CBSE Board 2023 Exams
Misc 16 (MCQ) Important Deleted for CBSE Board 2023 Exams
Misc 17 (MCQ) Deleted for CBSE Board 2023 Exams
Miscellaneous
Last updated at March 22, 2023 by Teachoo
Misc 1 Find the value of cos-1 (cos〖13π/6〗 ) Let y = cos-1 (cos〖13π/6〗 ) cos y = cos 13π/6 cos y = cos (390°) But, Range of cos−1 is [0, π] i.e. [0°, 180°] Hence, y = 390° not possible Now, cos y = cos (390°) cos y = cos (360° + 30°) cos y = cos (30°) cos y = cos (𝜋/6) ∴ y = 𝝅/𝟔 Which is in the range of cos-1 i.e. [0, π] Hence , cos −1 (cos 13π/6 " " ) = 𝝅/𝟔 (As cos (360 + θ) = cos θ)