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Example 9 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Feb. 13, 2020 by Teachoo
Transcript
Example 9 Find the value of cos (–1710°). cos(−1710°) = cos(1710°) = cos(1710 ×𝜋/180) = cos(19𝜋/2) = cos(9 1/2 𝜋 ) = cos(10𝜋−𝜋/2) Value of cos x repeats after an interval of 2π, Hence, ignoring 5 × (2π) i.e. 10π = cos((−𝜋)/2) = cos((−180)/2) = cos(–90°) = cos90° = 0 As cos (−x) = cos x
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Example 23 Not in Syllabus - CBSE Exams 2021
Example 24 Important Not in Syllabus - CBSE Exams 2021
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Example 26 Important
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Example 28 Important
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Chapter 3 Class 11 Trigonometric Functions
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.