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Example 11 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Aug. 10, 2020 by Teachoo
Transcript
Example 11 Find the value of sin 15°. sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° = 1/√2 × √3/2 −1/√2 × 1/2 = 1/√2 ((√3 − 1)/2) = (√𝟑 − 𝟏)/(𝟐√𝟐) sin (x – y) = sin x cos y – cos x sin y Putting x = 45° y = 30°
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Example 11 Important You are here
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Example 20 Not in Syllabus - CBSE Exams 2021
Example 21 Not in Syllabus - CBSE Exams 2021
Example 22 Important Not in Syllabus - CBSE Exams 2021
Example 23 Not in Syllabus - CBSE Exams 2021
Example 24 Important Not in Syllabus - CBSE Exams 2021
Example 25 Important
Example 26 Important
Example 27 Important
Example 28 Important
Example 29 Important
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.