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Example 11 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Nov. 17, 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°
Examples
Example 1
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6 Important
Example 7 Important
Example 8
Example 9 Important
Example 10
Example 11 Important You are here
Example 12
Example 13
Example 14
Example 15
Example 16 Important
Example 17 Important
Example 18
Example 19
Example 20 Deleted for CBSE Board 2022 Exams
Example 21 Deleted for CBSE Board 2022 Exams
Example 22 Important Deleted for CBSE Board 2022 Exams
Example 23 Deleted for CBSE Board 2022 Exams
Example 24 Important Deleted for CBSE Board 2022 Exams
Example 25 Important
Example 26
Example 27 Important
Example 28 Important
Example 29 Important
Chapter 3 Class 11 Trigonometric Functions (Term 2)
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Ex 3.4
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Examples
- Miscellaneous
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.