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Example 11 - Chapter 3 Class 11 Trigonometric Functions
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
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 Important
Example 20
Example 21
Example 22 Important
Example 23
Example 24 Important
Example 25 Important
Example 26 Important
Example 27 Important
Example 28 Important
Example 29 Important
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Ex 3.4
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- 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.