Example 11 - Chapter 3 Class 11 Trigonometric Functions
Last updated at May 29, 2018 by Teachoo
Transcript
Example, 11 Find the value of sin 15°. sin 15° = sin (45° – 30°) We know that sin (x – y) = sin x cos y – cos x sin y Hence putting x = 45° y = 30° 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)
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11 You are here
Example 12
Example 13
Example 14
Example 15
Example 16
Example 17
Example 18
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24 Important
Example 25
Example 26
Example 27 Important
Example 28 Important
Example 29
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.