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)
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Example 11 You are here
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Example 24 Important
Example 25
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Example 27 Important
Example 28 Important
Example 29
Chapter 3 Class 11 Trigonometric Functions
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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.