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  1. Chapter 3 Class 11 Trigonometric Functions
  2. Concept wise

Transcript

Ex 3.2,9 Find the value of the trigonometric function sin (–11"Ο€" /3) sin ((βˆ’11 πœ‹)/3) = –sin ((11 πœ‹)/3) = –sin ("3 " 2/3 " Ο€" ) = –sin ("4Ο€ – " 1/3 " Ο€" ) = –sin ((βˆ’1)/3 " Ο€" ) Values of sin x repeats after an interval of 2Ο€ , Hence ignoring 4Ο€ i.e. 2 Γ— (2Ο€) Rough So, 11/3 = 3 2/3 = 3 + 2/3 or 4 – 1/3 We take 4 – 1/3 as it is even = – ("–sin " (1/3 " Ο€" )) = sin (1/3 " Ο€" ) = sin 60Β° = βˆšπŸ‘/𝟐 ( As sin(βˆ’x) = βˆ’sin x )

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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.