Subscribe to our Youtube Channel - https://you.tube/teachoo
Ex 3.3, 24 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Jan. 7, 2020 by Teachoo
Transcript
Ex 3.3, 24 Prove that cos 4๐ฅ = 1 โ 8sin2 ๐ฅ cos2 ๐ฅ Taking L.H.S. cos 4x = 2(cos 2x)2 โ 1 = 2 ( 2 cos2 x โ 1)2 -1 Using (a โ b)2 = a2 + b2 โ 2ab = 2 [(2cos x)2 + (1)2 โ 2 ( 2cos2x ) ร 1] โ 1 = 2 (4cos4x + 1 โ 4 cos2x ) โ 1 = 2 ร 4cos4x + 2 ร 1โ 2 ร 4 cos2x โ 1 = 8cos4x + 2 โ 8 cos2x -1 = 8cos4x โ 8 cos2x + 2 โ 1 = 8cos4x โ 8 cos2x + 1 = 8cos2x (cos2x โ 1) + 1 = 8cos2x [โ (1 โ cos2x)] + 1 = โ8cos2x [(1 โ cos2x )] + 1 = โ 8cos2x sin2x + 1 = 1 โ 8 cos2x sin2x = R.H.S. Hence R.H.S. = L.H.S. Hence proved
Ex 3.3
Ex 3.3, 1
Ex 3.3, 2 Important
Ex 3.3, 3 Important
Ex 3.3, 4
Ex 3.3, 5 Important
Ex 3.3, 6 Important
Ex 3.3, 7
Ex 3.3, 8 Important
Ex 3.3, 9 Important
Ex 3.3, 10
Ex 3.3, 11 Important
Ex 3.3, 12
Ex 3.3, 13
Ex 3.3, 14
Ex 3.3, 15
Ex 3.3, 16
Ex 3.3, 17
Ex 3.3, 18 Important
Ex 3.3, 19
Ex 3.3, 20
Ex 3.3, 21 Important
Ex 3.3, 22 Important
Ex 3.3, 23 Important
Ex 3.3, 24 You are here
Ex 3.3, 25
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 9 years. He provides courses for Maths and Science at Teachoo.