Ex 3.3, 25 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 3.3, 25 Prove that: cos 6๐ฅ = 32 cos6 ๐ฅ โ 48 cos4 ๐ฅ + 18 cos2 ๐ฅ โ 1 Solving L.H.S. cos 6x = 2(cos 3x)2 โ 1 = 2 ( 4 cos3 x โ 3 cos x)2 โ 1 We know that cos 2x = 2 cos2 x โ 1 Replacing by 3x cos 2(3x) = 2 cos2 (3x) -1 cos 6x = 2 cos2 3x -1 Using (a โ b)2 = a2 + b2 โ 2ab = 2 [(4 cos3 x)2 + (3 cos x )2 โ 2 (4 cos3 x) ร (3 cos x)] โ 1 = 2 [(16 cos6x + 9 cos2 x โ 24 cos4x)] โ 1 = 2 ร 16 cos6x + 2 ร 9 cos2 x โ 2 ร 24 cos4x โ 1 = 32 cos6x โ 48 cos4x + 18 cos2x โ 1 = R.H.S. Hence L.H.S. = R.H.S Hence proved
Ex 3.3
Ex 3.3, 2 Important
Ex 3.3, 3 Important
Ex 3.3, 4
Ex 3.3, 5 (i) Important
Ex 3.3, 5 (ii)
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 Important
Ex 3.3, 14
Ex 3.3, 15
Ex 3.3, 16 Important
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
Ex 3.3, 25 You are here
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo