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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
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Ex 3.3, 11 Important
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Ex 3.3, 18 Important
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Ex 3.3, 24 You are here
Ex 3.3, 25
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
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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.