Ex 3.3, 24 - 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, 24 Prove that cos 4๐ฅ = 1 โ 8sin2 ๐ฅ cos2 ๐ฅ Solving L.H.S. cos 4x = 2(cos 2x)2 โ 1 = 2 ( 2 cos2 x โ 1)2 -1 We know that cos 2x = 2 cos2 x โ 1 Replacing by 2x cos 2(2x) = 2 cos2 (2x) โ 1 cos 4x = 2 cos2 2x โ 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 = 2 ร 4cos4x + 2 ร 1โ 2 ร 4 cos2x โ 1 = 8cos4x + 2 โ 8 cos2x โ 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
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