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Ex 3.3, 25 Prove that: cos 6𝑥 = 32 cos6 𝑥 – 48 cos4 𝑥 + 18 cos2 𝑥 – 1 Solving L.H.S. cos 6x = 2(cos 3x)2 – 1 = 2 ( 4 cos3 x – 3 cos x)2 – 1 We know that cos 2x = 2 cos2 x – 1 Replacing by 3x cos 2(3x) = 2 cos2 (3x) -1 cos 6x = 2 cos2 3x -1 Using (a – b)2 = a2 + b2 – 2ab = 2 [(4 cos3 x)2 + (3 cos x )2 – 2 (4 cos3 x) × (3 cos x)] – 1 = 2 [(16 cos6x + 9 cos2 x – 24 cos4x)] – 1 = 2 × 16 cos6x + 2 × 9 cos2 x – 2 × 24 cos4x – 1 = 32 cos6x – 48 cos4x + 18 cos2x – 1 = R.H.S. Hence L.H.S. = R.H.S Hence proved

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Davneet Singh

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