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Ex 3.3, 25 - Chapter 3 Class 11 Trigonometric Functions
Last updated at June 22, 2023 by Teachoo
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Ex 3.3, 25 You are here
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
Transcript
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
