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Ex 3.3, 24 - Chapter 3 Class 11 Trigonometric Functions
Last updated at May 29, 2023 by Teachoo
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Ex 3.3, 24 You are here
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Chapter 3 Class 11 Trigonometric Functions
Serial order wise
Transcript
Ex 3.3, 24 Prove that cos 4π₯ = 1 β 8sin2 π₯ cos2 π₯ Taking L.H.S. cos 4x = 2(cos 2x)2 β 1 = 2 ( 2 cos2 x β 1)2 -1 Using (a β b)2 = a2 + b2 β 2ab = 2 [(2cos x)2 + (1)2 β 2 ( 2cos2x ) Γ 1] β 1 = 2 (4cos4x + 1 β 4 cos2x ) β 1 = 2 Γ 4cos4x + 2 Γ 1β 2 Γ 4 cos2x β 1 = 8cos4x + 2 β 8 cos2x -1 = 8cos4x β 8 cos2x + 2 β 1 = 8cos4x β 8 cos2x + 1 = 8cos2x (cos2x β 1) + 1 = 8cos2x [β (1 β cos2x)] + 1 = β8cos2x [(1 β cos2x )] + 1 = β 8cos2x sin2x + 1 = 1 β 8 cos2x sin2x = R.H.S. Hence R.H.S. = L.H.S. Hence proved
