Ex 3.3
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Last updated at Dec. 13, 2024 by Teachoo
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Ex 3.3, 19 Prove that γsin x +γβ‘sinβ‘3x /(πππ β‘x + πππ β‘3x ) = tan 2x Solving L.H.S. γsin x +γβ‘sinβ‘3x /(πππ β‘x + πππ β‘3x ) We solve sin x + sin 3x & cos x + cos 3x seperately sin x + sin 3x = 2 sin ((x+3x)/2) cos ((xβ3x)/2) = 2 sin (4π₯/2) cos ((β2π₯)/2) = 2 sin 2x cos (βx) cos x + cos 3x = 2 cos ((x+3x)/2) cos ((5xβ3x)/2) = 2 cos (4π₯/2) cos ((β2π₯)/2) = 2 cos 2x cos (βx) Now π ππβ‘γπ₯ + π ππβ‘3π₯ γ/πππ β‘γπ₯ + πππ β‘3π₯ γ = (π γ πππ γβ‘γππ πππβ‘γ(βπ)γ γ)/(π πππβ‘γ ππ πππβ‘γ(βπ)γ γ ) = π ππβ‘γ 2xγ/cosβ‘γ 2xγ = tan 2x = R.H.S Hence L.H.S = R.H.S Hence proved
