Ex 3.3, 15 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 13, 2024 by Teachoo
Last updated at Dec. 13, 2024 by Teachoo
Ex 3.3, 15 Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x β sin 3x) Solving L.H.S. cot 4x (sin 5x + sin 3x) = cot 4x Γ [ 2 sin ((5π₯ + 3π₯)/2) cos ((5π₯ β 3π₯)/2) ] = cot 4x Γ [ 2sin (8π₯/2) cos (2π₯/2)] = 2 cot 4x sin 4x cos x = 2 πππ β‘4π₯/π ππβ‘4π₯ Γ sin 4x Γ cos x = 2 cos 4x cos x Solving R.H.S. cot x (sin 5x β sin 3x) = cot x ( 2 cos (5π₯ + 3π₯)/2 sin (5π₯ β 3π₯)/2 ) = cot x ( 2 cos (8π₯/2) sin (2π₯/2)] = cot x ( 2 cos 4x sin x) = 2 cos 4x sin x cot x = 2 cos 4x sin x Γ πππ β‘π₯/sinβ‘π₯ = 2 cos 4x cos x = L.H.S Hence L.H.S. = R.H.S. Hence Proved
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