Ex 3.3, 12 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 13, 2024 by Teachoo
Ex 3.3
Ex 3.3, 1
Important
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Chapter 3 Class 11 Trigonometric Functions
Serial order wise
Transcript
Ex 3.3, 12 Prove that sin2 6π₯ β sin2 4π₯ = sinβ‘2π₯ sinβ‘10π₯ Solving L.H.S. sin2 6x β sin2 4x = (sin 6x + sin 4x) (sin 6x β sin 4x) Lets calculate (sin 6x + sin 4x) and (sin 6x β sin 4x) separately sin 6x + sin 4x = 2 sin ((6x+4x)/2) cos ((6xβ4x)/2) = 2sin (10π₯/2) cos (2π₯/2) = 2sin 5x cos x sin 6x β sin 4x = 2 cos ((6x+4x)/2) sin((6xβ4x)/2) = 2 cos (10π₯/2) sin (2π₯/2) = 2 cos 5x sin x Hence sin2 6x β sin2 4x = (sin 6x + sin 4x) (sin 6x β sin 4x) = (2 sin 5x cos x) (2 cos 5x sin x) = (2 sin 5x cos 5x) . (2 sin x cos x) = (sin 10x) Γ (sin 2x) = R.H.S. Hence, L.H.S. = R.H.S. Hence proved
