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Misc 2 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at March 22, 2023 by Teachoo
Transcript
Misc 2 Prove that: (sin 3π₯ + sin π₯) sin π₯ + (cos 3π₯ β cos π₯) cos π₯ = 0 Lets calculate (sin 3x + sin x) and (cos 3x β cos x) separately We know that sin x + sin y = sin ((π₯ + π¦)/2) cos ((π₯ β π¦)/2) Replacing x with 3x and y with x sin 3x + sin x = 2sin ((3π₯ + π₯)/2) cos ((3π₯ β π₯)/2) sin 3x + sin x = 2 sin 2x cos x Similarly , We know that cos x β cos y = β2 sin ((π₯ + π¦)/2) sin ((π₯ β π¦)/2) Replacing x with 3x and y with x cos 3x β cos x = β2 sin ((3π₯ + π₯)/2) sin ((3π₯ β π₯)/2) cos 3x β cos x = β2 sin 2x sin x Now solving L.H.S (sin 3x + sin x) sin x + (cos 3x β cos x) cos x Putting values from (1) & (2) = (2 sin 2x cos x) (sin x) + (β2sin 2x) (sin x) (cos x) = 2 sin 2x cos x sin x β 2 sin 2x sin x cos x = 0 = R.H.S Hence L.H.S = R.H.S Hence proved
