Misc 2 - Prove (sin 3x + sin x) sin x + (cos 3x - cos x) - 2x 3x formula - Proving

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Misc, 2 Prove that: (sin 3๐‘ฅ + sin ๐‘ฅ) sin ๐‘ฅ + (cos 3๐‘ฅ โ€“ cos ๐‘ฅ) cos ๐‘ฅ = 0 Lets calculate (sin 3x + sin x) and (cos 3x โ€“ cos x) separately We know that sin x + sin y = sin ((๐‘ฅ + ๐‘ฆ)/2) cos ((๐‘ฅ โˆ’ ๐‘ฆ)/2) Replacing x with 3x and y with x Hence sin 3x + sin x = 2sin ((3๐‘ฅ + ๐‘ฅ)/2) cos ((3๐‘ฅ โˆ’ ๐‘ฅ)/2) sin 3x + sin x = 2 sin 2x cos x Similarly , we know that cos x โ€“ cos y = โ€“2 sin ((๐‘ฅ + ๐‘ฆ)/2) sin ((๐‘ฅ โˆ’ ๐‘ฆ)/2) Replacing x with 3x and y with x Hence cos 3x โ€“ cos x = โ€“ 2 sin ((3๐‘ฅ + ๐‘ฅ)/2) sin ((3๐‘ฅ โˆ’ ๐‘ฅ)/2) cos 3x โ€“ cos x = โ€“ 2 sin 2x sin x Now solving L.H.S (sin 3x + sin x) sin x + (cos 3x โ€“ cos x) cos x From (1) & (2) = (2 sin 2x cos x) (sin x) + (โ€“ 2sin 2x) (sin x) (cos x) = 2 sin 2x cos x sin x โ€“ 2 sin 2x sin x cos x = 0 =R.H.S Hence L.H.S = R.H.S Hence proved

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