Misc 7 - Prove sin 3x + sin2x - sin x = 4 sin x cos x/2 cos 3x/2 - cos x + cos y formula

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Misc 7 Prove that: sin 3x + sin2x โ€“ sin x = 4 sin x cos ๐‘ฅ/2 cos 3๐‘ฅ/2 Solving L.H.S sin 3x + sin 2x - sin x = sin 3x + (sin 2x โ€“ sin x ) = sin 3x + [2cos ((2๐‘ฅ + ๐‘ฅ)/2) . sin ((2๐‘ฅโˆ’๐‘ฅ)/2)] = sin 3x + [ 2cos (3๐‘ฅ/2) sin ๐‘ฅ/2] = 2sin 3๐‘ฅ/2 cos 3๐‘ฅ/2 + [ 2 cos 3๐‘ฅ/2 sin ๐‘ฅ/2 ] = 2cos 3๐‘ฅ/2 [ sin 3๐‘ฅ/2 + sin ๐‘ฅ/2 ] = 2 cos 3๐‘ฅ/2 [2 sin ("(" 3๐‘ฅ/2 " + " ๐‘ฅ/2 ")" )/2 . cos ("(" 3๐‘ฅ/2 " โˆ’ " ๐‘ฅ/2 ")" )/2] = 2 cos 3๐‘ฅ/2 [ 2 sin ("(" (3๐‘ฅ + ๐‘ฅ)/2 ")" )/2 . cos ("(" (3๐‘ฅ โˆ’ ๐‘ฅ)/2 ")" )/2] = 2 cos 3๐‘ฅ/2 [ 2 sin ("(" 4๐‘ฅ/2 ")" )/2 . cos ("(" 2๐‘ฅ/2 ")" )/2] = 2 cos 3๐‘ฅ/2 [ 2 sin ("(" 2๐‘ฅ/1 ")" )/2 . cos ("(" ๐‘ฅ/1 ")" )/2] = 2 cos 3๐‘ฅ/2 [ 2 sin 2๐‘ฅ/2 . cos ๐‘ฅ/2] = 2 cos 3๐‘ฅ/2 [ 2 sin ๐‘ฅ . cos ๐‘ฅ/2] = 4 cos 3๐‘ฅ/2 sin ๐‘ฅ cos ๐‘ฅ/2 = R.H.S Hence L.H.S = R.H.S Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.