Last updated at Jan. 7, 2020 by Teachoo

Transcript

Misc 7 Prove that: sin 3x + sin2x โ sin x = 4 sin x cos ๐ฅ/2 cos 3๐ฅ/2 Solving L.H.S sin 3x + sin 2x - sin x = sin 3x + (sin 2x โ sin x ) = sin 3x + [2cos ((2๐ฅ + ๐ฅ)/2) . sin ((2๐ฅโ๐ฅ)/2)] = sin 3x + [ 2cos (3๐ฅ/2) sin ๐ฅ/2] = 2sin 3๐ฅ/2 cos 3๐ฅ/2 + [ 2 cos 3๐ฅ/2 sin ๐ฅ/2 ] = 2cos 3๐ฅ/2 [ sin 3๐ฅ/2 + sin ๐ฅ/2 ] = 2 cos 3๐ฅ/2 [2 sin ("(" 3๐ฅ/2 " + " ๐ฅ/2 ")" )/2 . cos ("(" 3๐ฅ/2 " โ " ๐ฅ/2 ")" )/2] = 2 cos 3๐ฅ/2 [ 2 sin ("(" (3๐ฅ + ๐ฅ)/2 ")" )/2 . cos ("(" (3๐ฅ โ ๐ฅ)/2 ")" )/2] = 2 cos 3๐ฅ/2 [ 2 sin ("(" 4๐ฅ/2 ")" )/2 . cos ("(" 2๐ฅ/2 ")" )/2] = 2 cos 3๐ฅ/2 [ 2 sin ("(" 2๐ฅ/1 ")" )/2 . cos ("(" ๐ฅ/1 ")" )/2] = 2 cos 3๐ฅ/2 [ 2 sin 2๐ฅ/2 . cos ๐ฅ/2] = 2 cos 3๐ฅ/2 [ 2 sin ๐ฅ . cos ๐ฅ/2] = 4 cos 3๐ฅ/2 sin ๐ฅ cos ๐ฅ/2 = R.H.S Hence L.H.S = R.H.S Hence proved

Chapter 3 Class 11 Trigonometric Functions

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.