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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 โˆ’ 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 + 2 cos (๐Ÿ‘๐’™/๐Ÿ) sin ๐’™/๐Ÿ We know that sin 2x = 2 sin x cos x Divide by x by x/2 sin 2x/2 = 2 sin x/2 cos x/2 sin x = 2 sin x/2 cos x/2 Now Replace x by 3x sin 3x = 2 sin ๐Ÿ‘๐ฑ/๐Ÿ cos ๐Ÿ‘๐ฑ/๐Ÿ = 2 sin 3๐‘ฅ/2 cos 3๐‘ฅ/2 + ["2 cos " 3๐‘ฅ/2 " sin " ๐‘ฅ/2] = 2 cos 3๐‘ฅ/2 ["sin " ๐Ÿ‘๐’™/๐Ÿ " + sin " ๐’™/๐Ÿ] Using sin x + sin y = 2 sin (๐‘ฅ + ๐‘ฆ)/2 cos (๐‘ฅ โˆ’ ๐‘ฆ)/2 Putting x = 3๐‘ฅ/2 & y = 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 ๐Ÿ‘๐’™/๐Ÿ ["2 sin " ๐Ÿ๐’™/๐Ÿ " . cos " ๐’™/๐Ÿ] = 2 cos 3๐‘ฅ/2 ["2 sin " ๐‘ฅ" . cos " ๐‘ฅ/2] = 4 cos ๐Ÿ‘๐’™/๐Ÿ sin ๐’™ cos ๐’™/๐Ÿ = R.H.S Hence L.H.S = R.H.S Hence proved

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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo