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  1. Chapter 3 Class 11 Trigonometric Functions
  2. Concept wise

Transcript

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 + 2cos (3๐‘ฅ/2) sin ๐‘ฅ/2 Using sin x โ€“ sin y = 2 cos (๐‘ฅ + ๐‘ฆ)/2 sin (๐‘ฅ โˆ’ ๐‘ฆ)/2 Putting x = 2x & y = x , Rough As (3๐‘ฅ + ๐‘ฅ)/2 = 4๐‘ฅ/2 = 2x & (2๐‘ฅ + ๐‘ฅ)/2 = 3๐‘ฅ/2 As 3๐‘ฅ/2 is in R.H.S. , we take x & 2x 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 " 3๐‘ฅ/2 " + sin " ๐‘ฅ/2] 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 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
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.