Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts



Last updated at Feb. 13, 2020 by Teachoo
Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts
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 โ 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 Using sin x โ sin y = 2 cos (๐ฅ + ๐ฆ)/2 sin (๐ฅ โ ๐ฆ)/2 Putting x = 2x & y = x , Rough As (3๐ฅ + ๐ฅ)/2 = 4๐ฅ/2 = 2x & (2๐ฅ + ๐ฅ)/2 = 3๐ฅ/2 As 3๐ฅ/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 ๐๐ฑ/๐ cos ๐๐ฑ/๐ = 2 sin 3๐ฅ/2 cos 3๐ฅ/2 + ["2 cos " 3๐ฅ/2 " sin " ๐ฅ/2] = 2 cos 3๐ฅ/2 ["sin " 3๐ฅ/2 " + sin " ๐ฅ/2] Using sin x + sin y = 2 sin (๐ฅ + ๐ฆ)/2 cos (๐ฅ โ ๐ฆ)/2 Putting x = 3๐ฅ/2 & y = 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
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