Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
Last updated at May 29, 2023 by Teachoo
Misc 7 Prove that: sin 3x + sin2x β sin x = 4 sin x cos π₯/2 cos 3π₯/2 Solving L.H.S sin 3x + sin 2x β sin x = sin 3x + (sin 2x β sin x ) = sin 3x + 2cos ((2π₯ + π₯)/2) . sin ((2π₯βπ₯)/2) = sin 3x + 2cos (3π₯/2) sin π₯/2 Using sin x β sin y = 2 cos (π₯ + π¦)/2 sin (π₯ β π¦)/2 Putting x = 2x & y = x , Rough As (3π₯ + π₯)/2 = 4π₯/2 = 2x & (2π₯ + π₯)/2 = 3π₯/2 As 3π₯/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 ππ±/π cos ππ±/π = 2 sin 3π₯/2 cos 3π₯/2 + ["2 cos " 3π₯/2 " sin " π₯/2] = 2 cos 3π₯/2 ["sin " 3π₯/2 " + sin " π₯/2] Using sin x + sin y = 2 sin (π₯ + π¦)/2 cos (π₯ β π¦)/2 Putting x = 3π₯/2 & y = π₯/2 , = 2 cos 3π₯/2 ["2 sin " ((3π₯/2 " + " π₯/2))/2 " . cos " ((3π₯/2 " β " π₯/2))/2] = 2 cos 3π₯/2 ["2 sin " (((3π₯ + π₯)/2))/2 " . cos " (((3π₯ β π₯)/2))/2] = 2 cos 3π₯/2 ["2 sin " ((4π₯/2))/2 " . cos " ((2π₯/2))/2] = 2 cos 3π₯/2 ["2 sin " ((2π₯/1))/2 " . cos " ((π₯/1))/2] = 2 cos 3π₯/2 ["2 sin " 2π₯/2 " . cos " π₯/2] = 2 cos 3π₯/2 ["2 sin " π₯" . cos " π₯/2] = 4 cos 3π₯/2 sin π₯ cos π₯/2 = R.H.S Hence L.H.S = R.H.S Hence proved