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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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