Question

Prove that $sin 3x + sin 2x – sin x = 4sin \,x cos\, \frac{x}{2} cos \frac{3x}{2}$

Answer 1

L.H.S. = sin3x+sin2x-sinx

=sin3x+(sin2x-sinx)

$=sin3x+[2cos(\frac{2x+x}{2})sin(\frac{2x-x}{2})]\,\,\,[sinA-sinB=2cos(\frac{A+B}{2})sin(\frac{A-B}{2})]$

$=sin3x+[2cos(\frac{3x}{2})sin(\frac{x}{2})]$

$sin3x+2cos\frac{3x}{2}sin\frac{x}{2}$

$=2sin\frac{3x}{2}.cos\frac{3x}{2}+2cos\frac{3x}{2}sin\frac{x}{2}\,\,\,[sin2A=2sinA.cosB]$

$=2cos(\frac{3x}{2})[(\frac{3x}{2})+sin(\frac{x}{2})]$

$=2cos(\frac{3x}{2})[2sin\\{\frac{{(\frac{3x}{2})+(\frac{x}{2}}{})}{2}\\}cos\\{\frac{{(\frac{3x}{2})-(\frac{x}{2})}}{2}\\}]\,[sinA+sinB=2sin(\frac{A+B}{2})cos(\frac{A-B}{2})]$

$=2cos(\frac{3x}{2}).2sin\,x\,cos(\frac{x}{2})$

$=4sinxcos(\frac{x}{2})cos(\frac{3x}{2})=R.H.S.$