Proove : sinx+sin2x+sin4x+sin5x=4cos x/2.cos3x/2.sin3x Share with your friends Share 4 Vijay Kumar Gupta answered this Consider the following expression. sinx+sin2x+sin4x+sin5xGroup the terms as follows sinx+sin2x+sin4x+sin5x=sin5x+sinx+sin4x+sin2xUse the trigonometric identity sinA+sinB=2sinA+B2cosA-B2This gives, sinx+sin2x+sin4x+sin5x=2sin5x+x2cos5x-x2+2sin4x+2x2cos4x-2x2 =2sin3x cos2x+2sin3x cosx =2sin3x cos2x+ cosxUse the trigonometric identity cosA+cosB=2cosA+B2cosA-B2This implies that, sinx+sin2x+sin4x+sin5x=2sin3x 2cos2x+x2cos2x-x2 =2sin3x 2cos3x2cosx2 =4 cosx2 cos3x2sin3xHence it is proved that, sinx+sin2x+sin4x+sin5x=4 cosx2 cos3x2sin3x 15 View Full Answer Sagar Patel answered this In LHS, = sin x + sin 2x + sin 4x + sin 5x = sin x + sin 5x + sin 2x + sin 4x As sin C + sin D = 2 sin (C+D)/(2) cos (C+D)/(2) so here we get : = 2 sin 3x cos 2x + 2 sin 3x cosx = 2 sin 3x [(cos 2x) + (cos x)] = 2 sin 3x (2 cos 3x/2 .cos x/2) = 4 cos x/2 . cos 3x/2 . sin 3x = RHS Hence,proved. 5
