: Show that(3A. (11Asinsin+sinA. (7A= sin 2A-sin 5Aion:

1.	: Show that(3A. (11Asinsin+sinA. (7A= sin 2A-sin 5Aion:
Answer» sin(A/2).sin(7A/2) + sin(3A/2)sin(11A/2) =(1/2) [ 2sin(7A/2).sin(A/2) + 2sin(11A/2)sin(3A/2) ] =(1/2) [{ cos( 7A/2 - A/2) - cos( 7A/2 + A/2) } + { cos( 11A/2 - 3A/2)- cos( 11A/2 + 3A/2) } ( using formula, product to sum) =(1/2) [ ( cos3A - cos4A ) + (cos4A - cos7A) ] =(1/2) [ cos3A - cos7A ] =(1/2) 2 sin[(7A - 3A)/2 ] sin[(7A+3A)/2 ] ( using formula, sum to product) = sin2A sin5A ( Proved)

Answer»

sin(A/2).sin(7A/2) + sin(3A/2)sin(11A/2)

=(1/2) [ 2sin(7A/2).sin(A/2) + 2sin(11A/2)sin(3A/2) ]

=(1/2) [{ cos( 7A/2 - A/2) - cos( 7A/2 + A/2) } + { cos( 11A/2 - 3A/2)- cos( 11A/2 + 3A/2) } ( using formula, product to sum)

=(1/2) [ ( cos3A - cos4A ) + (cos4A - cos7A) ]

=(1/2) [ cos3A - cos7A ]

=(1/2) 2 sin[(7A - 3A)/2 ] sin[(7A+3A)/2 ] ( using formula, sum to product)

= sin2A sin5A ( Proved)

Discussion