Question

need help !! Thanks! =D

Prove:
sin A + sin 3A + sin 5A + sin 7A = 16 sin A cos^2 A cos^2 2A

Answer 1

LHS = sin A + sin 3A + sin 5A + sin 7A

=(sin A + sin 7A)+ (sin 3A + sin 5A) = 2*sin 4A * cos 3A +2*sin 4A * cos A

= 2*sin 4A *[cos 3A + cos A] =(2)[ 2*sin 2A*cos 2A]*[2*cos 2A *cos A]

=(2)[ 2*{2*sinA*cosA}*cos 2A]*[2*cos 2A *cos A]= 16*sin A*cos^2 A cos^ 2A = RHS
Aug 7 at 8:32

Show that sin A + sin 3A + sin 5A + sin 7A = 16 sin A cos 2 A cos 2 2A?

L.H.S. = sin A + sin 3A + sin 5A + sin 7A = sinA+sin7A + sin3A+sin5A

L.H.S. = 2sin[(A+7A)/2].cos[(A-7A)/2] +2sin[(3A+5A)/2].cos[(3A-5A)/2]

L.H.S. = 2sin4A.cos3A + 2sin4A.cosA

L.H.S. = 2sin4A[cos3A + cosA]

L.H.S. = 16 sin A cos 2 A cos 2 2A = R.H.S

Answer 2

Wao! thanks!!!