" Solution "=sin3A+sin PA-sin A=usin A cos(A)/(2)cos(3A)/(2)

sin3A+sin2A-sin A=4sin A*cos((A)/(2))*cos((3A)/(2))

Prove that: \ sin3A+sin2A-sin A=4sin A cos (A/2)cos((3A)/2)

Show that sin 3 A + sin 2A -sin A = 4 sin A cos (A)/(2) cos "" (3A)/( 2)

Prove that, sinA+sin2A+sin4A+sin5A= 4cos((A)/(2))cos((3A)/(2))sin3A .

Prove that : sin3A+ sin 2A- sinA= 4 sinA cos frac (A)(2) cos frac (3A)(2) .

Prove that: (i) sin 3x+"sin" 2x-"sin" x = 4 sin x cos((x)/(2))cos((3x)/(2)) (ii) ("sin" 3x+"sin" x)"sin" x+(cos 3x-cos x)cos x=0.

Prove that sinA+sin2A+sin4A+sin5A=4 cos(A/2)cos((3A)/2)sin3A

Prove that sin A+sin2A+sin4A+sin5A=4((cos A)/(2))(cos3(A)/(2))sin3A

Match list - I with list II ListI " " List II sin 2A " "cos^(2)A - sin^(2)A cos 2A " "2 sin A cos A sin 3A " "2 sin (A)/(2) cos (A)/(2) sin A " "3 sin A - 4 sin^(3) A

sin^(3)A sin3A+cos^(3)A cos3A=cos^(3)(2A)