Show that `4 sin alpha.sin (alpha + pi/3) sin (alpha + 2pi/3) = sin 3alpha`

prove that 4sin alpha*sin(alpha+(pi)/(3))*sin(alpha+(2 pi)/(3))=sin3 alpha

sin alpha+sin(alpha+((2pi)/3))+sin(alpha+((4pi)/3))=0

sin alpha * sin (60-alpha) sin (60 + alpha) = (1) / (4) * sin3 alpha

sin3 alpha=4sin alpha sin(x+alpha)sin(x-alpha)

If sin 3 alpha =4 sin alpha sin (x+alpha ) sin(x-alpha ) , then

The expression (1 + sin 2 alpha )/( cos (2 alpha - 2pi) tan (alpha - (3pi)/(4 ))) -1/4 sin 2 alpha [ cot "" (alpha )/(2) + (alpha )/(2))] when simplified reduces to

Sum the series: sin alpha - sin 2alpha + sin 3alpha - sin 4alpha + ... to n terms.

If 5tan alpha=4, show that (5sin alpha-3cos alpha)/(5sin alpha+2cos alpha)=(1)/(6)

show that :sin^(2)alpha+sin^(2)(alpha+(2 pi)/(n))+sin^(2)(alpha+(4 pi)/(n))+n terms =(n)/(2)