Hi Khalif27,

you have to apply (a+b)^3=a^3+3*a^2*b+3a*b^2+b^3

and i^2=-1, i^3=-i,

from the formula Moivre's you have:

(cosx+isinx)^3=(cos3x+isin3x)

(cosx+isinx)^3= cos^3(x)+3icos^2(x)*sin(x)-3cos(x)*sin^2(x) -icox^3(x)

= cos^3(x)-3cos(x)*sin^2(x) + i*[3cos^2(x)*sin(x)-sin^3(x)]

==> cos(3x)= cos^3(x)-3cos(x)*sin^2(x) = cos^3(x)-3cos(x)*(1-cos^2(x))=4cos^3(x)-3cos(x)

and sin(3x)= 3cos^2(x)*sin(x)-sin^3(x) =3(1-sin^2)*sin(x)-sin^3(x)=3sin(x)-4sin^3(x)