Question

give the expansion of sin^8@
as a series of cos@
plz help me i only know the normal way of doing that but i do not know how to apply complex number or exponential

Answer 1

not sure what you mean by a "series" of cos, but if you recall your binomial expansions,

(a+b)^4 = a^4+4a^3b+a^2b^2+4ab^3+b^4

and then let a=1 and b=cos^2(x), you have

sin^8(x)
= (sin^2(x))^4
= (1-cos^2(x))^4
= 1 - 4cos^2(x) + 6cos^4(x) - 4cos^6(x) + cos^8(x)