- #1

vabamyyr

- 66

- 0

for sin(x) Maclaurin series is

[tex]\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!} [/tex]

probably the key is to change (sin2x)^2 into new shape. I found that

(sin2x)^2=2sin(2x^2), but that coefficent 2 is bothering me, what to do?