The Taylor series formula is:
sum_(n=0)^N (f^((n))(a))/(n!)(x-a)^n
The Taylor series around a = 0 (not x = 0... the question is technically off) is also known as the Maclaurin series. You can write it then as:
sum_(n=0)^N (f^((n))(0))/(n!)x^n
= (f(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + (f''''(0))/(4!)x^4 + ...
So, you know you have to take some derivatives. sinx has cyclic derivatives that follow this pattern:
color(green)(sinx) = f^((0))(x) = color(green)(f(x))
d/(dx)[sinx] = color(green)(cosx = f'(x))
d/(dx)[cosx] = color(green)(-sinx = f''(x))
d/(dx)[-sinx] = color(green)(-cosx = f'''(x))
d/(dx)[-cosx] = color(green)(sinx = f''''(x))
Finally you can write the whole thing out, knowing that whenever trig(0) = 0, the whole term disappears. sinx appears in every even derivative. Hence, f(x), f''(x), and every even derivative disappears.
You have only odd terms to worry about, and those are all just 1 in the numerator and the signs alternate due to the alternating signs in front of cosx.
=> cancel((f(0))/(0!)x^0) + (f'(0))/(1!)x^1 + cancel((f''(0))/(2!)x^2) + (f'''(0))/(3!)x^3 + cancel((f''''(0))/(4!)x^4) + ...
= cos(0)x + ((-cos(0))x^3)/6 + (cos(0)x^5)/120 + ((-cos(0))x^7)/5040 + ...
= color(blue)(x - x^3/6+ x^5/120 - x^7/5040 + ...)
