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 + ...)#
