The general expression of the Taylor series of #f(x)# around #x=a# is:
#f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n#
To find the Taylor series of #sinx# we must therefore evaluate the derivatives of the function for all orders. The function #sinx# is indeed differentiable indefinitely in #RR#, and to obtain a synthetic notation for its derivatives we can use the following trigonometric equality:
#sin(x+pi/2) = cosx#
So we have:
#d/dx sinx = cosx = sin(x+pi/2)#
#d^2/dx^2 sinx = d/dx sin(x+pi/2) = cos(x+pi/2) = sin(x+pi)#
and clearly in general:
#d^n/dx^n sinx = sin(x+(npi)/2)#
For #x=pi/6# we have:
#[d^n/dx^n sinx]_(x=pi/6) = sin(pi/6+(npi)/2) = sin(((3n+1)pi)/6) #
The Taylor series is then:
#sinx = sum_(n=0)^oo sin(((3n+1)pi)/6)/(n!)(x-pi/6)^n#
