# How do you find the taylor series of 1 + sin(x)?

Apr 12, 2015

The general form of the Taylor series is
f(x) = Sigma_(n=0)^oo (f^(|n|) (a))/(n!)*(x-a)^n
where ${f}^{| n |}$ denotes the ${n}^{t h}$ derivative of $f$.

Set $f \left(x\right) = \sin \left(x\right)$
and use $a = 0$

Remember
$\frac{\mathrm{ds} \in \left(x\right)}{\mathrm{dx}} = \cos \left(x\right)$
$\frac{d \cos \left(x\right)}{\mathrm{dx}} = - \sin \left(x\right)$
$\sin \left(0\right) = 0$
$\cos \left(0\right) = 1$

Then the Taylor expansion of $1 + \sin \left(x\right)$
becomes
1 + x -(x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) ...