# How do you find the first three terms of the Taylor's series for sin(x^2) ?

Dec 29, 2016

$\sin \left({x}^{2}\right) \cong {x}^{2} - {x}^{6} / 6 + {x}^{10} / 120$

#### Explanation:

We take the MacLaurin series for $\sin t$:

sint = sum_(n=0)^oo (-1)^n(t^(2n+1))/((2n+1)!)

We substitute $t = {x}^{2}$

sin(x^2) = sum_(n=0)^oo (-1)^n(x^(2(2n+1)))/((2n+1)!)

and we take the terms for $n = 0 , 1 , 2$:

$\sin \left({x}^{2}\right) \cong {x}^{2} - {x}^{6} / 6 + {x}^{10} / 120$