3rd Order Taylor expansion of $e^x\cos(y)\sin(z)$ I'm looking for the 3rd.-order Taylor approximation of 
$(x,y,z) \mapsto e^x\cos(y)\sin(z)$  at $(x_0,y_0,z_0) = (0,0,0)$
I've got this piece of advice at hand: $\quad\textit{Use the Taylor series of known functions.}$
I can conclude that:
$\exp(x) = \sum_{n = 0}^\infty \frac{x^n}{n!}$
$\sin (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$
$\cos (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$
But how am I supposed to use this?
I get as a result by hiting it in Python (SymPy) :

$z + x*z + (x^2*z)/2 - (y^2*z)/2 - z^3/6$ 

Adding all three 2-dimensional taylor-approximations with n = 3 doesn't yield the same...
Any hints - not solutions at first- would be greatly appreciated.
 A: I used to ask essentially the same thing (I had $y$ and $z$ switched from what you have, and I only asked for the 2nd order approximation) on take-home tests I gave in the late 1990s. See problem #4 in the file named "LSMSA multi test 4b.pdf" in this 2 March 2010 math-teach post archived at Math Forum. In problem #4 I outlined two different approaches for students to use. One approach was a straightforward substitution into the 3-variable Taylor expansion, and the other approach was a way of reducing the algebra by making use of a trig. identity. Below is my hint for the method using the trig. identity. (I've switched $y$ and $z$ from my original version to match what you're asking about.)

Use $\cos{y}\sin{z} \; = \; \frac{1}{2}\sin (z-y) + \frac{1}{2}\sin(z+y)$ [see the appropriate trig. formula on p. 5 of your text's trig. appendix (or see equation (2) on p. 632)] and substitution into the one variable Taylor series for the SINE function to obtain an expansion for $\cos{y}\sin{z}$ (up to and including third degree terms is enough). Then multiply this expansion by the quadratic expansion for $e^{x},$ namely $1 + x + \frac{1}{2}x^{2},$ keeping only terms of degree $2$ or less.

A: To get what Python says try again by expanding $$\left(1 + x +x^2/2 + x^3/6\right)\left(1-y^2/2\right)\left(z - z^3/6\right)$$
and dropping monomials with degree more than $3$ (like $x^3z/6$ whose degree is $4$).
