Constructing an odd trigonometric function with no 1st-order term in Taylor expansion Is it possible to construct a polynomial $f$ of at least one of (the fewer, the better) $$\sin{x},\cos{x},\sin{y},\cos{y},\sin{z},\cos{z}$$ with the following properties?


*

*The smaller degree of $f$, the better. Hopefully a linear or quadratic one.

*it's odd. $f(x,y,z)=-f(-x,-y,-z)$

*the Taylor expansion of $f$ around $(0,0,z_0)$ (for some nonzero $z_0$) has a nonzero constant term but no 1st-order term.


For example, $f=a-b\cos{y}=a-b+O(y^2)$ satisfies 1.3. but not 2 while $f=\sin{z}$ satisfies 1.2. but not 3.
 A: In order for $f$ to be odd, each term must have odd total degree of the factors $\sin x, \sin y, \sin z$. Thus, any linear polynomial $f$ is a linear combination $a \sin x + b \sin y + c \sin z$, and this has Taylor expansion $$f \sim c \sin z_0 + a x + b y + (c \cos z_0) (z - z_0) + \textrm{(higher-order terms)}.$$ This series has nonzero constant term and zero linear term iff $a = b = \cos z_0 = 0, c \neq 0$. So, if $\cos z_0 \neq 0$, any minimal example is at least quadratic.
To produce a quadratic example, it's enough to consider functions of $z$, and by our first observation above, any such odd function is a linear combination $$f := A \sin z + B \sin z \cos z .$$

The series about $z = z_0$ is $$f \sim (A + B \cos z_0) \sin z_0 + [A \cos z_0 + B (\cos^2 z_0 - \sin^2 z_0)] (z - z_0) + O((z - z_0)^2) ,$$ so for $\sin z_0 \neq 0$ any solution of this form is a multiple of the one determined by $$A = \sin^2 z_0 - \cos^2 z_0, \qquad B = \cos z_0 .$$ On the other hand, if $\sin z_0 = 0$, the first observation in the answer shows that the Taylor series of $f$ has zero constant term, so in this case there is no function $f$ meeting the criteria.

A: The following polynomial of $\sin x$ functions only satisfies all the requirements:
$$
f(x,y,z)=\sin z(3\sin^2 z_0 -\sin^2z),
$$
for any $z_0\ne \pi n$.
A similar function using both $\sin x$ and $\cos x $ can be constructed similarly as:
$$
f(x,y,z)=\sin z (\cos 2z_0 - \cos z_0 \cos z).
$$
A: Quadratic polynomials meeting the requirement is in general possible.
Inspired by comments from @Travis, I got this answer.
It seems that working on $2\pi$-periodic functions of $z$ would suffice. Thus, for an odd $f(z)$, we can use the Fourier series $$f(z)=\sum_{n=1}a_n\sin{nz}.$$
The requirement is two-fold: nonzero constant term and vanishing 1st-order term, which means two tuning parameters (i.e., quadratic polynomial in the sense of the question) might be enough. Then one can pick out any two terms in the series above. Let's just use the first two $$f(z)=a_1\sin{z}+a_2\sin{2z}.$$
Its Taylor expansion around $z_0$ reads $$f(z)=(a_1\sin{z_0}+a_2\sin{2z_0})+(a_1\cos{z_0}+2a_2\cos{2z_0})(z-z_0)+O(z-z_0)^2.$$
To fulfill the two conditions, we need solve $$A\begin{bmatrix}
a_1  
\\a_2
\end{bmatrix}=\begin{bmatrix}
\lambda  
\\0
\end{bmatrix}$$ where $\lambda$ is any arbitrary constant (can be independent of $z_0$) we'd like to assign as the nonzero constant term and $$A=\begin{bmatrix}
\sin{z_0} & \sin{2z_0} \\
\cos{z_0} & 2\cos{2z_0} 
\end{bmatrix}.$$
Noting that $\mathrm{Det}{A}=-2\sin^3{z_0}$, it can be solved as long as $\sin{z_0}\ne0$
$$a_1=-\lambda  \cos (2 z_0) \csc ^3(z_0),a_2=\frac{1}{2} \lambda  \cot (z_0) \csc ^2(z_0).$$
Therefore, we finally reach a function $$f(z)=a_1\sin{z}+2a_2\sin{z}\cos{z}=\lambda+O(z-z_0)^2.$$
