Taylor's formula Taylor's Formula
Write taylor's formula for $F(x,y)= \sin(x)\sin(y)$ using $a=0$, $b=0$, and $n=2$.
$$\sin(h)\sin(k)=hk−\frac 16h(h^2+3k^2)\cos\theta h\sin\theta k−\frac 16 k(3h^2+k^2)\sin\theta h\cos\theta k$$
That's the answer, I don't get how to get it though.
I don't understand how to use the taylor's formula to find this answer. I've tried and failed, I have no idea where $h$,$k$, and $\theta$ comes from. Is anyone able to help me out?

It might be a slight overkill to use the general Taylor expansion in two variables for this particular example. In fact, since the function factors in two independent parts, you can proceed as follows:
$$\sin\left(h+\theta h\right)\sin\left(k+\theta k\right)=\left(\sin h\cos\theta h+\sin\theta h\cos h\right)\left(\sin k\cos\theta k+\sin\theta k\cos k\right)$$
$$=\left(\left(h-\frac{h^{3}}{6}+O\left(h^{5}\right)\right)\cos\theta h+\sin\theta h\left(1-\frac{h^{2}}{2}+O\left(h^{4}\right)\right)\right)\left(\left(k-\frac{k^{3}}{6}+O\left(k^{4}\right)\right)\cos\theta k+\sin\theta k\left(1-\frac{h^{2}}{2}+O\left(k^{4}\right)\right)\right)$$
Opening the brackets and gathering terms you should arrive at the same result. If this is not the kind of solution expected from you, it is still a good means of checking the answer
