You apparently only need the 1st-degree Taylor polynomial; note that the 1st-degree Taylor polynomial for a function $f(x)$ at $x=a$ is the same as the tangent line to the graph of $f$ at $a$, $y=f(a)+f'(a)\cdot(x-a)$.
Given your $g$ and $h$, you can write 1st-degree Taylor polynomials for each of them, then take the product, as you said. For $g$, $$y=g(1)+g'(1)\cdot(x-1)=3\sin(1)+3\cos(1)\cdot(x-1)\approx 1.621x+0.904.$$ I'll leave $h$ for you to do.
Now, you've got (for some numbers $a$ and $b$): $$(1.621x+0.904)(ax+b)=1.621ax^2+(0.904a+1.621b)x+0.904b,$$ in which you don't care about the quadratic term (it's not accurate anyway).
Given that you only need a 1st-degree Taylor polynomial, it might be easier to just differentiate $f$ as written and compute it directly (not bother with $g$ and $h$ and multiplying).