Taylor Series Expansions

Problem with this question

$$f(x)=3\sin x\cos \frac x 2 \text{ around } a=1$$

I have been breaking up $f(x)$ into $g(x) = 3\sin x$ and $h(x)= \cos \frac x 2$, and then multiplying them back ($g(x)\cdot h(x)$) this was suggested to my by the lecturer. but for the life of me I cant get the answer.

Answer is: $y = 0.817x+1.398$

any help would be greatly appreciated.

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What's a? I don't understand you first line. –  Noldorin Oct 27 '10 at 11:22
@basic: first of all what is ur question. –  anonymous Oct 27 '10 at 11:27
@basic: I think u want to expand $f(x)$ as a taylor series around $a$. –  anonymous Oct 27 '10 at 11:29
@Noldorin a = x, sorry its how the book has it. @Chandrui Sorry I wasn't clear I needed to do a taylor series expansion on the the question. –  PhilCK Oct 30 '10 at 12:13

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).