Approximate the sine function by a parabola I am looking at the following problem: 
Approximate the sine function in the region of the first maximum in the positive $x-$area by a parabola. 
Let $f(x)=\sin(x)$. 
The first maximum for $x>0$ is at $\frac{\pi}{2}$, isn't it? 
How can we find by which parabola we have to approximate the sine function? Do we have to consider the Taylor expansion of $\sin (x)$ at $x=\frac{\pi}{2}$?
But do we not get only odd powers of $x$? 
 A: You consider the Taylor Series centered at $a = \frac{\pi}{2}$. Using $\sin\left(\frac{\pi}{2}\right) = 1$, $\sin’\left(\frac{\pi}{2}\right) = 0$ and $\sin’’\left(\frac{\pi}{2}\right) = -1$, you get
$$f(x) \approx 1-\frac{1}{2!}\left(x-\frac{\pi}{2}\right)^2 = 1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2$$
As mentioned in the comments, it is also worth noting that this is equivalent to approximating cosine at $a = 0$, but shifted $\frac{\pi}{2}$ radians:
$$\cos(x) \approx 1-\frac{1}{2}x^2; \quad a = 0\implies \sin(x) \approx 1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2; \quad a = \frac{\pi}{2}$$
A: Just added for your curiosity.
As given to you, the question is slightly missing of context since, in particular, it not not precised for which range the approximation needs to be good.
If it is just around $x=\frac \pi 2$, then, as given in comments and answers, a Taylor expansion will be good.
Suppose now that you want the approximation to be good between $\theta$ and $(\pi-\theta)$, $\theta$ being given. Write the approximating function as
$$f(x)=a+b x+c x^2$$  and build the equation at three points
$$f(\theta)=a+b \theta +c \theta ^2=\sin(\theta)\tag 1$$
$$f\left(\frac{\pi }{2}\right)=a+\frac{\pi  b}{2}+\frac{\pi ^2 c}{4}=\sin\left(\frac{\pi }{2}\right)=1\tag 2$$
$$f(\pi-\theta)=a+b (\pi -\theta )+c (\pi -\theta )^2=\sin(\theta)\tag 3$$ Solving the system, you should get
$$a=\frac{4 \theta ^2-4 \pi  \theta +\pi ^2 \sin (\theta )}{(\pi -2 \theta )^2}\qquad b=\frac{4 \pi  (1-\sin (\theta ))}{(\pi -2 \theta )^2}\qquad c=-\frac{4 (1-\sin (\theta ))}{(\pi -2 \theta )^2}$$ If we take limits when $\theta \to \frac \pi 2$, this would give
$$a_{lim}=1-\frac{\pi ^2}{8}\qquad b_{lim}=\frac{\pi }{2}\qquad c_{lim}=-\frac{1}{2}$$ which, effectively, corresponds to
$$f(x)=1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2$$ obtained by Taylor expansion.
But we could do better and consider that, as in a least-square fit, we want to minimize with respect to $(a,b,c)$
$$\Phi(a,b,c)=\int_\theta^{\pi-\theta} \left(a+b x+c x^2-\sin (x)\right)^2\,dx$$ I shall skip the intermediate calculations and go to the results
$$a=\frac{12 \left(\left(-4 \left(\theta ^2-5\right) \theta ^2+4 \pi  \left(2 \theta
   ^2-5\right) \theta -\pi ^2 \left(\theta ^2+10\right)-3 \pi ^3 \theta +\pi
   ^4\right) \cos (\theta )+5 \left(4 \theta ^3-6 \pi  \theta ^2+\pi ^3\right) \sin
   (\theta )\right)}{(\pi -2 \theta )^5}$$
$$b=-\frac{60 \pi  \left(4 \theta ^2 \cos (\theta )-12 \theta  \sin (\theta )+6 \pi 
   \sin (\theta )-4 \pi  \theta  \cos (\theta )+\pi ^2 \cos (\theta )-12 \cos
   (\theta )\right)}{(\pi -2 \theta )^5}$$
$$c=\frac{60 \left(6 (\pi -2 \theta ) \sin (\theta )+\left((\pi -2 \theta )^2-12\right)
   \cos (\theta )\right)}{(\pi -2 \theta )^5}$$
If we take limits when $\theta \to  0$, this would give
$$a_{lim}=-\frac{12 \left(10-\pi ^2\right)}{\pi ^3}\qquad b_{lim}=\frac{60 \left(12-\pi ^2\right)}{\pi ^4}\qquad c_{lim}=-\frac{60 \left(12-\pi ^2\right)}{\pi ^5}$$ which will not reproduce the exact values of the since function at
$$\theta=0 \implies f(0)=-\frac{12 \left(10-\pi ^2\right)}{\pi ^3}\approx -0.05047$$
$$\theta=\frac{\pi }{2}\implies f\left(\frac{\pi }{2}\right)=\frac{60-3 \pi ^2}{\pi ^3}\approx 0.98016$$
$$\theta=\pi \implies f(\pi)=-\frac{12 \left(10-\pi ^2\right)}{\pi ^3}\approx -0.05047$$ but which would be the best over the whole range.
The answer to an apparently simple question is not so simple; it depends very much on what you plan to do with the approximation.
