Taylor Polynomial $\;y = e^{\sin x},\,$ $a = \frac{\pi}{2},\;$ $x = 1.5$ $y = e^{\sin x};\;$ $a = \frac{\pi}{2};\;$ $x = 1.5$
So first things get the derivative. I need to calculate the error and the $T_2$
$$f'(x) = \cos x e^{\sin x}$$
$$f''(x) = \cos ^2x e^{\sin x} - \sin x e^{\sin x}$$
So that now gives me
$$f(a) + f'(a)/1 (x-a) + f''(a)/2 (x-a)^2$$
$$e^ {\sin \frac{\pi}{2}} + \cos (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} * (1.5 - \frac{\pi}{2})^2 + \cos ^2 (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} - \sin {\frac{\pi}{2}} e^{\sin {\frac{\pi}{2}}} $$
This turns into $e$ I think.
So for the error
$$|f(x) - T_n (x)| < K \frac{|x - a|^{n+1}}{(n+1)!}$$
$$e^{\sin(1.5)} - e <  K \frac{(1.5 - \frac{\pi}{2})^3}{6}$$
That is a mess...where did I go wrong?
 A: Your derivatives $f'(x), f''(x)$ are fine. And the following is correct:
$$T_2(x) = f(a) + f'(a)(x-a) + \dfrac{f''(a)(x-a)^2}{2}\tag{1}$$
Where the problem seems to be is in your evaluation of $(1)$, given $f(a), f'(a), f''(a)$ with $a = \pi/2, x = 1.5$:

You wrote: $$e^ {\sin \frac{\pi}{2}} + \cos (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} * (1.5 - \frac{\pi}{2})^2 + \cos ^2 (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} - \sin {\frac{\pi}{2}} e^{\sin {\frac{\pi}{2}}} $$

which does not follow from $(1)$.
What you should obtain is:
$$T_2(1.5) = e^{\sin\frac{\pi}{2}} + \cos \left(\frac{\pi}{2}\right) e^{\sin {\frac{\pi}{2}}}\left(1.5 - \frac{\pi}{2}\right) + \dfrac{\left(\cos ^2 \left(\frac{\pi}{2}\right) e^{\sin {\frac{\pi}{2}}} - \sin \left({\frac{\pi}{2}}\right) e^{\sin {\frac{\pi}{2}}}\right)\left(1.5 - \frac{\pi}2\right)^2 }{2}$$
$$\begin{align}T_2(1.5) &= e^{1} + {(0)e^{1}}\left (1.5-\frac{\pi}{2}\right)+ \dfrac{\left((0)e^{1}-1\cdot e^{‌​1}\right)\left(1.5-\frac{\pi}{2}\right)^2}{2} \\ \\
& = e -{e}\cdot \dfrac{(1.5-\frac{\pi}{2})^2}{2}\\ \\
& = e\left(1 - \dfrac{\left(1.5 -\frac{\pi}{2}\right)^2}{2}\right)\end{align}$$
