## Thomas' Calculus 13th Edition

$A_{0}(x)=\dfrac{\sqrt 2}{2} \\A_{1}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4}) \\A_{2}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})}{2}-\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})^2}{4} \\A_{3}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2+(-\dfrac{\sqrt 2}{2})(\dfrac{1}{6})(x-\dfrac{\pi}{4})^3=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})}{2}-\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})^2}{4}-\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})^3}{12}$
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as: $A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$ Now, we get $f(\dfrac{\pi}{4})=\dfrac{\sqrt 2}{2}\\ f'(x)=\cos x \implies f'(\dfrac{\pi}{4})=\dfrac{\sqrt 2}{2}\\ f''(x)=-\sin x \implies f''(\dfrac{\pi}{4})=-\dfrac{\sqrt 2}{2} \\f'''(x)=-\cos x\implies f'''(\dfrac{\pi}{4})=-\dfrac{\sqrt 2}{2}$ This implies that $A_{0}(x)=\dfrac{\sqrt 2}{2} \\A_{1}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4}) \\A_{2}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})}{2}-\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})^2}{4} \\A_{3}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2+(-\dfrac{\sqrt 2}{2})(\dfrac{1}{6})(x-\dfrac{\pi}{4})^3=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})}{2}-\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})^2}{4}-\dfrac{\sqrt 2 (x-\dfrac{\pi}{4})^3}{12}$