Question

Find the third degree Taylor Polynomial for the function f(x) = cos x at a = −π/4.

Answer 1

We can use Taylor series formula

$T_3(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)$

a = −π/4.

f(x) = cos x

now, we can find values and plug in formula

$f(-\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

8) = -sin(1)

$f'(-\frac{\pi}{4})=-sin(\frac{-\pi}{4})=\frac{\sqrt{2}}{2}$

$f''(x)=-cos(x)$

$f''(-\frac{\pi}{4})=-cos(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

F"(x) = sin(x)

$f'''(-\frac{\pi}{4})=sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

now, we can plug values

and we get

$T_3(x)=\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{2}}\left(x+\frac{\pi }{4}\right)-\frac{1}{2\sqrt{2}}\left(x+\frac{\pi }{4}\right)^2-\frac{1}{6\sqrt{2}}\left(x+\frac{\pi }{4}\right)^3$...........Answer