Answer
$\dfrac{5}{4}+\dfrac{3}{4}(x-\ln 2)+\dfrac{5}{8}(x-\ln 2)^2+\dfrac{1}{8} (x-\ln 2)^3$
Work Step by Step
The Taylor approximation for degree $n$ centered at point $a$ can be written as:
$P_n(x)=f(a)+f'(a)(x-a)+\dfrac{1}{2}f''(a)(x-a)^2+.......+\dfrac{1}{n!}f^n(a)(x-a)^n ~~~~........(1)$----
We have: $f(x)=\cos h x \implies f(\ln 2)=\dfrac{5}{4}$ with $n=3$ and $a=\ln 2$
Further, $f'(x)= \sin h (x) \implies f'(\ln 2)=\dfrac{3}{4} \\ f''(x)= \cos h (x) \implies f''(\ln 2)=\dfrac{5}{4} \\f'''(x)= \sin h x \implies f'''(\ln 2)=\dfrac{3}{4}$
Now, plug these values in the equation (1) to obtain:
$P_3(x)=\dfrac{5}{4}+\dfrac{3}{4}\dfrac{(x-\ln 2) }{1!}+\dfrac{5}{4}\times \dfrac{1}{2!}(x-\ln 2)^2+\dfrac{3}{4} \times \dfrac{1}{3 \cdot 2 \cdot 1}(x-\ln 2)^3\\=\dfrac{5}{4}+\dfrac{3}{4}(x-\ln 2)+\dfrac{5}{8}(x-\ln 2)^2+\dfrac{1}{8} (x-\ln 2)^3$
