Write the first several terms of the Maclaurin polynomial

Write the first several terms of the Maclaurin polynomial for a) $y=\sinh x$, b) $y=\cosh x$.

Step 1: Part a): $y=\sinh x$

The derivatives of $\sinh x$ and $\cosh x$ alternate: $(\sinh x)'=\cosh x$, $(\cosh x)'=\sinh x$. Using $\sinh0=0,\ \cosh0=1$: \[f^{(k)}(0)=1\ (k\text{ odd}), \qquad f^{(k)}(0)=0\ (k\text{ even}). \tag{1}\]

Only odd-degree terms survive: \[\sinh x = x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots+\frac{x^{2n+1}}{(2n+1)!}+R_{2n+2}(x). \tag{2}\]

Since $f^{(2n+2)}(x)=\sinh x$, the Lagrange remainder is $R_{2n+2}(x)=\dfrac{\sinh\xi}{(2n+2)!}x^{2n+2}$, $\xi$ between $0$ and $x$; it tends to $0$ for every $x$, giving the everywhere-convergent series $\sinh x=\sum_{k=0}^\infty\dfrac{x^{2k+1}}{(2k+1)!}$.

Step 2: Part b): $y=\cosh x$

Here $f^{(k)}(0)=1$ for even $k$, $0$ for odd $k$. Only even-degree terms survive: \[\cosh x = 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots+\frac{x^{2n}}{(2n)!}+R_{2n+1}(x). \tag{3}\]

Since $f^{(2n+1)}(x)=\sinh x$, $R_{2n+1}(x)=\dfrac{\sinh\xi}{(2n+1)!}x^{2n+1}\to0$ for every $x$, giving $\cosh x=\sum_{k=0}^\infty\dfrac{x^{2k}}{(2k)!}$.

Step 3: Verification

\[x=1\text{: }1+\tfrac16+\tfrac1{120}+\tfrac1{5040}\approx1.17520,\ \sinh1\approx1.17520\ \checkmark\]

\[x=1\text{: }1+\tfrac12+\tfrac1{24}+\tfrac1{720}\approx1.54306,\ \cosh1\approx1.54308\ \checkmark\]