Using Maclaurin's series, find the first 4 (non zero) terms for the function $f(x)=\sin x$
$$
\begin{array}{cc}
f(x)=\sin x & f(0)=\sin 0=0 \\
f^{\prime \prime}(x)=-\sin x \quad f^{\prime \prime}(0)=-\sin 0=0 \\
f^{\prime \prime \prime}(x)=-\cos x \quad f^{\prime \prime \prime}(0)=-\cos 0=-1 \\
f^{\mathrm{iv}}(x)=\sin x \quad & f^{\mathrm{i}}(0)=\sin 0=0 \\
f^{\mathrm{v}}(x)=\cos x & f^{\mathrm{v}}(0)=\cos 0=1 \\
f^{\mathrm{vi}}(x)=-\sin x & f^{\mathrm{vi}}(0)=-\sin 0=0 \\
f^{\mathrm{vii}}(x)=-\cos x & f^{\mathrm{vii}}(0)=-\cos 0=-1
\end{array}
$$
Substituting the above values into Maclaurin's series of equation (5) gives:
$$
\begin{aligned}
\sin x=0+x(1)+\frac{x^{2}}{2 !}(0)+\frac{x^{3}}{3 !}(-1)+\frac{x^{4}}{4 !}(0) \\
+\frac{x^{5}}{5 !}(1)+\frac{x^{6}}{6 !}(0)+\frac{x^{7}}{7 !}(-1)+\cdots \\
\text { i.e. } \sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\cdots
\end{aligned}
$$