The Derivatives of the Complex Sine and Cosine Functions

# The Derivatives of the Complex Sine and Cosine Functions

We will now look at the derivatives of the complex sine and cosine functions which were introduced on The Complex Cosine and Sine Functions page.

Theorem 1: Let $f(z) = \sin z$. Then $f$ is analytic on all of $\mathbb{C}$ and $f'(z) = \cos z$. |

**Proof:**We have that:

\begin{align} \quad f(z) = \sin z = \frac{e^{iz} - e^{-iz}}{2i} \end{align}

- Note that $z$ is analytic on all of $\mathbb{C}$. We know that any constant multiplied by an analytic function is analytic. In particular, $iz$ and $-iz$ are also analytic on all of $\mathbb{C}$. We've already proven that $e^z$ is analytic on all of $\mathbb{C}$ so by the chain rule, $e^{iz}$ and $e^{-iz}$ are analytic on all of $\mathbb{C}$. By the sum rule and multiple rule we conclude that $\displaystyle{\frac{e^{iz} - e^{-iz}}{2i}}$ is analytic on all of $\mathbb{C}$, i.e., $f(z) = \sin z$ is analytic on all of $\mathbb{C}$. So:

\begin{align} \quad f'(z) = \frac{1}{2i} \cdot (ie^{iz} + ie^{-iz}) = \frac{e^{iz} + e^{-iz}}{2} = \cos z \quad \blacksquare \end{align}

Theorem 2: Let $f(z) = \cos z$. Then $f$ is analytic on all of $\mathbb{C}$ and $f'(z) = -\sin z$. |

**Proof:**We have that:

\begin{align} \quad f(z) = \cos z = \frac{e^{iz} + e^{-iz}}{2} \end{align}

- From the comments made in the preceeding theorem we can immediately conclude that $f(z) = \cos z$ is analytic on all of $\mathbb{C}$. So:

\begin{align} \quad f'(z) = \frac{1}{2} \cdot (ie^{iz} -ie^{-iz}) = i \frac{e^{iz} - e^{-iz}}{2} = -\frac{e^{iz} - e^{-iz}}{2i} = - \sin z \quad \blacksquare \end{align}