The cosine function, $\cos \colon \mathbb{C} \rightarrow \mathbb{C}$ is defined by the formula $$\cos(z)=\dfrac{e^{iz}-e^{-iz}}{2},$$ where $e^z$ is the exponential function.

Cosine.png

Graph of $\cos$ on $\mathbb{R}$.
Complex cos.jpg

Domain coloring of analytic continuation of $\cos$ to $\mathbb{C}$.

Properties

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) = -\sin(z),$$ where $\cos$ denotes the cosine and $\sin$ denotes the sine.

Proof

From the definition of cosine, $$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, the reciprocal of i, and the definition of the sine function, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) &= \dfrac{1}{2} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] + \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ &= \dfrac{1}{2} \left[ ie^{iz} - ie^{-iz} \right] \\ &= -\dfrac{e^{iz}-e^{-iz}}{2i} \\ &= -\sin(z), \end{array}$$ as was to be shown. █

References

1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.3.106$

Theorem

Let $z_0 \in \mathbb{C}$. The following Taylor series holds for all $z \in \mathbb{C}$: $$\cos(z)= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k}}{(2k)!},$$ where $\cos$ denotes the cosine function.

Proof

Using the Taylor series of the exponential function and the definition of $\cos$, $$\begin{array}{ll} \cos(z) &= \dfrac{e^{iz}+e^{-iz}}{2} \\ &= \dfrac{1}{2} \left[ \displaystyle\sum_{n=0}^{\infty} \dfrac{i^n (z-z_0)^n}{n!} + \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n i^n (z-z_0)^n}{n!} \right] \\ &= \dfrac{1}{2} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1+(-1)^n). \end{array}$$ Note that if $n=2k$ is a positive even integer, then $$i^n(1+(-1)^n)=i^{2k}(1+(-1)^{2k})=2(-1)^k,$$ and if $n=2k+1$ is a positive odd integer, then $$i^n(1+(-1)^n)=i^{2k+1}(1+(-1)^{2k+1})=0.$$ Hence we have derived $$\begin{array}{ll} \cos(z)&=\dfrac{1}{2} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1+(-1)^n) \\ &=\dfrac{1}{2} \displaystyle\sum_{n \mathrm{\hspace{2pt} even},n>0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n) \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k}}{(2k)!}, \end{array}$$ as was to be shown. █

References

Theorem

The Weierstrass factorization of $\cos(x)$ is $$\cos(x) = \displaystyle\prod_{k=1}^{\infty} \left( 1 - \dfrac{4x^2}{\pi^2 (2k-1)^2} \right).$$

Proof

References

Theorem

The following formula holds: $$B(x,y)=2 \displaystyle\int_0^{\frac{\pi}{2}} (\sin t)^{2x-1}(\cos t)^{2y-1} \mathrm{d}t,$$ where $B$ denotes the beta function, $\sin$ denotes the sine function, and $\cos$ denotes the cosine function.

Proof

From the definition, $$B(x,y)=\displaystyle\int_0^1 \xi^{x-1} (1-\xi)^{y-1} \mathrm{d}\xi.$$ Let $\xi=\sin^2(t)$. Then $d\xi = 2\sin(t)\cos(t)$. Also if $\xi=0$ then $0=\sin^2(t)$ implies that $t=\arcsin(0)=0$ and if $\xi=1$, then $1=\sin^2(t)$ implies $t=\arcsin(1)=\dfrac{\pi}{2}$. Therefore using substitution and the Pythagorean identity for sin and cos, $$\begin{array}{ll} B(x,y) &= \displaystyle\int_0^1 \xi^{x-1}(1-\xi)^{y-1} \mathrm{d}\xi \\ &= \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-2} (1-\sin^2(t))^{y-1} 2 \sin(t)\cos(t) \mathrm{d}t \\ &= 2 \displaystyle\int_0^{\frac{\pi}{2}} (\sin(t))^{2x-1} (\cos(t))^{2y-1} \mathrm{d}t, \end{array}$$ as was to be shown. █