In my Analysis class today I defined the trigonometric functions ${\sin}$ and ${\cos}$ by means of the complex exponential. As usual I noted that for real ${x}$ we have ${|\mathrm{e}^{\mathrm{i} x}| = 1}$, i.e. ${\mathrm{e}^{\mathrm{i} x}}$ lies on the complex unit circle. Then I drew the following picture: This was meant to show that the real part and the imaginary part of ${\mathrm{e}^{\mathrm{i} x}}$ are what is known as ${\cos(x)}$ and ${\sin(x)}$, respectively.

After the lecture a student came to me and noted that we could have started with ${a>1}$ and note that ${|a^{\mathrm{i} x}|=1}$ and could do the same thing. The question is: Does this work out? My initial reaction was: Yeah, that works, but you’ll get a different ${\pi}$

But then I wondered, if this would lead to something useful. At least for the logarithm one does a similar thing. We define ${a^x}$ for ${a>0}$ and real ${x}$ as ${a^x = \exp_a(x) = \exp(x\ln(a))}$, notes that this gives a bijection between ${{\mathbb R}}$ and ${]0,\infty[}$ and defines the inverse function as $\displaystyle \log_a = \exp_a^{-1}.$

So, nothing stops us from defining $\displaystyle \cos_a(x) = \Re(a^{\mathrm{i} x}),\qquad \sin_a(x) = \Im(a^{\mathrm{i} x}).$

Many identities are still valid, e.g. $\displaystyle \sin_a(x)^2 + \cos_a(x)^2 = 1$

or $\displaystyle \cos_a(x)^2 = \tfrac12(1 + \cos_a(2x)).$

For the derivative one has to be a bit more careful as it holds $\displaystyle \sin_a'(x) = \ln(a)\cos_a(x),\qquad \cos_a'(x) = -\ln(a)\sin_a(x).$

Coming back to “you’ll get a different ${\pi}$”: In the next lecture I am going to define ${\pi}$ by saying that ${\pi/2}$ is the smallest positive root of the functions ${\cos}$. Naturally this leads to a definition of “ ${\pi}$ in base ${a}$” as follows:

Definition 1 ${\pi_a/2}$ is the smallest positive root of ${\cos_a}$.

How is this related to the area of the unit circle (which is another definition for ${\pi}$)?

The usual analysis proof goes by calculating the area of a quarter the unit circle by integral ${\int_0^1 \sqrt{1-x^2} dx}$.

Doing this in base ${a}$ goes by substituting ${x = \sin_a(\theta)}$: $\displaystyle \begin{array}{rcl} \int\limits_0^1\sqrt{1-x^2}\, dx & = & \int\limits_0^{\pi_a/2}\sqrt{1-\sin_a(\theta)^2}\, \ln(a)\cos_a(\theta)\, d\theta\\ & = & \ln(a) \int\limits_0^{\pi_a/2} \cos_a(\theta)^2\, d\theta\\ & = & \ln(a) \frac12 \int\limits_0^{\pi_a/2}(1 + \cos_a(2\theta))\, d\theta\\ & = & \frac{\ln(a)}{2} \Big( \frac{\pi_a}{2} + \int\limits_0^{\pi_a/2}\cos_a(2\theta)\, d\theta\\ & = & \frac{\ln(a)\pi_a}{4} + 0. \end{array}$

Thus, the area of the unit circle is now ${\ln(a)\pi_a}$

Oh, and by the way, you’ll get the nice identity $\displaystyle \pi_{\mathrm{e}^\pi} = 1$

(and hence, the area of the unit circle is indeed ${\ln(\mathrm{e}^\pi)\pi_{\mathrm{e}^\pi} = \pi}$)…