In the last post we covered the traditional approach towards the theory of circular functions which is based on geometric notions related to a circle. In my opinion this approach is the easiest to understand and therefore commonly described in almost any trigonometry textbook (but without the theoretical justification of length (and area) of arcs (and sectors). However it is interesting to also have an approach which is independent of any geometrical notions. In this post we will introduce the circular functions as inverses to certain integrals.

When using integrals to define inverse circular functions there are two possibilities: either start with $\arcsin x$ as the integral of $(1 - x^{2})^{-1/2}$ or $\arctan x$ as the integral of $(1 + x^{2})^{-1}$. We prefer the second approach as it is technically easier to handle.

The challenge now is to define the functions $\sin x, \cos x$ for all values of $x$. One can first try to define $\tan x$ for all $x$ by making it periodic with period $\pi$. By making $\tan x$ periodic with period $\pi$ we ensure that $\tan x$ is defined for all $x$ except when $x$ is an odd multiple of $\pi/2$. Next we define $\sin x, \cos x$ for $x \in (-\pi/2, \pi/2)$ by the formulas $$\sin x = \frac{\tan x}{\sqrt{1 + \tan^{2}x}},\,\cos x = \frac{1}{\sqrt{1 + \tan^{2}x}}\tag{9}$$ We immediately obtain the fundamental identity $$\cos^{2}x + \sin^{2}x = 1\tag{10}$$ for all $x \in (-\pi/2, \pi/2)$. From the above definition it is obvious that $\sin x \to 1$ as $x \to (\pi/2)^{-}$ and $\cos x \to 0$ as $x \to (\pi/2)^{-}$. Similarly $\sin x \to -1$ as $x \to (-\pi/2)^{+}$ and $\cos x \to 0$ as $x \to (-\pi/2)^{+}$. We thus define $$\sin\left(\frac{\pi}{2}\right) = 1,\,\sin\left(-\frac{\pi}{2}\right) = -1,\,\cos\left(\frac{\pi}{2}\right) = 0,\,\cos\left(-\frac{\pi}{2}\right) = 0\tag{11}$$ in order to make $\sin x, \cos x$ continuous in interval $[-\pi/2, \pi/2]$. To extend the definition of $\sin x, \cos x$ for all $x$ we further define $$\sin (x + \pi) = -\sin x,\,\cos(x + \pi) = -\cos x\tag{12}$$ It now follows that $\sin x, \cos x$ are defined for all $x$ and are periodic with period $2\pi$ (while $\tan x$ is periodic with period $\pi$). With the definitions set up as above we can see that the equation $(8)$ above leads us to $$\sin\left(\frac{\pi}{2} - x\right) = \cos x,\,\cos\left(\frac{\pi}{2} - x\right) = \sin x\tag{13}$$ for $0 < x < \pi/2$. Further we also note that $\sin x$ is an odd function and $\cos x$ is an even function. It is easy to show that the functions $\sin x, \cos x$ are continuous everywhere and therefore from equation $(7)$ we get $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{14}$$

The circular functions defined above are differentiable wherever they are defined. The differentiability of $\tan x$ follows from that of $\arctan x$ and clearly if $\tan x = y$ then $x = \arctan y$ so that $$\frac{dx}{dy} = \frac{1}{1 + y^{2}} = \frac{1}{1 + \tan^{2}x} = \cos^{2}x$$ and therefore $$(\tan x)' = \frac{dy}{dx} = \frac{1}{\cos^{2}x}\tag{15}$$ for all $x \in (-\pi/2, \pi/2)$. Since $\tan (x + \pi) = \tan x$ and $\cos^{2}(x + \pi) = \cos^{2}x$ it follows that the formula $(15)$ holds for all $x$ where $\tan x$ is defined. It is now an easy matter to calculate the derivative of $\sin x$ and $\cos x$. Thus we have \begin{align} (\sin x)' &= \frac{d}{dx}\left(\frac{\tan x}{\sqrt{1 + \tan^{2}x}}\right)\notag\\ &= \dfrac{(1 + \tan^{2}x)\sqrt{1 + \tan^{2}x} - \tan x\cdot\dfrac{\tan x(1 + \tan^{2}x)}{\sqrt{1 + \tan^{2}x}}}{1 + \tan^{2}x}\notag\\ &= \frac{1}{\sqrt{1 + \tan^{2}x}}\notag\\ &= \cos x\notag \end{align} which is valid for all $x \in (-\pi/2, \pi/2)$. The same differentiation formula applies for $x = \pm\pi/2$ also. To check this let $f(x) = \sin x$ and we have \begin{align} f_{+}'(\pi/2) &= \lim_{h \to 0^{+}}\frac{\sin(\pi/2 + h) - \sin (\pi/2)}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{-\sin(\pi/2 + h - \pi) - 1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{-\sin(-\pi/2 + h) - 1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{\sin(\pi/2 - h) - 1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{\cos h - 1}{h}\notag\\ &= -\lim_{h \to 0^{+}}\frac{1 - \cos^{2}h}{h(1 + \cos h)}\notag\\ &= -\frac{1}{2}\lim_{h \to 0^{+}}\frac{\sin^{2}h}{h^{2}}\cdot h\notag\\ &= -\frac{1}{2}\cdot 1^{2}\cdot 0\notag\\ &= 0 = \cos (\pi/2)\notag \end{align} and similarly we can show that $f_{-}'(\pi/2) = 0$. Thus we see that the formula $(\sin x)' = \cos x$ holds for all $x \in (-\pi/2, \pi/2]$ and due to equation $(12)$ it holds for all values of $x$. Similarly we can establish that $(\cos x)' = -\sin x$ for all values of $x$.

When using integrals to define inverse circular functions there are two possibilities: either start with $\arcsin x$ as the integral of $(1 - x^{2})^{-1/2}$ or $\arctan x$ as the integral of $(1 + x^{2})^{-1}$. We prefer the second approach as it is technically easier to handle.

### $\arctan x$ as an Integral

Let us define**the inverse tangent function**denoted by symbol $\arctan$ (or $\tan^{-1}$) as follows $$\arctan x = \int_{0}^{x}\frac{dt}{1 + t^{2}}\tag{1}$$ From the above definition it is clear that $\arctan x$ is defined for all real $x$ and is a strictly increasing function of $x$. Since $$(\arctan x)' = \frac{1}{1 + x^{2}}$$ and $\arctan 0 = 0$ it follows that $$\lim_{x \to 0}\frac{\arctan x}{x} = 1\tag{2}$$ We introduce the number $\pi$ as $$\pi = 4\arctan 1 = 4\int_{0}^{1}\frac{dt}{1 + t^{2}}\tag{3}$$ We next show that $$\lim_{x \to \infty}\arctan x = \frac{\pi}{2}\tag{4}$$ Clearly we can see that \begin{align} \arctan x &= \int_{0}^{x}\frac{dt}{1 + t^{2}}\notag\\ &= \int_{0}^{1}\frac{dt}{1 + t^{2}} + \int_{1}^{x}\frac{dt}{1 + t^{2}}\notag\\ &= \frac{\pi}{4} + \int_{1/x}^{1}\frac{du}{1 + u^{2}}\text{ (by putting }u = 1/t)\notag\\ &= \frac{\pi}{4} + \int_{0}^{1}\frac{du}{1 + u^{2}} - \int_{0}^{1/x}\frac{du}{1 + u^{2}}\notag\\ &= \frac{\pi}{2} - \arctan(1/x)\tag{5} \end{align} Noting that $\arctan x$ is continuous and letting $x \to \infty$ we get $\lim_{x \to \infty}\arctan x = \dfrac{\pi}{2}$ and hence equation $(4)$ is established. Since $\arctan x$ is an odd function (because its derivative is even function and $\arctan 0 = 0$) it follows that $\lim_{x \to -\infty}\arctan x = -\dfrac{\pi}{2}$. It is now clear that $\arctan x$ maps the whole of $\mathbb{R}$ to the interval $(-\pi/2, \pi/2)$ and since it is strictly monotone, there is an inverse function denoted by symbol $\tan$ and we thus define the function $\tan x$ for $x \in (-\pi/2, \pi/2)$ by the equation $$\tan x = y \text{ if }\arctan y = x\tag{6}$$ It should be obvious now that the function $\tan x$ is continuous and differentiable for all $x \in (-\pi/2, \pi/2)$. Moreover $\tan x \to \infty$ as $x \to \pi/2$ and $\tan x \to -\infty$ as $x \to -\pi/2$. From equation $(2)$ we get $$\lim_{x \to 0}\frac{\tan x}{x} = 1\tag{7}$$ Also note that the equation $(5)$ can be recast in the form $$\tan\left(\dfrac{\pi}{2} - x\right) = \dfrac{1}{\tan x}\tag{8}$$ where $0 < x < \pi/2$.The challenge now is to define the functions $\sin x, \cos x$ for all values of $x$. One can first try to define $\tan x$ for all $x$ by making it periodic with period $\pi$. By making $\tan x$ periodic with period $\pi$ we ensure that $\tan x$ is defined for all $x$ except when $x$ is an odd multiple of $\pi/2$. Next we define $\sin x, \cos x$ for $x \in (-\pi/2, \pi/2)$ by the formulas $$\sin x = \frac{\tan x}{\sqrt{1 + \tan^{2}x}},\,\cos x = \frac{1}{\sqrt{1 + \tan^{2}x}}\tag{9}$$ We immediately obtain the fundamental identity $$\cos^{2}x + \sin^{2}x = 1\tag{10}$$ for all $x \in (-\pi/2, \pi/2)$. From the above definition it is obvious that $\sin x \to 1$ as $x \to (\pi/2)^{-}$ and $\cos x \to 0$ as $x \to (\pi/2)^{-}$. Similarly $\sin x \to -1$ as $x \to (-\pi/2)^{+}$ and $\cos x \to 0$ as $x \to (-\pi/2)^{+}$. We thus define $$\sin\left(\frac{\pi}{2}\right) = 1,\,\sin\left(-\frac{\pi}{2}\right) = -1,\,\cos\left(\frac{\pi}{2}\right) = 0,\,\cos\left(-\frac{\pi}{2}\right) = 0\tag{11}$$ in order to make $\sin x, \cos x$ continuous in interval $[-\pi/2, \pi/2]$. To extend the definition of $\sin x, \cos x$ for all $x$ we further define $$\sin (x + \pi) = -\sin x,\,\cos(x + \pi) = -\cos x\tag{12}$$ It now follows that $\sin x, \cos x$ are defined for all $x$ and are periodic with period $2\pi$ (while $\tan x$ is periodic with period $\pi$). With the definitions set up as above we can see that the equation $(8)$ above leads us to $$\sin\left(\frac{\pi}{2} - x\right) = \cos x,\,\cos\left(\frac{\pi}{2} - x\right) = \sin x\tag{13}$$ for $0 < x < \pi/2$. Further we also note that $\sin x$ is an odd function and $\cos x$ is an even function. It is easy to show that the functions $\sin x, \cos x$ are continuous everywhere and therefore from equation $(7)$ we get $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{14}$$

The circular functions defined above are differentiable wherever they are defined. The differentiability of $\tan x$ follows from that of $\arctan x$ and clearly if $\tan x = y$ then $x = \arctan y$ so that $$\frac{dx}{dy} = \frac{1}{1 + y^{2}} = \frac{1}{1 + \tan^{2}x} = \cos^{2}x$$ and therefore $$(\tan x)' = \frac{dy}{dx} = \frac{1}{\cos^{2}x}\tag{15}$$ for all $x \in (-\pi/2, \pi/2)$. Since $\tan (x + \pi) = \tan x$ and $\cos^{2}(x + \pi) = \cos^{2}x$ it follows that the formula $(15)$ holds for all $x$ where $\tan x$ is defined. It is now an easy matter to calculate the derivative of $\sin x$ and $\cos x$. Thus we have \begin{align} (\sin x)' &= \frac{d}{dx}\left(\frac{\tan x}{\sqrt{1 + \tan^{2}x}}\right)\notag\\ &= \dfrac{(1 + \tan^{2}x)\sqrt{1 + \tan^{2}x} - \tan x\cdot\dfrac{\tan x(1 + \tan^{2}x)}{\sqrt{1 + \tan^{2}x}}}{1 + \tan^{2}x}\notag\\ &= \frac{1}{\sqrt{1 + \tan^{2}x}}\notag\\ &= \cos x\notag \end{align} which is valid for all $x \in (-\pi/2, \pi/2)$. The same differentiation formula applies for $x = \pm\pi/2$ also. To check this let $f(x) = \sin x$ and we have \begin{align} f_{+}'(\pi/2) &= \lim_{h \to 0^{+}}\frac{\sin(\pi/2 + h) - \sin (\pi/2)}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{-\sin(\pi/2 + h - \pi) - 1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{-\sin(-\pi/2 + h) - 1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{\sin(\pi/2 - h) - 1}{h}\notag\\ &= \lim_{h \to 0^{+}}\frac{\cos h - 1}{h}\notag\\ &= -\lim_{h \to 0^{+}}\frac{1 - \cos^{2}h}{h(1 + \cos h)}\notag\\ &= -\frac{1}{2}\lim_{h \to 0^{+}}\frac{\sin^{2}h}{h^{2}}\cdot h\notag\\ &= -\frac{1}{2}\cdot 1^{2}\cdot 0\notag\\ &= 0 = \cos (\pi/2)\notag \end{align} and similarly we can show that $f_{-}'(\pi/2) = 0$. Thus we see that the formula $(\sin x)' = \cos x$ holds for all $x \in (-\pi/2, \pi/2]$ and due to equation $(12)$ it holds for all values of $x$. Similarly we can establish that $(\cos x)' = -\sin x$ for all values of $x$.

### Addition Formulas for Circular Functions

The addition formulas for circular functions are a cakewalk once we have the derivatives available for circular functions. Let $x, y$ be real variables connected by the equation $x + y = k$ so that their sum is a constant. Then it follows that $\dfrac{dy}{dx} = -1$. We now consider the function $$f(x) = \sin x\cos y + \cos x\sin y$$ and observe that $$f'(x) = \cos x\cos y + \sin x\sin y - \sin x\sin y - \cos x\cos y = 0$$ and hence $f(x)$ is a constant independent of $x$. Thus $$f(x) = f(k) = \sin k\cos 0 + \cos k\sin 0 = \sin k = \sin (x + y)$$ It follows that we have $$\sin (x + y) = \sin x\cos y + \cos x\sin y$$ and similarly we can establish the formula $$\cos (x + y) = \cos x\cos y - \sin x\sin y$$ Changing the sign of $y$ in these formulas we finally obtain \begin{align} \sin(x \pm y) &= \sin x\cos y \pm \cos x\sin y\tag{16a}\\ \cos(x \pm y) &= \cos x\cos y \mp \sin x\sin y\tag{16b} \end{align} With these addition formulas available all the formulas of trigonometry can be established and the theory of circular functions as inverses to certain integrals is now complete.**Print/PDF Version**
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