While answering certain questions on MSE in last few weeks it occurred to me that ample confusion is prevalent among students (and instructors alike) regarding a theoretically sound development of circular (or trigonometric) functions. In the past I had hinted at two usual approaches to trigonometry, but I guess that was not enough and hence I am writing this series on the development of circular functions (like I did for the exponential and logarithmic functions earlier).

Like the case of logarithmic and exponential functions I will restrict myself to the theory of circular functions of real variables only. Thus the functions $\sin x, \cos x$ will be defined for all real $x$ in a systematic and sound manner. Based on these definitions their elementary properties (like addition formulas) and analytic properties (continuity and differentiability) will also be derived.

In the above figure $A$ is the point where the circle meets the positive half of the $X$-axis and $P$ is an arbitrary point on the circle (and for purposes of illustration we have taken $P$ to be in the first quadrant). The introduction of the point $P$ and $A$ on the circle leads to two important things which we need here: the first is the arc $AP$ and second sector $AOP$. Using integral calculus it is easy to prove that the arc $AP$ possesses a length and the sector $AOP$ has an area. If the point $P$ has coordinates $(a, b)$ then it is easy to show that \begin{align} L &= \text{length of arc }AP = \int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}}\tag{1}\\ A &= \text{area of sector }AOP\notag\\ &= \text{area of }\Delta OPB + \text{area of region }APB\notag\\ &= \frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx\tag{2} \end{align} Note that the existence of the area of sector $AOP$ is dependent on the continuity of the function $f(x) = \sqrt{1 - x^{2}}$ on interval $[0, 1]$ and the existence of length of arc $AP$ follows from the fact that the function $f(x) = \sqrt{1 - x^{2}}$ is of bounded variation on $[0, 1]$ (because it is monotone on that interval). We now show that there is a direct relationship between the length of arc $AP$ and the area of sector $AOP$ namely $L = 2A$ so that the length of an arc of unit circle is twice the area of the corresponding sector. This fact is a direct consequence of the equation of unit circle (i.e. it is an inherent property of a circle) and it has nothing to do with the nature of circular functions (to be defined later).

Clearly we have via integration by parts \begin{align} \int\sqrt{1 - x^{2}}\,dx &= \int 1\cdot\sqrt{1 -x^{2}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int x\cdot\frac{-x}{\sqrt{1 - x^{2}}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int \frac{1 - x^{2} - 1}{\sqrt{1 - x^{2}}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int \sqrt{1 - x^{2}}\,dx + \int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow 2\int\sqrt{1 - x^{2}}\,dx &= x\sqrt{1 - x^{2}} + \int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int\sqrt{1 - x^{2}}\,dx &= \frac{x\sqrt{1 - x^{2}}}{2} + \frac{1}{2}\int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int_{a}^{1}\sqrt{1 - x^{2}}\,dx &= -\frac{a\sqrt{1 - a^{2}}}{2} + \frac{1}{2}\int_{a}^{1} \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}} &= 2\left(\frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx\right)\notag \end{align} From equations $(1)$ and $(2)$ it is now obvious that $L = 2A$.

Both $L$ and $A$ are dependent on the position of point $P$ on the circle. Let's make the

Thus although the dependence of $L$ on position of $P$ is direct, it does not match the definition of a functional dependence. The magic happens when we invert the dependence of $L$ on position of $P$ and start to think of the position of point $P$ as a function of the arc length $L$. We thus arrive at the definition of circular functions $\sin x, \cos x$. More formally,

This is the way some elementary textbooks define the circular functions, but they typically miss the justification of area of sectors and length of arcs. Also the fundamental relation between the area of a sector and the length of corresponding arc is not established directly, but rather this link is shown via the use of analytic properties of circular functions (for more details see my answer on MSE). Further as we had discussed earlier, the length of arc $AP$ is also taken to be the measure of angle $AOP$ and we sometimes say that $\sin x$ is the sine of angle $AOP$. A definition of circular functions is not complete unless we introduce the number $\pi$ (like the definition of logarithm is not complete without the introduction of number $e$).

where $P(\cos x, \sin x)$ is a point on the unit circle and $P$ lies in the first quadrant so that $0 < x < \pi/2$. Let $AT$ be the tangent to circle at point $A$ and let $OP$ intersect $AT$ in $T$. Since $PB$ is perpendicular to $OA$ it follows that the triangles $OPB$ and $OTA$ are similar and hence $$TA = \frac{TA}{OA} = \frac{PB}{OB} = \frac{\sin x}{\cos x} = \tan x$$ Now consider the areas of $\Delta OAP, \Delta OAT$ and sector $OAP$. From the figure it is obvious that $$\text{area of }\Delta OAP < \text{area of sector }OAP < \text{area of }\Delta OAT$$ or $$\frac{1}{2}\cdot OA\cdot PB < \frac{x}{2} < \frac{1}{2}\cdot OA \cdot TA$$ or $$\sin x < x < \tan x$$ Thus we have $0 < \sin x < x$ for $0 < x < \pi/2$ and taking limits when $x \to 0^{+}$ we get $\lim_{x \to 0^{+}}\sin x = 0$. Since $\sin (-x) = -\sin x$ it follows now that $$\lim_{x \to 0}\sin x = 0\tag{11a}$$ Further we can see that when $-\pi/2 < x < \pi/2$ then $\cos x$ is positive and thus $\cos x = \sqrt{1 - \sin^{2}x}$ and hence $$\lim_{x \to 0}\cos x = 1\tag{11b}$$ We can now prove that $\sin x, \cos x$ are continuous everywhere. Clearly we have \begin{align} \lim_{x \to a}\sin x &= \lim_{h \to 0}\sin (a + h)\notag\\ &= \lim_{h \to 0}\sin a\cos h + \cos a\sin h\notag\\ &= \sin a \cdot 1 + \cos a\cdot 0\notag\\ &= \sin a\notag \end{align} so that $\sin x$ is continuous everywhere and continuity of $\cos x$ is proved in exactly the same manner. To obtain derivatives of circular functions we establish the fundamental limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{12}$$ From inequality $(10)$ we get $$\sin x < x < \frac{\sin x}{\cos x}$$ or $$\cos x < \frac{\sin x}{x} < 1$$ for $0 < x < \pi/2$. Taking limits when $x \to 0^{+}$ we get $\lim\limits_{x \to 0^{+}}\dfrac{\sin x}{x} = 1$. Also since $(\sin (-x))/(-x) = (\sin x)/x$, it follows that the same result holds when $x \to 0^{-}$. Thus the limit formula $(12)$ is established. We now calculate the derivative of $\sin x$ as follows \begin{align} (\sin x)' &= \lim_{h \to 0}\frac{\sin (x + h) - \sin x}{h}\notag\\ &= \lim_{h \to 0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h}\notag\\ &= \lim_{h \to 0}\sin x \cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}\notag\\ &= \sin x\lim_{h \to 0}\frac{\cos^{2} h - 1}{h(\cos h + 1)} + \cos x\cdot 1\notag\\ &= \cos x - \sin x\lim_{h \to 0}\frac{\sin^{2}h}{h^{2}}\cdot\frac{h}{1 + \cos h}\notag\\ &= \cos x - \sin x \cdot 1\cdot\frac{0}{1 + 1}\notag\\ &= \cos x\notag \end{align} Similarly we can show that $(\cos x)' = -\sin x$. We have now established both the algebraic and analytic properties of circular functions and the development of theory of circular functions is complete. In the next post we will discuss other approaches (not based on geometric notions) to develop the theory of circular functions.

Like the case of logarithmic and exponential functions I will restrict myself to the theory of circular functions of real variables only. Thus the functions $\sin x, \cos x$ will be defined for all real $x$ in a systematic and sound manner. Based on these definitions their elementary properties (like addition formulas) and analytic properties (continuity and differentiability) will also be derived.

### What is the $x$ in $\sin x$?

While dealing with functions of a real variable $x$ it is expected that given a real number $x$, it should be possible to find the value of the function at point $x$ in a definite manner so that for each given $x$ in a certain domain there is a unique value of the function. Sometimes the value of the function is obtained by the use of algebraic manipulations of $x$ with other numbers, but when dealing with circular functions like $\sin x, \cos x$ the way to obtain $\sin x, \cos x$ given the value of $x$ is very very very (!) indirect. And that is the whole problem with these functions. Beginners are almost always kept in the dark about "What is the $x$ in $\sin x$?". It appears that the only way to figure out the $x$ in $\sin x$ is to talk something about circles first (that's the reason for the name circular functions!) and then define $\sin x, \cos x$ in terms of quantities related to a circle. Thus we start off with a unit circle with center at origin whose equation is given by $x^{2} + y^{2} = 1$.In the above figure $A$ is the point where the circle meets the positive half of the $X$-axis and $P$ is an arbitrary point on the circle (and for purposes of illustration we have taken $P$ to be in the first quadrant). The introduction of the point $P$ and $A$ on the circle leads to two important things which we need here: the first is the arc $AP$ and second sector $AOP$. Using integral calculus it is easy to prove that the arc $AP$ possesses a length and the sector $AOP$ has an area. If the point $P$ has coordinates $(a, b)$ then it is easy to show that \begin{align} L &= \text{length of arc }AP = \int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}}\tag{1}\\ A &= \text{area of sector }AOP\notag\\ &= \text{area of }\Delta OPB + \text{area of region }APB\notag\\ &= \frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx\tag{2} \end{align} Note that the existence of the area of sector $AOP$ is dependent on the continuity of the function $f(x) = \sqrt{1 - x^{2}}$ on interval $[0, 1]$ and the existence of length of arc $AP$ follows from the fact that the function $f(x) = \sqrt{1 - x^{2}}$ is of bounded variation on $[0, 1]$ (because it is monotone on that interval). We now show that there is a direct relationship between the length of arc $AP$ and the area of sector $AOP$ namely $L = 2A$ so that the length of an arc of unit circle is twice the area of the corresponding sector. This fact is a direct consequence of the equation of unit circle (i.e. it is an inherent property of a circle) and it has nothing to do with the nature of circular functions (to be defined later).

Clearly we have via integration by parts \begin{align} \int\sqrt{1 - x^{2}}\,dx &= \int 1\cdot\sqrt{1 -x^{2}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int x\cdot\frac{-x}{\sqrt{1 - x^{2}}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int \frac{1 - x^{2} - 1}{\sqrt{1 - x^{2}}}\,dx\notag\\ &= x\sqrt{1 - x^{2}} - \int \sqrt{1 - x^{2}}\,dx + \int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow 2\int\sqrt{1 - x^{2}}\,dx &= x\sqrt{1 - x^{2}} + \int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int\sqrt{1 - x^{2}}\,dx &= \frac{x\sqrt{1 - x^{2}}}{2} + \frac{1}{2}\int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int_{a}^{1}\sqrt{1 - x^{2}}\,dx &= -\frac{a\sqrt{1 - a^{2}}}{2} + \frac{1}{2}\int_{a}^{1} \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\ \Rightarrow \int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}} &= 2\left(\frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx\right)\notag \end{align} From equations $(1)$ and $(2)$ it is now obvious that $L = 2A$.

Both $L$ and $A$ are dependent on the position of point $P$ on the circle. Let's make the

*convention*that when the length of arc $AP$ is measured anti-clockwise starting from $A$ to $P$ then the arc length $L$ is positive. If the length of arc $AP$ is measured clockwise from starting from $A$ to $P$ then the arc length $L$ is negative. Also let the same convention hold for the area of sector $AOP$ so that the relation $L = 2A$ holds for both positive and negative values of $L$ and $A$. Further we can see that as the point $P$ moves in anti-clockwise direction from point $A$ the corresponding arc length $L$ increases and once the point $P$ completes one revolution and reaches back to the starting point $A$ the arc length $L$ reaches its maximum value namely the circumference of the circle. Let's again make the convention that the point $P$ is allowed to make as many revolutions as required so that $P$ can traverse the full circumference a multiple number of times. This will ensure that we can have the value of arc length $L$ as large as we want. Similarly these revolutions can be made in clockwise direction also to ensure that the arc length $L$ can have large negative values. Thus by convention we allow the length $L$ of arc $AP$ (or area $A$ of sector $AOP$) to take any real value based on position of point $P$ on unit circle. In so doing we also observe that there can be multiple values of $L$ or $A$ for the same position of $P$ based on the number of revolutions of point $P$ (in clockwise or anti-clockwise direction).Thus although the dependence of $L$ on position of $P$ is direct, it does not match the definition of a functional dependence. The magic happens when we invert the dependence of $L$ on position of $P$ and start to think of the position of point $P$ as a function of the arc length $L$. We thus arrive at the definition of circular functions $\sin x, \cos x$. More formally,

**let $x$ be a real number and based on the sign of $x$ let point $P$ move on unit circle (in anti-clockwise direction if $x > 0$ and in clockwise direction if $x < 0$) starting with point $A(1, 0)$ and reach a position such that the length of arc $AP$ is $x$ (or the area of sector $AOP$ is $x/2$). Then by definition the point $P$ has the coordinates $(\cos x, \sin x)$.**This is the way some elementary textbooks define the circular functions, but they typically miss the justification of area of sectors and length of arcs. Also the fundamental relation between the area of a sector and the length of corresponding arc is not established directly, but rather this link is shown via the use of analytic properties of circular functions (for more details see my answer on MSE). Further as we had discussed earlier, the length of arc $AP$ is also taken to be the measure of angle $AOP$ and we sometimes say that $\sin x$ is the sine of angle $AOP$. A definition of circular functions is not complete unless we introduce the number $\pi$ (like the definition of logarithm is not complete without the introduction of number $e$).

*We define $\pi$ as the area of the unit circle*so that the circumference of the unit circle is by definition $2\pi$. This is equivalent to the following integral formulas $$\pi = 4\int_{0}^{1}\sqrt{1 - x^{2}}\,dx = 2\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}\tag{3}$$### Elementary Properties of Circular Functions

Since the point $P(\cos x, \sin x)$ lies on the unit circle it follows from the equation of the circle that $$\cos^{2}x + \sin^{2}x = 1\tag{4}$$ With the definition of $\pi$ available we have the following results available immediately $$\cos 0 = 1, \sin 0 = 0, \cos \frac{\pi}{2} = 0, \sin \frac{\pi}{2} = 1\tag{5a}$$ $$\cos \pi = -1, \sin \pi = 0, \cos \frac{3\pi}{2} = 0, \sin\frac{3\pi}{2} = -1\tag{5b}$$ $$\cos 2\pi = 1, \sin 2\pi = 0\tag{5c}$$ Further note that if the point $P$ moves in anti-clockwise direction on unit circle and reaches a position such that length of arc $AP$ is $x$ and if $P'$ is the position when point $P$ moves the same distance in clockwise direction so that length of arc $AP'$ is $-x$ then both the points $P$ and $P'$ have same $x$-coordinate, but their $y$-coordinates are of opposite signs. Hence we have $$\sin (-x) = -\sin x, \cos (-x) = \cos x\tag{6}$$ Further note that if point $P$ on unit circle has coordinates $(a, b)$ and we rotate $P$ by a right angle in anti-clockwise direction then $P$'s coordinates change to $(-b, a)$. It thus follows that $$\cos\left(\frac{\pi}{2} + x\right) = -\sin x, \sin\left(\frac{\pi}{2} + x\right) = \cos x\tag{7}$$ Using $(6)$ and $(7)$ together we get $$\cos\left(\frac{\pi}{2} - x\right) = \sin x, \sin\left(\frac{\pi}{2} - x\right) = \cos x\tag{8}$$ Using these formulas repeatedly it is possible to compute $\cos(n(\pi/2) \pm x), \sin (n(\pi/2) \pm x)$ where $n$ is an integer. Moreover we should note that $\cos x$ is positive if $x$ lies in first or fourth quadrant and $\sin x$ is positive if $x$ lies in first or second quadrant. From the above formulas we also conclude that $\sin x, \cos x$ are both periodic with period $2\pi$. Next we establish the addition formulas for the circular functions mentioned below \begin{align} \cos(a \pm b) &= \cos a\cos b \mp \sin a\sin b\tag{9a}\\ \sin(a \pm b) &= \sin a\cos b \pm \cos a\sin b\tag{9b} \end{align} Consider points $P_{1} (\cos a, \sin a), P_{2}(\cos b, \sin b)$ on the unit circle. Then from elementary geometry we know that $$P_{1}P_{2} = \sqrt{2 - 2\cos (a - b)}$$ and from distance formula we know that $$P_{1}P_{2} = \sqrt{(\cos a - \cos b)^{2} + (\sin a - \sin b)^{2}}$$ Equating the above expressions for $P_{1}P_{2}$ we get $$2 - 2\cos(a - b) = \cos^{2}a + \sin^{2}a + \cos^{2}b + \sin^{2}b - 2\{\cos a\cos b + \sin a\sin b\}$$ or $$\cos (a - b) = \cos a\cos b + \sin a\sin b$$ Changing the sign of $b$ and using equation $(6)$ we get the formula for $\cos(a + b)$. Replacing $a$ with $(\pi/2) - a$ in equation $(9a)$ we get the equation $(9b)$ and thus the addition formulas for the circular functions are established.### Analytic Properties of Circular Functions

Next we establish that the circular functions $\cos x, \sin x$ are continuous and differentiable everywhere and in so doing we will also find their derivatives. In order to proceed further we establish the inequality $$\sin x < x < \tan x\tag{10}$$ for $0 < x < \pi/2$ where $\tan x$ is the tangent function defined by $\tan x = \dfrac{\sin x}{\cos x}$. Note that the tangent function is defined for all values of $x$ for which $\cos x \neq 0$. For values of $x$ where $\cos x = 0$ the function $\tan x$ is not defined. Consider the following figurewhere $P(\cos x, \sin x)$ is a point on the unit circle and $P$ lies in the first quadrant so that $0 < x < \pi/2$. Let $AT$ be the tangent to circle at point $A$ and let $OP$ intersect $AT$ in $T$. Since $PB$ is perpendicular to $OA$ it follows that the triangles $OPB$ and $OTA$ are similar and hence $$TA = \frac{TA}{OA} = \frac{PB}{OB} = \frac{\sin x}{\cos x} = \tan x$$ Now consider the areas of $\Delta OAP, \Delta OAT$ and sector $OAP$. From the figure it is obvious that $$\text{area of }\Delta OAP < \text{area of sector }OAP < \text{area of }\Delta OAT$$ or $$\frac{1}{2}\cdot OA\cdot PB < \frac{x}{2} < \frac{1}{2}\cdot OA \cdot TA$$ or $$\sin x < x < \tan x$$ Thus we have $0 < \sin x < x$ for $0 < x < \pi/2$ and taking limits when $x \to 0^{+}$ we get $\lim_{x \to 0^{+}}\sin x = 0$. Since $\sin (-x) = -\sin x$ it follows now that $$\lim_{x \to 0}\sin x = 0\tag{11a}$$ Further we can see that when $-\pi/2 < x < \pi/2$ then $\cos x$ is positive and thus $\cos x = \sqrt{1 - \sin^{2}x}$ and hence $$\lim_{x \to 0}\cos x = 1\tag{11b}$$ We can now prove that $\sin x, \cos x$ are continuous everywhere. Clearly we have \begin{align} \lim_{x \to a}\sin x &= \lim_{h \to 0}\sin (a + h)\notag\\ &= \lim_{h \to 0}\sin a\cos h + \cos a\sin h\notag\\ &= \sin a \cdot 1 + \cos a\cdot 0\notag\\ &= \sin a\notag \end{align} so that $\sin x$ is continuous everywhere and continuity of $\cos x$ is proved in exactly the same manner. To obtain derivatives of circular functions we establish the fundamental limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{12}$$ From inequality $(10)$ we get $$\sin x < x < \frac{\sin x}{\cos x}$$ or $$\cos x < \frac{\sin x}{x} < 1$$ for $0 < x < \pi/2$. Taking limits when $x \to 0^{+}$ we get $\lim\limits_{x \to 0^{+}}\dfrac{\sin x}{x} = 1$. Also since $(\sin (-x))/(-x) = (\sin x)/x$, it follows that the same result holds when $x \to 0^{-}$. Thus the limit formula $(12)$ is established. We now calculate the derivative of $\sin x$ as follows \begin{align} (\sin x)' &= \lim_{h \to 0}\frac{\sin (x + h) - \sin x}{h}\notag\\ &= \lim_{h \to 0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h}\notag\\ &= \lim_{h \to 0}\sin x \cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}\notag\\ &= \sin x\lim_{h \to 0}\frac{\cos^{2} h - 1}{h(\cos h + 1)} + \cos x\cdot 1\notag\\ &= \cos x - \sin x\lim_{h \to 0}\frac{\sin^{2}h}{h^{2}}\cdot\frac{h}{1 + \cos h}\notag\\ &= \cos x - \sin x \cdot 1\cdot\frac{0}{1 + 1}\notag\\ &= \cos x\notag \end{align} Similarly we can show that $(\cos x)' = -\sin x$. We have now established both the algebraic and analytic properties of circular functions and the development of theory of circular functions is complete. In the next post we will discuss other approaches (not based on geometric notions) to develop the theory of circular functions.

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Circular functions actually live naturally on S^1 and not on the reals but in school we only introduce the reals hence the trouble. But I don't know why we teach it like that in school since we are intuitively aware of at least (and this is enough) two one-dimensional 'spaces': one that extends indefinitely in either 'direction' (actually, orientation) and one that returns to the start. In fact, in certain Brahmin cosmologies time was modelled using what we would now think of S^1.

But we are lucky that S^1 does have the reals as its universal cover.

Btw, let me say that its great that you have kept your interest in math alive despite working in the industry.

Anonymous

May 5, 2016 at 12:08 PM