Exercise 5.1 · Q22: Continuity of $\cos$⁠, $\operatorname{cosec}$⁠, $\sec$ and $\cot$

Basics you need

Solution

1. Cosine is continuous at every real number

   $\cos x$ is defined for every real $x$⁠. Fix any real $c$ and put $x=c+h$⁠, so $x\to c$ means $h\to0$⁠:

   $$\begin{aligned} \lim_{x\to c}\cos x&=\lim_{h\to0}\left(\cos c\cos h-\sin c\sin h\right)\\ &=\cos c\cdot1-\sin c\cdot0=\cos c. \end{aligned}$$

   So cosine is continuous at every real number; it is a continuous function on $\R$⁠.

   This is the "similar proof" promised in the Remark after Example 17. The limits $\cos h\to1$ and $\sin h\to0$ (Basic 4) do all the work; $\cos c$ and $\sin c$ are fixed numbers.
   $\cos(A+B)=\cos A\cos B-\sin A\sin B$⁠. Basics: trigonometric identities · Example 17

2. Where $\sin x=0$ and where $\cos x=0$

   On the unit circle, $\sin x$ is the $y$-coordinate of the point at angle $x$⁠. It is $0$ exactly at the two points on the $x$-axis, angles $0$ and $\pi$⁠, plus whole turns:

   $$\sin x=0\iff x=n\pi,\quad n\in\Z.$$

   $\cos x$ is the $x$-coordinate, which is $0$ exactly at the points on the $y$-axis, angles $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ plus whole turns, that is, the odd multiples of $\frac{\pi}{2}$⁠:

   $$\cos x=0\iff x=(2n+1)\tfrac{\pi}{2},\quad n\in\Z.$$
   These zeros are the only places where a quotient with $\sin x$ or $\cos x$ in the denominator can fail, so we need the complete list, not just the zeros in $[0,2\pi)$⁠.
   Zeros of sine and cosine (Class XI). Basics: trigonometric identities

3. The three domains

   A quotient is defined exactly where its denominator is not zero. So, by Step 2:

   Function	Formula	Domain
   cosecant	$\frac{1}{\sin x}$	all real $x$ except $x=n\pi$
   secant	$\frac{1}{\cos x}$	all real $x$ except $x=(2n+1)\frac{\pi}{2}$
   cotangent	$\frac{\cos x}{\sin x}$	all real $x$ except $x=n\pi$

   ($n$ runs over all integers in each case.)

   Continuity is only discussed at points of the domain, so we must know the domain first. For cotangent, the numerator $\cos x$ is never $0$ at the same time as $\sin x$ (since $\sin^2x+\cos^2x=1$⁠), but that does not matter: $\sin x=0$ alone makes $\frac{\cos x}{\sin x}$ undefined.

4. Cosecant is continuous at every point of its domain

   Let $c$ be in the domain, so $\sin c\ne0$⁠. Sine is continuous at $c$ (Example 17), so by Theorem 1 (4) (Remark (ii) with $\lambda=1$⁠), $\frac{1}{\sin x}$ is continuous at $c$⁠:

   $$\lim_{x\to c}\operatorname{cosec}x=\frac{1}{\sin c}=\operatorname{cosec}c.$$
   The condition "$g(c)\ne0$" in Theorem 1 (4) is exactly the condition for $c$ to be in the domain. So the theorem covers every point where the question makes sense.
   Theorem 1 (4) and Remark (ii). Lesson: algebra of continuous functions

5. Secant is continuous at every point of its domain

   Let $c$ be in the domain, so $\cos c\ne0$⁠. Cosine is continuous at $c$ (Step 1), so by Theorem 1 (4), $\frac{1}{\cos x}$ is continuous at $c$⁠:

   $$\lim_{x\to c}\sec x=\frac{1}{\cos c}=\sec c.$$
   Same argument as Step 4 with cosine in the denominator; the excluded points are now the odd multiples of $\frac{\pi}{2}$⁠.

6. Cotangent is continuous at every point of its domain

   Let $c$ be in the domain, so $\sin c\ne0$⁠. Both $\cos$ and $\sin$ are continuous at $c$ (Step 1, Example 17), so by Theorem 1 (4):

   $$\lim_{x\to c}\cot x=\frac{\cos c}{\sin c}=\cot c.$$
   This is the same reasoning the textbook uses for $\tan x=\frac{\sin x}{\cos x}$ in Example 18, with the roles of sine and cosine swapped.
   Compare Example 18 (tangent). Example 18

7. The excluded points, and how to read the answer key

   At $x=n\pi$ the cosecant and cotangent are not defined (division by $0$⁠), and at $x=(2n+1)\frac{\pi}{2}$ the secant is not defined. These points are not in the domains, so by Definition 1 there is nothing to test there: there is no value $f(c)$ to compare with a limit. Moreover, no value could be assigned to repair them, because near such a point the function grows without bound; for example $\operatorname{cosec}(0.001)\approx1000.0002$ and $\operatorname{cosec}(-0.001)\approx-1000.0002$⁠.

   So, in the sense of Definition 2, cosecant, secant and cotangent are continuous functions: each is continuous at every point of its domain. The answer key's wording "continuous except for $x=n\pi$" (and so on) says the same thing: the listed points are exactly the points where the function is not defined.

   The textbook treats $\frac1x$ the same way in Example 9: it is called a continuous function, even though the graph breaks at $0$⁠, because $0$ is not in its domain. A break in the graph at a point outside the domain is not a discontinuity in this book's sense.
   Definition 2 and Example 9, Section 5.2. Lesson: continuity at a point

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