Discontinuity of Trigonometric Functions

Trigonometric functions are an important type of function in calculus, and some of them are discontinuous at certain points.

Understanding the discontinuity of trigonometric functions is essential for mastering the concept of discontinuity.

Basic Properties

Trigonometric functions have the following basic properties:

  • Periodicity: have periodic features
  • Continuity: sin⁡x\sin x and cos⁡x\cos x are continuous on R\mathbb{R}
  • Discontinuity: tan⁡x\tan x and cot⁡x\cot x are discontinuous at certain points
  • Asymptotes: there is a vertical asymptote at the points of discontinuity

Continuous Trigonometric Functions

The Sine and Cosine Functions

Sine function: f(x)=sin⁡xf(x) = \sin x

Properties:

  • Domain: R\mathbb{R}
  • Continuous everywhere on R\mathbb{R}
  • Period 2π2\pi
  • Range: [−1,1][-1, 1]

Cosine function: f(x)=cos⁡xf(x) = \cos x

Properties:

  • Domain: R\mathbb{R}
  • Continuous everywhere on R\mathbb{R}
  • Period 2π2\pi
  • Range: [−1,1][-1, 1]

Discontinuous Trigonometric Functions

The Tangent Function

Tangent Function

f(x)=tan⁡xf(x) = \tan x

Properties:

  • Domain: x≠π2+kπx \neq \frac{\pi}{2} + k\pi (kk an integer)
  • Continuous on its domain
  • Period π\pi
  • Range: R\mathbb{R}
  • Has a vertical asymptote at every point of discontinuity

Analysis of the points of discontinuity:

  • Points of discontinuity: x=π2+kπx = \frac{\pi}{2} + k\pi (kk an integer)
  • Left limit: lim⁡x→(π2+kπ)−tan⁡x=+∞\lim_{x \to (\frac{\pi}{2} + k\pi)^-} \tan x = +\infty
  • Right limit: lim⁡x→(π2+kπ)+tan⁡x=−∞\lim_{x \to (\frac{\pi}{2} + k\pi)^+} \tan x = -\infty
  • These are discontinuities of the second kind

The Cotangent Function

Cotangent Function

f(x)=cot⁡xf(x) = \cot x

Properties:

  • Domain: x≠kπx \neq k\pi (kk an integer)
  • Continuous on its domain
  • Period π\pi
  • Range: R\mathbb{R}
  • Has a vertical asymptote at every point of discontinuity

Analysis of the points of discontinuity:

  • Points of discontinuity: x=kπx = k\pi (kk an integer)
  • Left limit: lim⁡x→(kπ)−cot⁡x=−∞\lim_{x \to (k\pi)^-} \cot x = -\infty
  • Right limit: lim⁡x→(kπ)+cot⁡x=+∞\lim_{x \to (k\pi)^+} \cot x = +\infty
  • These are discontinuities of the second kind

The Secant and Cosecant Functions

Secant function: f(x)=sec⁡x=1cos⁡xf(x) = \sec x = \frac{1}{\cos x}

Properties:

  • Domain: x≠π2+kπx \neq \frac{\pi}{2} + k\pi (kk an integer)
  • Continuous on its domain
  • Period 2π2\pi
  • Has a vertical asymptote at every point of discontinuity

Cosecant function: f(x)=csc⁡x=1sin⁡xf(x) = \csc x = \frac{1}{\sin x}

Properties:

  • Domain: x≠kπx \neq k\pi (kk an integer)
  • Continuous on its domain
  • Period 2π2\pi
  • Has a vertical asymptote at every point of discontinuity

Analysis of the Discontinuity of Trigonometric Functions

Types of Point of Discontinuity

The points of discontinuity of trigonometric functions mainly fall into the following types:

  1. Discontinuities of the second kind: the limit tends to infinity
  2. Vertical asymptotes: there is a vertical asymptote at the point of discontinuity
  3. Periodic discontinuities: the points of discontinuity have periodic features

Determining Continuity

  1. sin⁡x\sin x and cos⁡x\cos x: continuous everywhere on R\mathbb{R}
  2. tan⁡x\tan x and cot⁡x\cot x: continuous on their domains, with a vertical asymptote at every point of discontinuity
  3. sec⁡x\sec x and csc⁡x\csc x: continuous on their domains, with a vertical asymptote at every point of discontinuity

Discontinuity of Composite Trigonometric Functions

Basic Theorem

If the function f(x)f(x) is continuous at x0x_0, then sin⁡(f(x))\sin(f(x)) and cos⁡(f(x))\cos(f(x)) are continuous at x0x_0.

Examples

Example 1: f(x)=sin⁡(1x)f(x) = \sin(\frac{1}{x})

Analysis:

  • The inner function h(x)=1xh(x) = \frac{1}{x} is continuous at x≠0x \neq 0
  • The outer function g(x)=sin⁡xg(x) = \sin x is continuous on R\mathbb{R}
  • Therefore f(x)=g(h(x))f(x) = g(h(x)) is continuous at x≠0x \neq 0
  • It is discontinuous at x=0x = 0 (the inner function is undefined at that point)

Example 2: f(x)=tan⁡(x2)f(x) = \tan(x^2)

Analysis:

  • The inner function h(x)=x2h(x) = x^2 is continuous on R\mathbb{R}
  • The outer function g(x)=tan⁡xg(x) = \tan x is continuous on its domain
  • Therefore f(x)=g(h(x))f(x) = g(h(x)) is continuous on its domain
  • Points of discontinuity: those where x2=π2+kπx^2 = \frac{\pi}{2} + k\pi, that is, x=±π2+kπx = \pm\sqrt{\frac{\pi}{2} + k\pi}

Graph Characteristics of Trigonometric Functions

Summary of Graph Features

  1. Periodicity: trigonometric functions are periodic, so the graph repeats itself
  2. Smoothness: on its domain, the graph of a trigonometric function is a smooth curve
  3. Asymptotes: there is a vertical asymptote at every point of discontinuity
  4. Symmetry: sin⁡x\sin x is an odd function and cos⁡x\cos x is an even function

Exercises

Exercise 1

Determine the continuity of f(x)=tan⁡(x2)f(x) = \tan(x^2) at x=0x = 0.

trigonometric functioncontinuityfunction continuity

Approach: Use the continuity of composite functions.

Steps:

  1. The inner function g(x)=x2g(x) = x^2 is continuous on R\mathbb{R}
  2. The outer function h(x)=tan⁡xh(x) = \tan x is continuous on its domain
  3. Since x2≥0x^2 \geq 0, and tan⁡x\tan x is continuous at x=0x = 0
  4. Therefore f(x)=h(g(x))f(x) = h(g(x)) is continuous at x=0x = 0

Answer: The function is continuous at x=0x = 0.

Exercise 2

Determine the continuity of f(x)=sin⁡(1x)f(x) = \sin(\frac{1}{x}) at x=0x = 0.

trigonometric functioncontinuityfunction continuity

Approach: Check whether the function is defined at that point.

Steps:

  1. The function is undefined at x=0x = 0
  2. Therefore the function is discontinuous at x=0x = 0

Answer: The function is discontinuous at x=0x = 0, because the function is undefined at that point.

Exercise 3

Determine the intervals of continuity of f(x)=cot⁡(x2)f(x) = \cot(x^2).

trigonometric functioncontinuityfunction continuity

Approach: Analyse the domain and the continuity of the composite function.

Steps:

  1. The inner function g(x)=x2g(x) = x^2 is continuous on R\mathbb{R}
  2. The outer function h(x)=cot⁡xh(x) = \cot x is continuous on its domain
  3. We need x2≠kπx^2 \neq k\pi, that is, x≠±kπx \neq \pm\sqrt{k\pi} (kk a non-negative integer)
  4. Therefore the interval of continuity is all real numbers except x=±kπx = \pm\sqrt{k\pi}

Answer: The interval of continuity is all real numbers except x=±kπx = \pm\sqrt{k\pi}.

Exercise 4

Determine the continuity of f(x)=tan⁡(sin⁡x)f(x) = \tan(\sin x) on R\mathbb{R}.

trigonometric functioncontinuityfunction continuity

Approach: Use the continuity of composite functions.

Steps:

  1. The inner function g(x)=sin⁡xg(x) = \sin x is continuous on R\mathbb{R}
  2. The outer function h(x)=tan⁡xh(x) = \tan x is continuous on its domain
  3. Since sin⁡x∈[−1,1]\sin x \in [-1, 1], and tan⁡x\tan x is continuous on [−1,1][-1, 1]
  4. Therefore f(x)=h(g(x))f(x) = h(g(x)) is continuous on R\mathbb{R}

Answer: The function is continuous on R\mathbb{R}.

Exercise 5

Determine the continuity of f(x)=cos⁡(ln⁡x)f(x) = \cos(\ln x) at x=1x = 1.

trigonometric functioncontinuityfunction continuity

Approach: Use the continuity of composite functions.

Steps:

  1. The inner function g(x)=ln⁡xg(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  2. The outer function h(x)=cos⁡xh(x) = \cos x is continuous on R\mathbb{R}
  3. Therefore f(x)=h(g(x))f(x) = h(g(x)) is continuous on (0,+∞)(0, +\infty)
  4. In particular, f(x)f(x) is continuous at x=1x = 1

Answer: The function is continuous at x=1x = 1.

Exercise 6

Let f(x)={sin⁡x,x≤0x+1,x>0f(x) = \begin{cases} \sin x, & x \leq 0 \\ x + 1, & x > 0 \end{cases}. Determine the continuity of f(x)f(x) at x=0x = 0.

trigonometric functioncontinuityfunction continuity

Approach: Compute the left limit, the right limit and the function value separately, and check whether the three are equal.

Steps:

  1. Left limit: lim⁡x→0−f(x)=lim⁡x→0−sin⁡x=0\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin x = 0
  2. Right limit: lim⁡x→0+f(x)=lim⁡x→0+(x+1)=1\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1
  3. Function value: f(0)=sin⁡0=0f(0) = \sin 0 = 0
  4. The left limit and the right limit are not equal, so the function is discontinuous at x=0x = 0

Answer: The function is discontinuous at x=0x = 0; this is a jump discontinuity.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
R\mathbb{R}Mathematical symbolBlackboard bold R (Real numbers)The set of all real numbers
π\piGreek letterPi (pi)The ratio of a circle’s circumference to its diameter, used to denote angles (such as π\pi, kπ\sqrt{k\pi})

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