krit.club logo

Trigonometric Functions - Domain and Range of Trigonometric Functions

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The Domain of a trigonometric function refers to the set of all real numbers (angles) for which the function is defined. Visually, this is represented by the horizontal spread of the function's graph along the xx-axis. For functions like sin⁡x\sin x and cos⁡x\cos x, the domain is the entire set of real numbers R\mathbb{R} because they are defined for every possible angle.

•

The Range of a trigonometric function is the set of all output values (resultant ratios) the function can take. On a graph, this is the vertical interval between the lowest and highest points on the yy-axis. For sin⁡x\sin x and cos⁡x\cos x, the graph oscillates between the horizontal lines y=−1y = -1 and y=1y = 1, making their range [−1,1][-1, 1].

•

Vertical Asymptotes and Domain Restrictions: Functions such as tan⁡x\tan x and sec⁡x\sec x are undefined where the denominator cos⁡x\cos x equals zero. This occurs at odd multiples of π2\frac{\pi}{2}, i.e., x=(2n+1)π2x = (2n+1)\frac{\pi}{2}. Visually, the graphs of these functions feature vertical dashed lines (asymptotes) at these xx-values, and the curve never touches these lines.

•

Sine and Cosecant Relationship: Since csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}, it is undefined whenever sin⁡x=0\sin x = 0 (at x=nπx = n\pi). Because the values of sin⁡x\sin x lie between −1-1 and 11, the values of its reciprocal csc⁡x\csc x must be outside that range. Visually, the csc⁡x\csc x graph consists of alternating U-shaped curves that stay above y=1y = 1 and below y=−1y = -1, resulting in a range of (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty).

•

Tangent and Cotangent Ranges: Unlike sine and cosine, the tan⁡x\tan x and cot⁡x\cot x functions do not have a maximum or minimum value. Their graphs approach ∞\infty or −∞-\infty as they get closer to their respective vertical asymptotes. Consequently, their range covers all real numbers, represented as (−∞,∞)(-\infty, \infty) or R\mathbb{R}.

•

Periodicity: Trigonometric functions are periodic, meaning their values repeat at regular intervals. sin⁡x\sin x, cos⁡x\cos x, sec⁡x\sec x, and csc⁡x\csc x have a period of 2π2\pi, while tan⁡x\tan x and cot⁡x\cot x have a period of π\pi. This periodic nature means that the domain and range patterns observed in one cycle repeat infinitely across the entire number line.

📐Formulae

f(x)=sin⁡x:Domain =R,Range =[−1,1]f(x) = \sin x: \text{Domain } = \mathbb{R}, \text{Range } = [-1, 1]

f(x)=cos⁡x:Domain =R,Range =[−1,1]f(x) = \cos x: \text{Domain } = \mathbb{R}, \text{Range } = [-1, 1]

f(x)=tan⁡x:Domain =R−{(2n+1)π2:n∈Z},Range =Rf(x) = \tan x: \text{Domain } = \mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}, \text{Range } = \mathbb{R}

f(x)=cot⁡x:Domain =R−{nπ:n∈Z},Range =Rf(x) = \cot x: \text{Domain } = \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}, \text{Range } = \mathbb{R}

f(x)=sec⁡x:Domain =R−{(2n+1)π2:n∈Z},Range =(−∞,−1]∪[1,∞)f(x) = \sec x: \text{Domain } = \mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}, \text{Range } = (-\infty, -1] \cup [1, \infty)

f(x)=csc⁡x:Domain =R−{nπ:n∈Z},Range =(−∞,−1]∪[1,∞)f(x) = \csc x: \text{Domain } = \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}, \text{Range } = (-\infty, -1] \cup [1, \infty)

💡Examples

Problem 1:

Find the range of the function f(x)=2−5cos⁡xf(x) = 2 - 5\cos x.

Solution:

  1. We know the fundamental range of cos⁡x\cos x is −1≤cos⁡x≤1-1 \leq \cos x \leq 1.
  2. Multiply the inequality by −5-5. Note that multiplying by a negative number reverses the inequality: −5(1)≤−5cos⁡x≤−5(−1)-5(1) \leq -5\cos x \leq -5(-1), which simplifies to −5≤−5cos⁡x≤5-5 \leq -5\cos x \leq 5.
  3. Add 22 to all parts of the inequality: 2−5≤2−5cos⁡x≤2+52 - 5 \leq 2 - 5\cos x \leq 2 + 5.
  4. This yields −3≤f(x)≤7-3 \leq f(x) \leq 7.
  5. Therefore, the range is [−3,7][-3, 7].

Explanation:

The range of a transformed cosine function is determined by scaling the basic range [−1,1][-1, 1] by the amplitude and then shifting it vertically by the constant term.

Problem 2:

Find the domain of the function f(x)=1sin⁡2xf(x) = \frac{1}{\sin 2x}.

Solution:

  1. The function is undefined when the denominator is zero: sin⁡2x=0\sin 2x = 0.
  2. The general solution for sin⁡θ=0\sin \theta = 0 is θ=nπ\theta = n\pi, where n∈Zn \in \mathbb{Z}.
  3. Here, 2x=nπ2x = n\pi.
  4. Dividing by 22, we get x=nπ2x = \frac{n\pi}{2}.
  5. The domain is the set of all real numbers excluding these values: Domain=R−{nπ2:n∈Z}\text{Domain} = \mathbb{R} - \{\frac{n\pi}{2} : n \in \mathbb{Z}\}.

Explanation:

For rational trigonometric functions, we must exclude any values from the domain that cause the denominator to become zero. We solve the trigonometric equation sin⁡2x=0\sin 2x = 0 to find those excluded points.