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Trigonometric Functions - Domain and Range of Trigonometric Functions

Grade 11CBSE

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

🔑Concepts

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The sine function, f(x)=sin⁡xf(x) = \sin x, is defined for all real numbers. Geometrically, on a unit circle, it represents the yy-coordinate of a point. Its values oscillate between −1-1 and 11, inclusive. This is visualized by its wave-like periodic graph.

Graph of the sine function showing domain as all real numbers and range from -1 to 1.
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The tangent function, f(x)=tan⁡xf(x) = \tan x, is undefined where cos⁡x=0\cos x = 0. These points occur at odd multiples of π2\frac{\pi}{2}, such as ±π2,±3π2,…\pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \dots. Consequently, the domain is R−{(2n+1)π2:n∈Z}\mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}.

Graph of tangent function showing vertical asymptotes at odd multiples of pi/2.
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The range of sec⁡x\sec x and csc⁡x\csc x is determined by the reciprocal relationship with cos⁡x\cos x and sin⁡x\sin x. Since ∣"sinx∣≤1|"sin x| \le 1 and ∣cos⁡x∣≤1|\cos x| \le 1, the absolute values of their reciprocals must be greater than or equal to 11. Thus, Range =(−∞,−1]∪[1,∞)= (-\infty, -1] \cup [1, \infty).

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Domain constraints for f(x)=cot⁡xf(x) = \cot x arise where sin⁡x=0\sin x = 0. This occurs at integral multiples of π\pi, specifically x=nπx = n\pi. The graph shows that as xx approaches these values, the function tends toward infinity.

📐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.

Problem 3:

Find the range of the function f(x)=3sin⁡x+4f(x) = 3\sin x + 4.

Graph of 3sin(x) + 4 oscillating between y=1 and y=7.

Solution:

  1. We know the range of sin⁡x\sin x is [−1,1][-1, 1].
  2. Multiplying by 33: −1×3≤3sin⁡x≤1×3-1 \times 3 \le 3\sin x \le 1 \times 3, so −3≤3sin⁡x≤3-3 \le 3\sin x \le 3.
  3. Adding 44 to all parts: −3+4≤3sin⁡x+4≤3+4-3 + 4 \le 3\sin x + 4 \le 3 + 4.
  4. This simplifies to 1≤f(x)≤71 \le f(x) \le 7.
  5. Range =[1,7]= [1, 7].

Explanation:

The range is found by applying transformations to the basic sine function. The amplitude increases to 33, and the entire graph shifts upward by 44 units.

Problem 4:

Determine the domain of the function f(x)=11−cos⁡xf(x) = \frac{1}{1 - \cos x}.

Graph showing vertical asymptotes at even multiples of pi for the function 1/(1-cos x).

Solution:

  1. The function is undefined when the denominator is zero: 1−cos⁡x=01 - \cos x = 0.
  2. This implies cos⁡x=1\cos x = 1.
  3. The cosine function equals 11 at x=0,±2π,±4π,…x = 0, \pm 2\pi, \pm 4\pi, \dots.
  4. In general form, x=2nπx = 2n\pi where n∈Zn \in \mathbb{Z}.
  5. Therefore, Domain =R−{2nπ:n∈Z}= \mathbb{R} - \{2n\pi : n \in \mathbb{Z}\}.

Explanation:

The domain excludes points where the denominator vanishes. For 1−cos⁡x1 - \cos x, these are the peaks of the cosine wave.