Lesson Video: Domain and Range of Trigonometric Functions Mathematics

In this video, we will learn how to determine the domain and range of a trigonometric function.

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Video Transcript

In this video, we will learn how to determine the domain and range of trigonometric functions. We will begin by recalling the definitions of domain and range of a function. The domain of a function 𝑓 of π‘₯ is a set of all possible values of π‘₯ such that the expression 𝑓 of π‘₯ is defined. The range of a function 𝑓 of π‘₯ is a set of all possible values that the expression 𝑓 of π‘₯ can take, where π‘₯ is any number from the domain of the function. In particular, we can find the domain and range of a function from its graph. Given the graph of a function, the domain is the part of the horizontal axis where the graph exists and the range is the part of the vertical axis where the graph exists.

Let’s begin by considering the graph of 𝑦 equals sin π‘₯ for values of π‘₯ from negative 360 to 360 degrees. We can see that the function is well defined for every π‘₯-value. This means that the domain of sin π‘₯ are all real numbers. This can also be written as the set of numbers on the open interval from negative ∞ to ∞. We see that the graph oscillates between negative one and one. The maximum value of the graph is one and the minimum value is negative one. This means that the possible values of sin π‘₯ are between these two values. And the range of this function is the set of values on the closed interval from negative one to one. The same is true of the cosine function. This, once again, has a domain of all real numbers and a range from negative one to one inclusive.

We can summarize this as follows. The domain of the functions sin π‘₯ and cos π‘₯ is all real numbers denoted as shown. Note that these are often written as sin πœƒ and cos πœƒ, where the function would be 𝑓 of πœƒ. The range of the functions sin π‘₯ and cos π‘₯ is the set of numbers on the closed interval negative one to one. Let’s now consider how we can find the domain and range of any periodic function from its graph.

The following graph shows the function 𝑓 of πœƒ. Assume the function has a period of two πœ‹. What is the domain of 𝑓 of πœƒ? What is the range of 𝑓 of πœƒ?

We know that all the characteristics of a periodic function are contained over an interval of this length. In this question, we are told the period is equal to two πœ‹. Therefore, we only need to consider the graph between zero and two πœ‹. The domain of any function is a set of all possible input values. And we can see from the graph that the function is well defined at all values of πœƒ. We can therefore conclude that the domain of 𝑓 of πœƒ is all real numbers written as the open interval from negative ∞ to ∞.

The range of any function is the set of all output values. From the graph, we see that the function oscillates and is continuous between negative seven and three. The maximum value of the graph is three, and the minimum value is negative seven. We can therefore conclude that the range of 𝑓 of πœƒ is the set of values on the closed interval from negative seven to three. The two answers to this question are the open interval from negative ∞ to ∞ and the closed interval from negative seven to three.

We will now consider how the transformation of trigonometric functions affects the domain and range. We recall that the sine function had domain and range as shown. Any transformation to this function would not alter its domain. However, certain transformations will affect the range of our function.

Let’s consider the function 𝑓 of π‘₯, which is equal to π‘Ž sin π‘₯ plus 𝑏, where π‘Ž and 𝑏 are real constants. Multiplying a function by a positive constant π‘Ž results in a vertical dilation or stretch by the scale factor π‘Ž. This would change the range of a function from the closed interval negative one, one to the closed interval negative π‘Ž, π‘Ž. However, multiplying a function by a negative constant results in a reflection over the π‘₯-axis and a dilation by the scale factor of the absolute value of π‘Ž. This means that the range of the function π‘Ž sin π‘₯ is equal to the closed interval from the negative absolute value of π‘Ž to the positive absolute value of π‘Ž.

Next, we know that adding 𝑏 to a function results in a vertical shift upwards if 𝑏 is greater than zero and downwards if 𝑏 is less than zero. We can therefore conclude that the range of the function π‘Ž sin π‘₯ plus 𝑏 is the closed interval from the negative absolute value of π‘Ž plus 𝑏 to the positive absolute value of π‘Ž plus 𝑏. Let’s now consider how this works in practice.

Consider the function 𝑓 of π‘₯ is equal to four cos of seven π‘₯ plus πœ‹ plus five. What is the domain of 𝑓 of π‘₯? What is the range of 𝑓 of π‘₯?

We begin by recalling that the domain of any function is the set of all possible input values and that the domain of the cos of πœƒ is all real values. In this question, the expression seven π‘₯ plus πœ‹ is inside the cosine function. As this expression is well defined for any real number, the domain of 𝑓 of π‘₯ is all real numbers, which can be written as the open interval from negative ∞ to ∞. We know that the range is the set of output values. Since the range of seven π‘₯ plus πœ‹ is all real numbers, this expression can take any real value. We will therefore let this equal πœƒ so that we have four cos πœƒ plus five.

We know that cos of πœƒ has a range on the closed interval from negative one to one. So we need to consider how the transformations of this function to four cos πœƒ plus five affect the range. Firstly, we have multiplied the function by four, which results in stretching the range vertically by the factor of four. This gives us the closed interval from negative four to four. Adding five to this expression shifts the function up by five. Negative four plus five is equal to one, and four plus five is equal to nine. This means that the range of 𝑓 of π‘₯ is the closed interval from one to nine.

We could have solved the second part algebraically using our knowledge of inequalities. We know that cos of πœƒ is greater than or equal to negative one and less than or equal to one. Multiplying through by four, we have four cos πœƒ is greater than or equal to negative four and less than or equal to four. Adding five to each term in the inequality, we have four cos πœƒ plus five is greater than or equal to one and less than or equal to nine. This corresponds to the closed interval from one to nine. The domain of the function four cos of seven π‘₯ plus πœ‹ plus five is the open interval from negative ∞ to ∞, and its range is the closed interval from one to nine.

Before moving on to one final example, let’s consider the domain and range of the tangent function. Unlike the sine and cosine functions, the tangent function has domain restrictions. Considering the graph of tan πœƒ over the interval from negative 360 degrees to 360 degrees or negative two πœ‹ radians to two πœ‹ radians, we note that the graph is undefined at 90 degrees, 270 degrees, negative 90 degrees, and negative 270 degrees. Since the tangent function is periodic, this behavior repeats indefinitely every 180 degrees. We can therefore conclude that π‘₯ is not defined and the graph has an asymptote at values of π‘₯ equal to 90 degrees plus 180 degrees multiplied by 𝑛, where 𝑛 is any integer.

The domain of tan π‘₯ can therefore be written as shown. It is all real numbers except for π‘₯ is equal to 90 degrees plus 180 degrees multiplied by 𝑛. This can also be written in radians as π‘₯ is equal to πœ‹ over two plus πœ‹π‘›. Once again, 𝑛 is any integer value. The range of the tangent function is all real values, which can be written as the open interval from negative ∞ to ∞. In our final example, we will identify the input values where a tangent function is undefined.

Find the values of πœƒ in radians such that the function 𝑓 of πœƒ equals tan of three πœƒ is undefined.

We begin by recalling that the domain of the tangent function in radians excludes values of the form πœƒ is equal to πœ‹ over two plus π‘›πœ‹, where 𝑛 is an integer. In this question, we are given the function 𝑓 of πœƒ is equal to the tan of three πœƒ, and we wish to find the values where this is undefined. We can therefore let three πœƒ equal πœ‹ over two plus π‘›πœ‹, where 𝑛 is an integer value. Dividing both sides of this equation by three, we have πœƒ is equal to πœ‹ over six plus π‘›πœ‹ over three. Once again, this is for all integer values of 𝑛. The tan of three πœƒ is therefore undefined for all values of πœƒ equal to πœ‹ over six plus π‘›πœ‹ over three where 𝑛 is an integer.

We will now finish this video by summarizing the key points. The domain of the functions sin πœƒ and cos πœƒ is all real numbers. And the range of these functions is the set of numbers on the closed interval from negative one to one. For any constants π‘Ž and 𝑏, the range of the functions π‘Ž sin πœƒ plus 𝑏 or π‘Ž cos πœƒ plus 𝑏 is the closed interval from the negative absolute value of π‘Ž plus 𝑏 to the absolute value of π‘Ž plus 𝑏. The domain of the tan of πœƒ written in radians is all real numbers except for πœƒ is equal to πœ‹ over two plus π‘›πœ‹, where 𝑛 is an integer. We can also write this in degrees, where πœ‹ over two is equal to 90 degrees and πœ‹ is 180 degrees. Finally, the range of the tangent function, tan πœƒ, is all real numbers.

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