In this explainer, we will learn how to find angle measures given interval and function values.
A trigonometric equation is an equation that involves at least one of the following: a trigonometric function, such as sine, cosine, or tangent; a reciprocal trigonometric function, such as cosecant, secant, or cotangent; or an inverse of any of these. Some of the simpler examples of such equations can be solved without using a calculator; however, for the majority of these, it would be tedious to try and remember specific trigonometric function values. In these cases, we make use of the inverse of the functions alongside knowledge of the symmetry and periodicity of their graphs to calculate additional solutions.
Before we demonstrate how to use the symmetry of the graph of a trigonometric function to find all solutions in a given interval, we will first recall the exact values for the sine, cosine, and tangent of a number of special angles.
Consider a right isosceles triangle with two sides of length 1 cm, as shown in the figure.
Using the Pythagorean theorem, we can calculate the length of the hypotenuse to be . Then, by using trigonometric convention to label the sides with respect to the angle at the top of the diagram, we can calculate the exact value of . 
Using this isosceles triangle and the equilateral triangle below, you should be able to derive the following exact values for sine, cosine, and tangent (as shown in the table).
The exact values for the given values of are as follows.
| 0 | 1 | ||||
| 1 | 0 | ||||
| 0 | 1 | Undefined |
It is important that we are able to recall these exact values without feeling the need to reach for a calculator, and, as such, we should take time to familiarize ourselves with the tables given.
In the first example, we will demonstrate how to use the symmetry of the graph of the sine function, alongside the table of exact values, to find all solutions to a simple trigonometric equation.
Example 1: Identifying the General Solution to a Trigonometric Equation
What is the general solution of ?
Answer
In order to find the general solution to a trigonometric equation, we begin by finding a particular solution. In this case, the table of exact trigonometric values can help.
For an angle given in radian measure, the exact values of the sine function are as follows.
| 0 | |||||
|---|---|---|---|---|---|
| 0 | 1 |
We observe that , so is a particular solution to the equation .
To find further solutions, we sketch the graph of as shown in the following figure.
The solutions to are found by adding the line to the diagram. The values of at the points of intersection of this line with the sine curve are the solutions to the equation.
Since the sine curve has symmetry about in the interval , the second solution is found by subtracting from :
Remember, the sine function is periodic with a period of radians, so further solutions can be found by adding or subtracting radians from the particular solutions.
In other words, the solution is , , for integer values of .
Alternatively, this can be represented using set notation as , , where .
In the previous example, we demonstrated how to interpret the symmetry of the graph of the sine function to find all solutions to an equation. Another powerful tool that can be used to extend the domain of the sine function is the unit circle. Remember, this is a circle centered at the origin with a radius of 1 unit. To use the unit circle to determine the sine of any angle , we start at the point and travel along the circumference of the circle in a counterclockwise direction until the angle that is formed between this point, the origin, and the positive -axis is equal to . Then, if this point has coordinates , is equal to the value of .
Notice that the value of the -coordinate is positive in both the 1st and 2nd quadrants; hence, the value of will also be positive in these quadrants. Also, since the unit circle has reflectional symmetry about the -axis, we can see that .
By continuing to move along the circumference of the unit circle, we can see that for all values of .
These results can be generalized as follows.
How To: Finding Solutions to Simple Equations Involving the Sine Function
If is a solution to the equation , for some constant , a second solution is given by . Then, the set of all solutions to is
If is measured in degrees, the solutions are
While we might feel inclined to memorize these formulas, in practice, it can be much more effective to sketch the graph of the sine function or the unit circle to help us deduce the set of solutions to an equation involving the sine function. Letβs demonstrate this in our next example.
Example 2: Finding the Solutions to a Trigonometric Equation in a Specified Range
Find the set of values satisfying , where .
Answer
To solve this simple trigonometric equation, we begin by rearranging to make the sine function the subject:
Care must be taken with the next stage of calculations. We must take both the positive and negative square roots of . This will have the effect of creating a pair of equations in terms of :
Note that we might have also chosen to factor the expression as and then solve the equation to achieve the same result.
Next, to solve the first equation, we recall the exact values for , where is measured in degrees, as follows.
| 0 | 1 |
Hence, a solution to the equation is . Note that this is outside of the specified interval , so we will consider the symmetry of the unit circle to find further solutions.
We can see that the only solution in the given interval is .
Now, we will solve the second equation by, once again, using the symmetry of the unit circle. Since for all and the value of is negative in the 3rd and 4th quadrants, we add right triangles here as shown.
Hence, the next two solutions are and . At this stage, we might feel inclined to find further solutions by adding integer multiples of to these, but doing so would result in values outside of the required interval.
The set of values satisfying , where , is
In the previous example, we demonstrated how to use the symmetry of the unit circle with information about exact values to solve trignometric equations where is not an acute angle. It is worth noting that in order to solve the second equation , we could have alternatively used the inverse sine function and a calculator to find
If we had chosen this method, we would have then needed to consider the regions in the unit circle where and applied the rules about symmetry and periodicity as normal.
In our next example, we will look at how to use the symmetry of the graph of the cosine function to solve a trignometric equation.
Example 3: Solving a Trigonometric Equation Involving a Shifted Angle in a Specified Range
Find the set of values satisfying , where .
Answer
In order to find the solutions to a trigonometric equation in a given interval, we begin by finding a particular solution. In this case, the table of exact trigonometric values can help.
We will first redefine the argument of the function by letting such that
Amending the interval over which our solutions are valid by adding gives . Then, the exact values for , where is measured in degrees, are as follows.
| 1 | 0 |
We can see that ; however, there are no values of given in the table such that . By sketching the graph of the cosine function and the lines and , we can find the associated value of .
The graph has rotational symmetry between about , so the first solution to is
Using the symmetry of the curve, the next solution is
There appears to be a further solution obtained by ; however, this is outside the interval .
Hence, the solutions to are and .
Since we defined , the solutions to are
Therefore, the solution set is
Note that an alternative technique to finding the particular solution to is to use the inverse cosine function such that
At this stage, the remaining steps to find the other solutions are the same.
Remember, we can also use the symmetry of the unit circle to achieve the same result. Just as the value of the -coordinate of the point of intersection of the terminal side of an angle with the unit circle tells us the value of , the value of the -coordinate tells us the value of .
Since the value of the -coordinate is positive in the 1st and 4th quadrants, by symmetry of the unit circle, .
Using the symmetry of the unit circle and periodicity of the cosine function, we can quote formulas for the general solution to equations involving this function.
How To: Solving Simple Equations Involving the Cosine Function
If is a solution to the equation , for some constant , a second solution is given by . Then, the set of all solutions to is
If is measured in degrees, the solutions are
In the previous example, we demonstrated how to solve a trigonometric equation where the argument of the trigonometric function has been transformed in some way. In such cases, we begin by redefining the argument and amending the interval over which we are solving, allowing us to use the symmetry of the standard trigonometric curves, before finally solving the resulting equations in . This is, generally, a more sensible route to take than attempting to apply transformations to the curves of the trigonometric functions. This is summarized below.
How To: Solving a Simple Trigonometric Equation Involving a Transformed Argument
- Redefine the argument of the trigonometric function (e.g., let ).
- Amend the given interval over which solutions are to be found using the same definition (e.g., means ).
- Find all solutions to the resulting equation over this new interval.
- Convert these solutions back into the original variable using the original definition (e.g., ).
We will now demonstrate how to use this technique to solve an equation involving a triple angle.
Example 4: Solving a Trigonometric Equation Involving a Triple Angle in a Specified Range
Find the set of values satisfying , where .
Answer
To solve , we will begin by redefining the argument. Letting so that , the equation becomes , where . To simplify this interval, we multiply by 3 to get .
Now, we can solve the equation over this new interval. For an angle given in radian measure, the exact values of the sine function are as follows.
| 0 | |||||
|---|---|---|---|---|---|
| 0 | 1 |
Hence, a solution to the equation is . In order to find further solutions, we will sketch the graph of over the interval , remembering that the sine function is periodic with a period of radians.
We can see that the solutions occur every radians, so our additional solutions are and .
Hence, the solution set of over the required interval is
Since , we can find the solution set of by dividing each of these values by 3. The set of solutions is then
We will now demonstrate how to apply this process to solve equations involving the tangent function.
Example 5: Solving a Trigonometric Equation Involving a Shifted Double Angle in a Specified Range
Find the set of values satisfying , where .
Answer
To solve this equation, we will begin by redefining the argument to allow us to use the symmetry of the tangent function. Let , so over .
Then, we can use the table of exact values and knowledge about the periodicity of the tangent function to solve this new equation.
Remember, for measured in radians, the exact values of are as follows.
| 0 | |||||
|---|---|---|---|---|---|
| 0 | 1 | Undefined |
We can see that , so letβs use this to find the value of , where . The graph of is sketched over the interval below.
We can see that the graph of the tangent function has rotational symmetry about , where . Hence, the first solution to is . Similarly, since the function is periodic with a period of radians, the remaining solutions are found by adding multiples of to this value:
We now have four solutions to over the required interval. Since we defined , we can find the value of by setting in each case and solving for . In the case of the first value of , this gives
In a similar way, the remaining values of are , and . Hence, the set of values satisfying , where , is
In the previous example, we considered the periodicity of the tangent function. As we have done for the sine and cosine functions, we can quote the general solutions to equations involving this function.
How To: Solving Simple Equations Involving the Tangent Function
If is a solution to the equation , for some constant , the set of all solutions to is
If is measured in degrees, the solution is
So far, we have only considered the βstandardβ trigonometric functions: sine, cosine, and tangent. It is important to understand that the process holds for the reciprocal functions: cosecant, secant, and cotangent. We will demonstrate this in the next example.
Example 6: Finding the General Solution to a Reciprocal Trigonometric Equation
Find the general solution to the equation .
Answer
Recall that the secant function is the reciprocal of the cosine function. In other words,
Hence, the equation can be rewritten as
We know that for an angle measured in radians, the following exact values of the cosine function apply.
| 0 | |||||
|---|---|---|---|---|---|
| 0 | 0 |
While we observe the value of to be , there is no value for . Instead, we will sketch the graph of to deduce the relevant solutions.
The graph has rotational symmetry between about , so the first solution to is
Similarly, the next solution is given by
Since the cosine function is periodic with a period of radians, further solutions are found by adding integer multiples to either of these solutions. For instance, we could find another solution by subtracting from :
In other words, the general solution to is
Alternatively,
In this explainer, we have demonstrated how to use the symmetry and periodicity of the sine, cosine, and tangent functions to find all solutions to a trigonometric equation over a given interval or its general solution. We will now recap the key concepts.
Key Points
- We can solve simple trigonometric equations using tables of exact values or inverse trigonometric functions.
- To help us calculate all solutions to a given equation in a specified range, we can draw the graph of the necessary trigonometric function or use the unit circle.
- The symmetry and periodicity of the sine, cosine, and tangent functions allow us to calculate further solutions to trigonometric equations or general solutions involving integer multiples of or for sine and cosine and and for tangent.
