In this explainer, we will learn how to evaluate the trigonometric functions with special angles and how to use them to evaluate trigonometric expressions.
We will begin by recalling special angles, together with the sine, cosine, and tangent values of these angles.
Let us consider a unit circle. This enables us to calculate , , and between and or 0 and radians. All three functions have key values at , , , , and . We know that , so , , and . Knowing these conversions enables us to solve trigonometric problems when the angles are given in either degrees or radians.
| 0 | 1 | 0 | 0 | ||
| 1 | 0 | 0 | 1 | ||
| 0 | undefined | 0 | undefined | 0 | |
As the functions are periodic, we can calculate the sine, cosine, and tangent of angles outside of this range by adding multiples of for and , or by adding multiples of for .
Next, we recall that the special angles are , , and . The sine, cosine, and tangent of these angles are given below.
| 1 | |||
While we will not consider the derivation or proof of these results in this explainer, it is worth recalling the identity . This enables us to calculate the tangent of any angle if we are given the sine and cosine of that angle.
For example, since and , then .
It will also be necessary for us to recall the definitions of the reciprocal trigonometric functions.
Definition: Reciprocal Trigonometric Functions
The reciprocal trigonometric functions cosecant , secant , and cotangent are the reciprocal of sine , cosine , and tangent such that
We can use these identities to calculate the cosecant, secant, and cotangent of , , and .
| 2 | |||
| 2 | |||
| 1 | |||
We will now recall the related angles of the trigonometric functions:
One way of recalling whether the sine, cosine, and tangent of any angle between and are positive or negative is using the CAST diagram. This is a memory device that we use to remember the signs of the trigonometric ratios in each of the four quadrants.
| The Quadrant in Which the Terminal Side of the Angle Lies | The Interval in Which the Measure of the Angle Belongs | Signs of Trigonometric Functions | ||
|---|---|---|---|---|
| , | , | , | ||
| First | + | + | + | |
| Second | + | |||
| Third | + | |||
| Fourth | + | |||
We note that the angles are measured, from to or from 0 to radians in a counterclockwise direction, where the positive -axis is the initial side of the angle. The terminal side is where the angle stops. Any angle between and lies in the first quadrant. Any angle between and lies in the second quadrant. Any angle between and lies in the third quadrant. Any angle between and lies in the fourth quadrant.
Let us consider an example where we have to evaluate the cosine of an angle using the property of related angles to relate it to a special angle.
Example 1: Using Periodic Identities to Find the Value of a Trigonometric Function Involving Special Angles
Find the value of .
Answer
We begin by recalling that .
So,
We therefore need to calculate .
Let us recall the property of related angles
If , then
From our knowledge of special angles, we know that .
So,
We can therefore conclude that .
Let us now consider a similar example where, this time, we need to evaluate the sine and tangent of given angles.
Example 2: Evaluating Trigonometric Expressions Involving Special Angles
Evaluate .
Answer
We begin by recalling that .
So,
Also,
We therefore need to calculate .
From our knowledge of special angles, we know that and .
By considering the CAST diagram, as shown below, we see that lies in the 3rd quadrant and the sine of any angle here is negative.
Since then
So,
Substituting the values of and into our expression, we have
Therefore, .
In the remaining examples in this explainer, we will also need to use reciprocal trigonometric functions.
Example 3: Using Periodic Identities to Evaluate a Trigonometric Expression Involving Special Angles
Find the value of .
Answer
In order to answer this question, we need to recall reciprocal trigonometric identities and special angles.
From our knowledge of special angles, , , and .
Using the CAST diagram, we know that both the cosine and tangent of any angle in the second quadrant are negative. The sine and cosine of any angle in the third quadrant are also both negative.
The properties of related angles state that
So,
The properties of related angles also state that
So,
Therefore,
They also state that
So,
Therefore,
They also state that
So,
Therefore,
Since then
Substituting these values into our expression, we have
So, our final answer is .
For the next example, we will need to evaluate a second-degree trigonometric expression.
Example 4: Evaluating Trigonometric Expressions Involving Special Angles
Evaluate .
Answer
Let us recall the property of related angles
If , then
From our knowledge of special angles, we know that .
So,
Substituting this value into our expression gives us
So, our final answer is 0.
In the next example, we will investigate how different angles can satisfy a trigonometric equation involving multiple angles and the power of a trigonometric function.
Example 5: Using Periodic Identities to Evaluate a Trigonometric Expression Involving Special Angles
Which of the following values of does not satisfy the equation ?
Answer
In this example, we have been given several values of and are being asked which does not satisfy the given equation. We note that since the given equation is a trigonometric one, it can possibly have an infinite number of solutions due to the properties of related angles.
The easiest way to check which value of does not satisfy the equation is to substitute the given values into the left-hand side one by one and see whether we get the right-hand side.
First, for , we have
To evaluate this, we first recall that is a special angle, and so we can look up the value of the first term directly to find that . For the second term, that is, , we can use the related angle property:
Taking , we have
Putting this together, we have
Thus, since this is equal to the right-hand side, satisfies the equation.
Let us repeat this process for the other angles, continuing to use properties of related angles to help us. For , we have
We use the identities and , with , to get
As this is not equal to the right-hand side, this shows that does not satisfy the equation.
We have already found the answer, but for completeness, let us check the remaining options. For , we have
The first term can be simplified using (by taking ), to get . The second can be simplified by first using the periodicity of cosine, in other words, , to get , and then using to get . Recall that we have already found that . Putting this together, we have
Thus, works. The fourth option is . For this, we have
For both the first and second terms, we can use the periodicity of sine and cosine to simplify these terms. That is, and to get
This is the same left-hand side as we had for , so we can conclude that this value satisfies the equation.
Finally, we have , which gives us a left-hand side of
We can use the periodic identities and to get
Once more, we have the same left-hand side as we had for and , so we can conclude that this value satisfies the equation.
In conclusion, option B does not satisfy the equation.
In the previous example, let us note that three of the possible solutions, namely, , , and , ended up with the exact same expression on the left-hand side. This was a result of the periodicity of trigonometric functions, which can be seen by considering the positions of these angles on a CAST diagram.
That is to say, the values of and are equal for these values of because they share the same place on the CAST diagram. By extension, the values of and will also be the same.
It is important to be aware, however, that we cannot always use this approach of a CAST diagram. In particular, we should be careful since one of the terms in the equation (i.e., ) has a multiple angle. Recall that for trigonometric functions with multiple angles, the periodicity is different. This difference is illustrated in the graphs below.
Specifically, the period of is (or ), but the period of is (or ). Nevertheless, since the periods of and overlap, it turns out that angles that are apart will still have the same value of . We can see this by considering the periodic property of , specifically that since it has a period of , we have where is an integer. Since is three times , if we let , where is also an integer, we have
This shows us that the value of will be the same for any angles that differ by , so , , and will be the same too.
For our final example, let us find the value of an expression involving the product of multiple trigonometric functions.
Example 6: Evaluating Trigonometric Expressions Involving Special Angles
Find the value of .
Answer
In order to answer this question, we need to recall reciprocal trigonometric identities and special angles.
From our knowledge of special angles, we know that , , and .
From the sine and cosine graphs below, we see that and .
Since then
We can now substitute all of these values into our expression:
We will finish this explainer by recapping some of the key points.
Key Points
- We can evaluate trigonometric functions and expressions using our knowledge of special angles:
- We can use reciprocal trigonometric identities to solve more complicated problems:
- We can also use properties of related angles to evaluate trigonometric expressions:
