Lesson Explainer: Trigonometric Functions’ Values with Reference Angles | Nagwa Lesson Explainer: Trigonometric Functions’ Values with Reference Angles | Nagwa

Lesson Explainer: Trigonometric Functions’ Values with Reference Angles Mathematics

In this explainer, we will learn how to find reference angles and how to use them to find the values of trigonometric functions.

We recall that we can evaluate trigonometric functions by sketching the argument in standard position and then determining the coordinates of the point of intersection between the terminal side of the argument and the unit circle centered at the origin. To sketch an angle in standard position, we measure in the counterclockwise direction from the positive 𝑥-axis when the angle is positive and in the clockwise direction if the angle is negative.

For example, we can find the value of sin150 from the following diagram.

The coordinates of the point of intersection between the unit circle centered at the origin and the terminal side of the 150 angle in standard position are (150,150)cossin. We can evaluate the trigonometric expression for each coordinate by using the diagram and trigonometry. First, we note that the angles in a straight line sum to 180, so we can add the 30 angle to the diagram as follows.

Second, since the unit circle centered at the origin is the locus of all points a distance of 1 from the origin, we know that the line segment between (150,150)cossin and the origin is of length 1. If we drop a perpendicular from the point (150,150)cossin to the 𝑥-axis, we get the following right triangle.

We use the absolute value of the coordinates of the point (150,150)cossin since we want the lengths of the sides of the triangle rather than the coordinates. Finally, we can determine these values using trigonometry; the sine of an angle is the ratio between the lengths of the side opposite the angle and the hypotenuse. Hence, sinsinsin30=|150|1=|150|.

Similarly, the cosine of an angle is the ratio between the lengths of the side adjacent to the angle and the hypotenuse. Hence, coscoscos30=|150|1=|150|.

We know that sinandcos30=1230=32.

Therefore, by considering the fact that the point (150,150)cossin lies in the second quadrant, we have sinandcos150=12150=32.

In a similar manner, we can evaluate trigonometric values of angles beyond 360. For example, we can evaluate sin(405) by sketching the angle in standard position and noting it is the same as the 405+2×360=315 angle in standard position.

Since the terminal side of an angle in standard position is invariant under full revolutions clockwise or counterclockwise, this means that the sine and cosine of the angle are periodic about 360. We can then evaluate sin(405) by finding the measure of the angle between the terminal side and the positive 𝑥-axis; in this case, we note that a full revolution is 360, so its measure is 360315=45. Adding this to the diagram and the unit circle centered at the origin gives us the following.

We can then determine the value of sin(405) by dropping a perpendicular to the 𝑥-axis and applying trigonometry.

Since the point lies in the fourth quadrant, we see that sin(405) is negative. Hence, by applying trigonometry to the right triangle, we have sinsin(405)=(45)=22.

In the above examples, we were able to evaluate trigonometric functions for any angle by first finding an equivalent positive angle in standard position and then determining the measure of the acute angle the terminal side makes with 𝑥-axis. We call the equivalent positive angle the principal angle, and the measure of the acute angle that the terminal side makes with the 𝑥-axis the reference angle; we define these formally as follows.

Definition: Principal Angle

If 𝜃 is an angle in standard position, then the counterclockwise angle between the initial and terminal side of 𝜃 (less than a full turn) is called the principal angle of 𝜃.

Definition: Reference Angle

If 𝜃 is an angle in standard position, and not a quadrantal angle (integer multiple of a right angle), then the measure of the acute angle the terminal side makes with the 𝑥-axis is called the reference angle of 𝜃.

There are four different possibilities for the principal and reference angles based on which of the four quadrants the terminal side lies in, which we can see below.

Let’s see some examples of how to determine the principal angle of various angles given in radians.

Example 1: Identifying the Principal Angle for a Negative Angle

Given the angle 2𝜋3, find the principal angle.

Answer

We recall that to find the principal angle of 𝜃, we sketch the angle 𝜃 in standard position and then find the counterclockwise angle between the initial and terminal side, which is less than a full turn, [0,2𝜋]. To sketch 2𝜋3 in standard position, we note the value is negative, so we measure the angle in a clockwise direction from the positive 𝑥-axis to give us the following, where we label the principal angle 𝛼.

While the directed angle 𝜃 is negative, if we take the magnitude of this angle, then together with 𝛼 we have a full angle of 2𝜋, giving us 𝛼+2𝜋3=2𝜋𝛼=2𝜋2𝜋3𝛼=4𝜋3.

Hence, the principal angle of 2𝜋3 is 4𝜋3.

Example 2: Identifying the Principal Angle for an Angle Greater than 2𝜋

Given the angle 39𝜋4, find the principal angle.

Answer

We recall that to find the principal angle of 𝜃, we sketch the angle 𝜃 in standard position and then find the counterclockwise angle between the initial and terminal side, which is less than a full turn.

To sketch 39𝜋4 in standard position, we note that 39𝜋4>2𝜋, so it represents more than a full turn. This means we need to remove integer multiples of 2𝜋 to find an equivalent angle in standard position. Since 394=8+74, we subtract 8𝜋 as follows: 39𝜋44×2𝜋=39𝜋48𝜋=39𝜋432𝜋4=7𝜋4.

We can see that 39𝜋4 and 7𝜋4 have the same terminal angle when drawn in standard position in the following diagrams.

Since this value is between 0 and 2𝜋, we can conclude the principal angle of 39𝜋4 is 7𝜋4.

Let’s now see an example of how to use the principal value to evaluate a trigonometric expression without a calculator.

Example 3: Evaluating the Cosine of a Negative Angle by Finding the Principal Angle

Find cos(960) without using a calculator.

Answer

We recall that we can evaluate trigonometric functions by sketching angles in standard position and then determining the coordinates of the intersection between the terminal side of the angle and the unit circle centered at the origin. We could do this by sketching the 960 angle in standard position. However, we can also find the principal angle as this will give the same terminal side.

The principal angle of 960 will be between 0 and 360, and the terminal side of the angle in standard position will be the same. We can find the principal angle by adding integer multiples of 360 to the 960 angle. When we do, we obtain 960+3×360=120.

We can then determine the value of cos(960) by sketching 120 in standard position along with the unit circle centered at the origin. The 𝑥-coordinate of the point of intersection between the circle and terminal side will be cos(960).

To determine the 𝑥-coordinate of this point, we note that the angles in a straight line sum to 180, so the angle between the negative 𝑥-axis and the terminal side measures 180120=60.

Dropping a perpendicular from the point of intersection to the 𝑥-axis then gives us the following.

By applying right triangle trigonometry, the side adjacent to the 60 angle will have a length of cos(60).

Since the point of intersection has a negative 𝑥-coordinate, we see that coscos(960)=60=12.

In the above example, we evaluated a trigonometric expression by finding the principal angle of its argument. In actual fact, we also used the reference angle of this principal value.

Let’s see another example of this.

Example 4: Evaluating the Sine of an Angle by Finding the Reference Angle

Find the value of sin11𝜋6.

Answer

We recall that we can evaluate trigonometric functions by sketching the argument in standard position and then determining the coordinates of the intersection between the terminal side of the angle and the unit circle. We start by sketching 11𝜋6 in standard position, noting that it is positive, so the angle is measured counterclockwise from the positive 𝑥-axis. Since 3𝜋2<11𝜋6<2𝜋, the terminal side will lie in the fourth quadrant, as shown in the following diagram.

To determine the value of sin11𝜋6, we need to find the 𝑦-coordinate of the point of intersection. We do this by finding the reference angle of 11𝜋6; that is, the measure of the acute angle between the terminal side and the 𝑥-axis when 11𝜋6 is drawn in standard position. We can see in the diagram that the angle between the terminal side and the 𝑥-axis and 11𝜋6 add together to make 2𝜋. Labeling the reference angle as 𝜃, we have 𝜃+11𝜋6=2𝜋𝜃=2𝜋11𝜋6𝜃=𝜋6.

We then add this to our diagram and drop a perpendicular from the intersection point to the 𝑥-axis, as shown in the following diagram.

We can add a direction to the reference angle to see that this is the angle 𝜋6 in standard position. The coordinates of the point of intersection can also be written as 𝜋6,𝜋6cossin. Equating the two expressions for the 𝑦-coordinate of the point of intersection and evaluating, we have sinsin11𝜋6=𝜋6=12.

Hence, sin11𝜋6=12.

In our next example, we will use the reference angle of an argument to evaluate a reciprocal trigonometric function.

Example 5: Evaluating the Secant of an Angle by Finding the Reference Angle

Find sec300 without using a calculator.

Answer

To determine the secant of an angle, we first need to recall that the secant function is the reciprocal of the cosine function. This gives us seccos300=1300.

We can determine the cosine of 300 by sketching the angle in standard position and then finding the 𝑥-coordinate of the point of intersection between the terminal side of the angle and the unit circle. Since 300 is positive, the angle is measured in a counterclockwise direction, and we note that 270<300<360, so the terminal side will lie in the fourth quadrant. This gives us the following sketch.

To determine the 𝑥-coordinate of the point of intersection, we find the reference angle for 300, which is the measure of the acute angle between the terminal side and the 𝑥-axis in the diagram above. Labeling the reference angle 𝜃 and noting that these two angles make a full revolution, we have 360=300+𝜃𝜃=360300=60.

Adding this angle into our diagram and dropping a perpendicular from the point of intersection to the 𝑥-axis gives us the following.

Since the terminal side lies in the fourth quadrant, the 𝑥-coordinate of the point of intersection is positive, which means that cos300 is also positive. We can determine the exact value by applying trigonometry to the right triangle in the diagram; the side adjacent to the 60 angle has a length of cos300, and the hypotenuse has a length of 1.

The cosine of 60 is then the ratio of the lengths of the side adjacent to the angle and the hypotenuse, giving us coscoscoscos60=300112=300300=12.

Finally, we take the reciprocal of both sides of this equation to see that 1300=1300=2.cossec

In our final example, we will evaluate the tangent function by first finding the reference angle of its argument.

Example 6: Evaluating the Tangent of an Angle by Finding the Reference Angle

Find the exact value of tan7𝜋6 without using a calculator.

Answer

To determine the tangent of an angle without a calculator, we first recall that the tangent is the quotient of the sine and cosine of the same angle. Applying this to the angle 7𝜋6 gives us tansincos7𝜋6=.

We can determine the sine and cosine of an angle by sketching it in standard position and then finding the coordinates of the point of intersection between the terminal side of the angle and the unit circle centered at the origin. Since 7𝜋6 is positive, the angle is measured in a counterclockwise direction, and we note that 𝜋<7𝜋6<3𝜋2. So, the terminal side will lie in the third quadrant. This gives us the following sketch.

To determine the values of the 𝑥- and 𝑦-coordinates of the point of intersection, we need to find the measure of the angle between the terminal side and the 𝑥-axis (called the reference angle). The angle between the positive 𝑥-axis and negative 𝑥-axis is 𝜋, so the reference angle is 7𝜋6𝜋=𝜋6.

We can add this to our diagram and drop the perpendicular from the point of intersection to the 𝑥-axis to give us the following.

Since the point of intersection is in the third quadrant, both the 𝑥- and 𝑦-coordinates of the point of intersection will be negative. Therefore, the base of the right triangle will have a length of 7𝜋6cos and a height of 7𝜋6sin, giving us the following right triangle.

We can find expressions for cos7𝜋6 and sin7𝜋6 by applying trigonometry to the right triangle. First, by taking the ratio of the lengths of the side opposite to 𝜋6 and the hypotenuse, we have sinsinsinsin𝜋6=112=7𝜋67𝜋6=12.

Second, by taking the ratio of the lengths of the side adjacent to 𝜋6 and the hypotenuse, we have coscoscoscos𝜋6=132=7𝜋67𝜋6=32.

Finally, we can take the quotient of these values to find the tangent of the argument as follows: tansincos7𝜋6===12×23=13=33.

Hence, tan7𝜋6=33.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • If 𝜃 is an angle in standard position, then the counterclockwise angle between the initial and terminal side of 𝜃 (less than a full turn) is call the principal angle.
  • Taking a principal value does not change the value of the sine or cosine functions.
  • If 𝜃 is an angle in standard position, and not a quadrantal angle, then the measure of the acute angle the terminal side makes with the 𝑥-axis is called the reference angle.
  • By sketching an angle in standard position and using the principal and reference angles, we can find equivalent arguments to evaluate trigonometric functions.

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