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Lesson Explainer: Evaluating Reciprocal Trigonometric Functions Mathematics

In this explainer, we will learn how to evaluate reciprocal trigonometric functions at given values in both degrees and radians.

Recall that the unit circle is a circle with a radius of 1 whose center lies at the origin of a coordinate plane. For any point (π‘₯,𝑦) on the unit circle, a right triangle can be formed as in the following diagram.

The hypotenuse of the right triangle forms an acute angle, which we label πœƒ, with the positive π‘₯-axis. Its length is 1, the same as the radius of the unit circle, and it goes from the origin to point (π‘₯,𝑦). This point then defines the lengths of the two shorter sides. The side opposite the angle has length |𝑦|. The adjacent side has length |π‘₯|.

In questions on this topic, we will meet some angles that are measured in radians instead of degrees. We can convert between radians and degrees, as needed, using the fact that 180=πœ‹βˆ˜radians.

Using right triangle trigonometry, we can define the trigonometric functions in terms of the unit circle: sinoppositehypotenusesosincosadjacenthypotenusesocostanoppositehypotenusesotanπœƒ==𝑦1,πœƒ=𝑦,πœƒ==π‘₯1,πœƒ=π‘₯,πœƒ==𝑦π‘₯,πœƒ=𝑦π‘₯.

Since sinπœƒ=𝑦 and cosπœƒ=π‘₯, the tangent function can also be written in terms of sine and cosine: tansincosπœƒ=𝑦π‘₯=πœƒπœƒ.

On the unit circle, the standard angle πœƒ is measured from 0∘ to 360∘ in a counterclockwise direction, starting from the positive π‘₯-axis, which we refer to as the initial side of the angle. We refer to the side where the standard angle stops as the terminal side.

We find πœƒ satisfying 0<πœƒ<90∘∘ in quadrant I of the unit circle; that is, 0<πœƒ<πœ‹2 in radians.

We find πœƒ satisfying 90<πœƒ<180∘∘ in quadrant II of the unit circle; that is, πœ‹2<πœƒ<πœ‹ in radians.

We find πœƒ satisfying 180<πœƒ<270∘∘ in quadrant III of the unit circle; that is, πœ‹<πœƒ<3πœ‹2 in radians.

We find πœƒ satisfying 270<πœƒ<360∘∘ in quadrant IV of the unit circle; that is, 3πœ‹2<πœƒ<2πœ‹ in radians.

Quadrantal angles terminate along the π‘₯- or 𝑦-axis, specifically where πœƒ equals 0∘, 90∘, 180∘, 270∘, or 360∘. We should note that 0∘ is coterminal with 360∘, since both of these angles terminate on the positive π‘₯-axis. In radians, these quadrantal angles are 0, πœ‹2, πœ‹, 3πœ‹2, and 2πœ‹.

What we have looked at so far appears to be a departure from foundational right triangle trigonometry, where πœƒ is always an acute angle and triangle sides are measured in positive units. But when we refer to the unit circle, we can still make use of right triangle trigonometry concepts if we think of π‘₯ and 𝑦 as the adjacent and opposite sides of a reference triangle. A reference triangle is a right triangle created by dropping a perpendicular segment from point (π‘₯,𝑦) on the unit circle back to the π‘₯-axis. It is important to recognize that on the unit circle, π‘₯ and 𝑦 can take any real number value including zero or a negative.

Let’s examine the angle whose vertex is at (0,0) and whose sides are the π‘₯-axis and the hypotenuse of a reference triangle. This is called a reference angle πœƒβ€². Therefore, even if πœƒβ‰₯90∘, πœƒβ€² will be an acute angle. The following diagram shows the placement of reference triangles for all four quadrants.

Given a standard angle πœƒ on the unit circle, the corresponding reference angle πœƒβ€² can be calculated for any quadrant as follows:

In quadrant I, πœƒβ€²=πœƒ.

In quadrant II, πœƒβ€²=180βˆ’πœƒ.∘

In quadrant III, πœƒβ€²=πœƒβˆ’180.∘

In quadrant IV, πœƒβ€²=360βˆ’πœƒ.∘

In the diagram below, we show the unit circle with every multiple of 30∘ from 0∘ to 360∘. The radian conversions are included in pink.

Now, recall the properties of 30∘-60∘-90∘ special right triangles, as shown in the next diagram. According to these properties, a 30∘-60∘-90∘ right triangle with a hypotenuse of 1 will have shorter sides of length 12 and √32. The hypotenuse of 1 corresponds to the radius of a unit circle.

Dividing the unit circle into 30∘ increments results in eight 30∘-60∘-90∘ reference triangles.

Thus, each coordinate value is 12, βˆ’12, √32, or βˆ’βˆš32, depending on which quadrant we are in. The exact coordinates are written as ordered pairs (π‘₯,𝑦) for each multiple of 30∘.

We can also find every multiple of 45∘ from 0∘ to 360∘ on the unit circle, in degrees or radians.

Now, recall the properties of 45∘-45∘-90∘ special right triangles, as shown in the diagram below. According to these properties, a 45∘-45∘-90∘ right triangle with a hypotenuse of 1 will have shorter sides, both of length √22.

Dividing the unit circle into 45∘ increments gives us four 45∘-45∘-90∘ reference triangles.

Thus, each coordinate is either √22 or βˆ’βˆš22, depending on which quadrant we are in. The exact coordinates are included in the following diagram.

We can determine the sign of our trigonometric function value based on where the angle terminates. One way of recalling whether the sine, cosine, and tangent of any angle between 0∘ and 360∘ are positive or negative is by using the CAST diagram. This is a memory device that we use to remember the signs of the trigonometric functions in each of the four quadrants.

  • In quadrant I, All values are positive.
  • In quadrant II, Sine is positive.
  • In quadrant III, Tangent is positive.
  • In quadrant IV, Cosine is positive.

Now that we have reviewed properties of standard trigonometric functions in relation to the unit circle, we will turn our attention to the reciprocals of these functions. Note that the sign of the value does not change when we take the reciprocal.

Definition: Reciprocal Trigonometric Functions

The reciprocal trigonometric functions, cosecant, secant, and cotangent, are defined as cscsinseccoscottancossinπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=1πœƒ=πœƒπœƒ.

We will use the earlier established sine, cosine, and tangent definitions, described in terms of the unit circle, to write new definitions of their reciprocals.

Given the trigonometric function sinπœƒ=𝑦, its reciprocal function, cosecant, can be written as cscsinwhereπœƒ=1πœƒ=1𝑦,𝑦≠0.

Given the trigonometric function cosπœƒ=π‘₯, its reciprocal function, secant, can be written as seccoswhereπœƒ=1πœƒ=1π‘₯,π‘₯β‰ 0.

Given the trigonometric function tanπœƒ=𝑦π‘₯, its reciprocal function, cotangent, can be written as cotcossinwhereπœƒ=πœƒπœƒ=π‘₯𝑦,𝑦≠0.

Now, we will practice using the CAST diagram to determine the sign of various trigonometric functions.

Example 1: Identifying the Sign of a Reciprocal Trigonometric Ratio

Which of the following trigonometric functions is positive at 7πœ‹6?

  1. Secant
  2. Cosecant
  3. Cotangent
  4. None of the answers are correct.

Answer

To answer this question, we will need to find 7πœ‹6 on the unit circle and identify the signs of standard trigonometric functions in that particular quadrant. If desired, we can convert radians into degrees, using the fact that 180=πœ‹βˆ˜radians: 7πœ‹6β‹…180πœ‹=210.∘∘

In quadrant I, all trigonometric functions are positive, due to the positive π‘₯- and π‘₯-coordinates. But we find 210∘ in quadrant III, where only tangent and its reciprocal, cotangent, are positive. This information comes from the CAST diagram, which is a memory device to help us recall which standard trigonometric functions are positive in each quadrant.

  • In quadrant I, All values are positive.
  • In quadrant II, Sine is positive.
  • In quadrant III, Tangent is positive.
  • In quadrant IV, Cosine is positive.

The reciprocal trigonometric functions are defined in terms of the standard trigonometric functions as follows: cscsinseccoscottanπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=1πœƒ.

The sign of the value does not change when we take the reciprocal, and we know that only tangent and its reciprocal are positive in quadrant III, where 210∘ (equivalently, 7πœ‹6 radians) is found.

Therefore, of the three reciprocal trigonometric functions, only cotangent is positive at 7πœ‹6.

This means that choice C is correct.

Next, we will look at how to find the exact value of a reciprocal trigonometric expression when it involves an angle in radians that can be found on the unit circle.

Example 2: Finding the Exact Value of a Reciprocal Trigonometric Expression

Find the exact value of cot5πœ‹6.

Answer

To begin, we find the standard angle 5πœ‹6 in quadrant II of the unit circle. If desired, we can convert radians into degrees, using the fact that 180=πœ‹βˆ˜radians: 5πœ‹6β‹…180πœ‹=150.∘∘

We recall that cotangent is the reciprocal of the trigonometric function tangent. Therefore, cottanπœƒ=1πœƒ. For any point on the unit circle, tanπœƒ=𝑦π‘₯ and cotπœƒ=π‘₯𝑦. So, we need to find both the π‘₯- and 𝑦-coordinates of the point where the standard angle 150∘ intersects the unit circle.

Next, we calculate the reference angle that 150∘ makes with the negative π‘₯-axis. This 30∘ angle is found by subtracting 150∘ from 180∘. We can then sketch a special 30∘-60∘-90∘ right reference triangle, as shown in the diagram below.

Now, we recall the properties of 30∘-60∘-90∘ special right triangles, as shown in the next diagram. According to these properties, a 30∘-60∘-90∘ right triangle with a hypotenuse of 1 will have shorter sides of length 12 and √32. The hypotenuse of 1 corresponds to the radius of a unit circle.

On the unit circle, the π‘₯-coordinate of the point we desire corresponds to the length of the side adjacent to the 30∘ reference angle. The 𝑦-coordinate corresponds to the length of the side opposite the 30∘ reference angle. However, because of the location of the reference triangle in quadrant II, the π‘₯-coordinate must be negative. Thus, the terminal side of the 150∘ angle intersects the point ο€Ώβˆ’βˆš32,12.

Since cotπœƒ=π‘₯𝑦, we can substitute the π‘₯- and 𝑦-coordinates to find the exact value.

This means we will need to simplify the resulting complex fraction: cot150=βˆ’.∘√

Recall how taking the quotient of two fractions is more simply calculated by multiplying the first fraction by the reciprocal of the second fraction. In this case, we multiply βˆ’βˆš32 by the reciprocal of 12 and simplify. So, cot150=βˆ’βˆš32β‹…21=βˆ’βˆš3.∘

In conclusion, the exact value of cot150∘ or cot5πœ‹6 is βˆ’βˆš3.

As we reviewed earlier in this explainer, the unit circle only provides the π‘₯- and 𝑦-coordinates for angles with measurements equal to multiples of 30∘ or 45∘ between 0∘ and 360∘. For other angle measures not found on the unit circle, we can use a calculator to find the approximate value of trigonometric functions. In cases like this, we will usually be asked to round to a specific number of significant figures. We will need a calculator that can switch between degree mode and radian mode.

How To: Evaluating Reciprocal Trigonometric Functions with a Calculator

If the angle is given in degrees, make sure your calculator is in degree mode.

If the angle is given in radians, make sure your calculator is in radian mode.

  • To evaluate the cosecant of an angle πœƒ, carry out the following operations on your calculator: [1][Γ·][][]sinanglemeasure.
  • To evaluate the secant of an angle πœƒ, carry out the following operations on your calculator: [1][Γ·][][]cosanglemeasure.
  • To evaluate the cotangent of an angle πœƒ, carry out the following operations on your calculator: [1][Γ·][][]tananglemeasure.

For example, we can place 130∘ in standard position on a coordinate plane in the second quadrant. But it would not match any of the designated angles on the unit circle because it is not a multiple of 30∘ or 45∘. The following diagrams show how 130∘ lies between two angles given on the unit circle: 120∘ and 135∘.

Now, let’s see how to find the approximate secant value at 130∘ using our calculator.

Example 3: Finding the Value of a Reciprocal Trigonometric Expression Using a Calculator

Use a calculator to find the value of sec130∘ to 3 significant figures.

Answer

First, we must determine whether the given angle is in degrees or radians. Since we see a degree symbol, we will put our calculator into degree mode.

Then, we recall that secant is the reciprocal trigonometric function of cosine. Therefore, seccosπœƒ=1πœƒ. To evaluate secant of 130∘, we will carry out the following operations on our calculator: [1][Γ·][][130]cos.

The calculator should show the first few figures of the value, including βˆ’1.555724….

This is, in fact, an irrational number with infinitely many digits. We are asked to round to 3 significant figures, so we consider the first 3 nonzero digits from left to right. We cut off all the digits after the 3rd significant digit, as shown below: βˆ’1.55∣5724….

The digit to the right of the 3rd significant digit fits the criteria of being greater than or equal to five, so we round the third significant digit up from 5 to 6.

Finally, we conclude that the value of sec130∘ to 3 significant figures is βˆ’1.56.

Next, let’s consider how we would evaluate trigonometric functions at quadrantal angles, which terminate along the π‘₯- or 𝑦-axis, as shown in the next diagram.

We need to pay special attention to coordinate points containing zero. This presents a unique challenge because we cannot possibly draw a reference triangle with a side length of zero. This means some trigonometric functions are not defined at specific quadrantal angles. This occurs whenever a coordinate of 0 appears in the denominator of the function. The angles where the trigonometric functions are undefined are highlighted in the table below.

Trig FunctionUnit Circle Ratioπœƒ=0/360=0/2πœ‹βˆ˜βˆ˜radπœƒ=90=πœ‹2∘radπœƒ=180=πœ‹βˆ˜radπœƒ=270=3πœ‹2∘rad
sinπœƒπ‘¦101=011=101=0βˆ’11=βˆ’1
cosπœƒπ‘₯111=101=0βˆ’11=βˆ’101=0
tanπœƒπ‘¦π‘₯01=010 is undefined.0βˆ’1=0βˆ’10 is undefined.
cscπœƒ1𝑦10 is undefined.11=110 is undefined.1βˆ’1=βˆ’1
secπœƒ1π‘₯11=110 is undefined.1βˆ’1=βˆ’110 is undefined.
cotπœƒπ‘₯𝑦10 is undefined.01=0βˆ’10 is undefined.0βˆ’1=0

In the following example, we will look at how to find the exact value of a reciprocal trigonometric function of a negative quadrantal angle. Although negative angles are not shown on the unit circle, we will use our knowledge of coterminal angles to find an equivalent principal angle. Recall that coterminal angles share the same initial and terminal sides. Performing any number of full 360∘ rotations, clockwise or counterclockwise, will lead us to a coterminal angle.

Example 4: Finding the Exact Value of a Reciprocal Trigonometric Expressio

Find the exact value of csc(βˆ’90)∘.

Answer

Let’s begin by imagining that the standard angle βˆ’90∘ is the angle between two rays 𝑂𝐴 and οƒͺ𝑂𝐡, where 𝑂𝐴 is the initial side and οƒͺ𝑂𝐡 is the terminal side of the angle.

As our angle is negative, we need to measure the angle in a clockwise direction from the initial side 𝑂𝐴, as shown.

We recall that coterminal angles share the same initial and terminal sides. There is an infinite number of equivalent coterminal angles. However, we need to find a principal angle equivalent to βˆ’90∘, which should terminate between 0∘ and 360∘. To find this angle, we need to measure in the counterclockwise positive direction.

We recall that angles at a point sum to 360∘ (or 2πœ‹ if in radians), so a positive equivalent of βˆ’90∘ is found by subtracting 90∘ from 360∘: 360βˆ’90=270.∘∘∘

Therefore, the principal angle equivalent to βˆ’90∘ is 270∘. As seen on the unit circle, 270∘ is a quadrantal angle intersecting the point (0,βˆ’1).

We are looking for csc(βˆ’90)∘, which we now know is equivalent to csc(270)∘. We recall that cosecant is the reciprocal of the trigonometric function sine. For any point on the unit circle, sinπœƒ=𝑦; therefore, cscπœƒ=1𝑦. Using the point (0,βˆ’1), we have csc(270)=1βˆ’1=βˆ’1∘.

Since we have seen that βˆ’90∘ is coterminal to 270∘, then csccsc(270)=(βˆ’90)∘∘.

In conclusion, we have shown that csc(βˆ’90)=βˆ’1∘.

Now that we have practiced evaluating various reciprocal trigonometric functions in degrees and radians, let’s try evaluating a product.

Example 5: Evaluating a Reciprocal Trigonometric Expression

Evaluate csccot240240∘∘.

Answer

To evaluate csccot240240∘∘, we need to find the cosecant and cotangent values first and then work out their product. We recall that cosecant is the reciprocal of sine, and cotangent is the reciprocal of tangent.

We find that 240∘ lies in the third quadrant of the unit circle, as shown above. The terminal side of the angle intersects the point ο€Ώβˆ’12,βˆ’βˆš32. Since sine is defined as the 𝑦-coordinate, we have sin240=𝑦=βˆ’βˆš32.∘

Next, we will take the reciprocal of sin240∘ to find csc240∘. The reciprocal of βˆ’βˆš32 is found by swapping the numerator and denominator, so csc240=βˆ’2√3.∘

We do not need to be concerned about rationalizing the denominator at this point, since this is not our final answer. Next, we need to find cot240∘.

Cotangent is the reciprocal of tangent. Since tanπœƒ=𝑦π‘₯, we will be using cotπœƒ=π‘₯𝑦. We refer to the point ο€Ώβˆ’12,βˆ’βˆš32 we found earlier and substitute the π‘₯- and 𝑦-coordinates into the cotangent formula: cot240=π‘₯𝑦=βˆ’βˆ’.∘√

Recall how taking the quotient of two fractions is more simply calculated by multiplying the first fraction by the reciprocal of the second fraction. In this case, we multiply βˆ’12 by the reciprocal of βˆ’βˆš32 and simplify. So, cot240=βˆ’12β‹…ο€Ώβˆ’2√3=1√3.∘

Finally, we use substitution to evaluate the original expression and work out the product: csccot240240=ο€Ώβˆ’2√31√3=βˆ’23.∘∘

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

Key Points

  • The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. The standard angles found on the unit circle are multiples of 30∘ and 45∘ between 0∘ and 360∘. The coordinate pairs corresponding to these angles are derived from the sides of special right triangles with a hypotenuse length of 1.
  • The reciprocal trigonometric functions, cosecant, secant, and cotangent, are defined in terms of the standard trigonometric functions as follows: cscsinseccoscottanπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=1πœƒ.
  • All trigonometric functions can also be defined using points (π‘₯,𝑦) from the unit circle.
sinπœƒ=𝑦cscπœƒ=1𝑦, where 𝑦≠0
cosπœƒ=π‘₯secπœƒ=1π‘₯, where π‘₯β‰ 0
tanπœƒ=𝑦π‘₯, where π‘₯β‰ 0cotπœƒ=π‘₯𝑦, where 𝑦≠0

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