High School Trigonometry/Relating Trigonometric Functions

In previous lessons we defined and worked with the six trig functions individually. In this lesson, we will consider relationships among the functions. In particular, we will develop several identities involving the trig functions. An identity is an equation that is true for all values of the variables, as long as the expressions or functions involved are defined. For example, x + x = 2x is an identity. In this lesson we will develop several identities involving trig functions. Because of these identities, the same function can have very many different algebraic representations. These identities will allow us to relate the trig functions’ domains and ranges, and the identities will be useful in solving problems in later chapters.

Learning Objectives

 * State the reciprocal relationships between trig functions, and use these identities to find values of trig functions.
 * State quotient relationships between trig functions, and use quotient identities to find values of trig functions.
 * State the domain and range of each trig function.
 * State the sign of a trig function, given the quadrant in which an angle lies.
 * State the Pythagorean identities and use these identities to find values of trig functions.

Reciprocal Identities
The first set of identities we will establish are the reciprocal identities. A reciprocal of a fraction $$\tfrac{a}{b}$$ is the fraction $$\tfrac{b}{a}$$. That is, we find the reciprocal of a fraction by interchanging the numerator and the denominator, or flipping the fraction. The six trig functions can be grouped in pairs as reciprocals.

First, consider the definition of the sine function for angles of rotation: sin θ = $$\tfrac{y}{r}$$. Now consider the cosecant function: csc θ = $$\tfrac{r}{y}$$. In the unit circle, these values are sin θ = $$\tfrac{y}{1}$$ = y and csc θ = $$\tfrac{1}{y}$$. These two functions, by definition, are reciprocals. Therefore, the sine value of an angle is always the reciprocal of the cosecant value, and vice versa. For example, if sin θ = $$\tfrac{1}{2}$$, then csc θ = $$\tfrac{2}{1}$$ = 2.

Analogously, the cosine function and the secant function are reciprocals, and the tangent and cotangent function are reciprocals:


 * $$\sec \theta = \frac{1}{\cos \theta}\ \ \text{or}\ \cos \theta = \frac{1}{\sec \theta}$$


 * $$\cot \theta = \frac{1}{\tan \theta}\ \ \text{or}\ \tan \theta = \frac{1}{\cot \theta}$$

We can use these reciprocal relationships to find values of trig functions. The fundamental identity stemming from the Pythagorean Theorem 1 = sin2 x + cos2 x can take a great many new forms.

Find the value of each expression using a reciprocal identity.

a. cos θ = .3, sec θ = ?

b. cot θ = $$\tfrac{4}{3}$$, cot θ = ?

Solution:

a. sec θ = $$\tfrac{10}{3}$$

These functions are reciprocals, so if cos θ = .3, then sec θ = $$\tfrac{1}{.3}$$. It is easier to find the reciprocal if we express the values as fractions: cos θ = .3 = $$\tfrac{3}{10}$$ → sec θ = $$\tfrac{10}{3}$$.

b. tan θ = $$\tfrac{3}{4}$$

These functions are reciprocals, and the reciprocal of $$\tfrac{4}{3}$$ is $$\tfrac{3}{4}$$.

We can also use the reciprocal relationships to determine the domain and range of functions.

Domain, Range, and Signs of Functions
While the trigonometric functions may seem quite different from other functions you have worked with, they are in fact just like any other function. We can think of a trig function in terms of "input" and "output". The input is always an angle. The output is a ratio of sides of a triangle. If you think about the trig functions in this way, you can define the domain and range of each function.

Let's first consider the sine and cosine functions. The input of each of these functions is always an angle, and as you learned in the previous chapter, these angles can take on any real number value. Therefore, the sine and cosine function have the same domain, the set of all real numbers, R. We can determine the range of the functions if we think about the fact that the sine of an angle is the y-coordinate of the point where the terminal side of the angle intersects the unit circle. The cosine is the x-coordinate of that point. Now recall that in the unit circle, we defined the trig functions in terms of a triangle with hypotenuse 1.


 * Defining Trigonometric Functions Figure 4.svg

In this right triangle, x and y are the lengths of the legs of the triangle, which must have lengths less than 1, the length of the hypotenuse. Therefore, the ranges of the sine and cosine function do not include values greater than one. The ranges do, however, contain negative values. Any angle whose terminal side is in the third or fourth quadrant will have a negative y−coordinate, and any angle whose terminal side is in the second or third quadrant will have a negative x−coordinate.


 * Relating Trigonometric Functions Figure 2.svg

In either case, the minimum value is −1. For example, cos(180°) = −1 and sin(270°) = −1. Therefore, the sine and cosine function both have range from −1 to 1.

The table below summarizes the domains and ranges of these functions:


 * {| class="wikitable"

! scope="col" | ! scope="col" | Domain ! scope="col" | Range
 * + Table 1.12
 * Sine || θϵ° || −1 ≤ sin θ ≤ 1
 * Cosine || θϵ° || −1 ≤ cos θ ≤ 1
 * }
 * Cosine || θϵ° || −1 ≤ cos θ ≤ 1
 * }

Knowing the domain and range of the cosine and sine function can help us determine the domain and range of the secant and cosecant function. First consider the sine and cosecant functions, which as we showed above, are reciprocals. The cosecant function will be defined as long as the sine value is not 0. Therefore, the domain of the cosecant function excludes all angles with sine value 0, which are 0°, 180°, 360°, etc.

In Chapter 2 you will analyze the graphs of these functions, which will help you see why the reciprocal relationship results in a particular range for the cosecant function. Here we will state this range, and in the review questions you will explore values of the sine and cosecant function or order to begin to verify this range, as well as the domain and range of the secant function.


 * {| class="wikitable"

! scope="col" | ! scope="col" | Domain ! scope="col" | Range
 * + Table 1.13
 * Cosecant || θϵ°, θ ≠ 0, 180, 360… || csc θ ≤ −1 or csc θ ≥ 1
 * Secant || θϵ°, θ ≠ 90, 270, 450… || sec θ ≤ −1 or sec θ ≥ 1
 * }
 * Secant || θϵ°, θ ≠ 90, 270, 450… || sec θ ≤ −1 or sec θ ≥ 1
 * }

Now let's consider the tangent and cotangent functions. The tangent function is defined as $$\tan \theta = \tfrac{y}{x}$$. Therefore, the domain of this function excludes angles for which the ordered pair has an x-coordinate of 0: 90°, 270°, etc. The cotangent function is defined as $$\cot \theta = \tfrac{x}{y}$$, so this function's domain will exclude angles for which the ordered pair has a y-coordinate of 0: 0°, 180°, 360°, etc. As you will learn in chapter 3 when you study the graphs of these functions, there are no restrictions on the ranges.


 * {| class="wikitable"

! scope="col" | Function ! scope="col" | Domain ! scope="col" | Range
 * + Table 1.14
 * Tangent || θϵ°, θ ≠ 90, 270, 450… || $$\infty\,\!$$
 * Cotangent || θϵ°, θ ≠ 0, 180, 360… || $$\infty\,\!$$
 * }
 * Cotangent || θϵ°, θ ≠ 0, 180, 360… || $$\infty\,\!$$
 * }

Knowing the ranges of these functions tells you the values you should expect when you determine the value of a trig function of an angle. However, for many problems you will need to identify the sign of the function of an angle: Is it positive or negative?

In determining the ranges of the sine and cosine functions above, we began to categorize the signs of these functions in terms of the quadrants in which angles lie. The figure below summarizes the signs for angles in all 4 quadrants.


 * Relating Trigonometric Functions Figure 3.svg

State the sign of each expression.

a. cos(100°)

b. csc(220°)

c. tan(370°)

Solution:

a. The angle 100° is in the second quadrant. Therefore, the x-coordinate is negative and so cos(100°) is negative.

b. The angle 220° is in the third quadrant. Therefore, the y-coordinate is negative. So the sine and the cosecant are negative.

c. The angle 370° is in the first quadrant. Therefore, the tangent value is positive.

So far we have considered relationships between pairs of functions: the six trig functions can be grouped in pairs as reciprocals. Now we will consider relationships among three trig functions.

Quotient Identities
The definitions of the trig functions led us to the reciprocal identities above. They also lead us to another set of identities, the quotient identities.

Consider first the sine, cosine, and tangent functions. For angles of rotation (not necessarily in the unit circle) these functions are defined as follows:


 * $$\sin \theta = \frac{y}{r}$$


 * $$\cos \theta = \frac{x}{r}$$


 * $$\tan \theta = \frac{y}{x}$$

Given these definitions, we can show that tan θ = $$\tfrac{\sin \theta}{\cos \theta}$$, as long as cos θ ≠ 0:


 * $$\frac{\sin \theta}{\cos \theta} = \frac{\frac{y}{r}}{\frac{x}{r}} = \frac{y}{r} \cdot \frac{r}{x} = \frac{y}{x} = \tan \theta$$

The equation tan θ = $$\tfrac{\sin \theta}{\cos \theta}$$ is therefore an identity that we can use to find the value of the tangent function, given the value of the sine and cosine.

If cos θ = $$\tfrac{5}{13}$$ and sin θ = $$\tfrac{12}{13}$$, what is the value of tan θ?

Solution:

$$\tan \theta = \tfrac{12}{5}$$


 * $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{13} \cdot \frac{13}{5} = \frac{12}{5}$$

Show that cot θ = $$\tfrac{\cos \theta}{\sin \theta}$$

Solution:

$$\frac{\cos \theta}{\sin \theta} = \frac{\frac{x}{r}}{\frac{y}{r}} = \frac{x}{r} \cdot \frac{r}{y} = \frac{x}{y} = \cot \theta$$

This is also an identity that you can use to find the value of the cotangent function, given values of sine and cosine. Both of the quotient identities will also be useful in chapter 3, in which you will prove other identities.

Pythagorean Identities
The final set of identities that we will examine in this lesson are called the Pythagorean identities because they rely on the Pythagorean Theorem. In previous lessons we used the Pythagorean theorem to find the sides of right triangles. Consider once again the way that we defined the trig functions in lesson 4. Let's look at the unit circle:


 * Defining Trigonometric Functions Figure 4.svg

The legs of the right triangle are x, and y. The hypotenuse is 1. Therefore, the following equation is true for all x and y on the unit circle:


 * $$x^2 + y^2 = 1\,\!$$

Now remember that on the unit circle, cos θ = x and sin θ = y. Therefore, the following equation is an identity:


 * $$\cos^2 \theta + \sin^2 \theta = 1\,\!$$

''Note: Writing the exponent 2 after the cos and sin is the standard way of writing exponents. Just keep in mind that cos2 θ means (cos θ)2 and sin2 θ means (sin θ'')2.

We can use this identity to find the value of the sine function, given the value of the cosine, and vice versa. We can also use it to find other identities.

If cos θ = $$\tfrac{1}{4}$$ what is the value of sin θ? Assume that θ is an angle in the first quadrant.

Solution:

$$\sin \theta = \frac{\sqrt{15}}{4}$$


 * {| cellpadding="3" style="background:transparent;"


 * style="text-align:right;" | $$\cos^2 \theta + \sin^2 \theta =\,\!$$ || $$1\,\!$$
 * style="text-align:right;" | $$\left ( \frac{1}{4} \right )^2 + \sin^2 \theta =$$ || $$1\,\!$$
 * style="text-align:right;" | $$\frac{1}{16} + \sin^2 \theta =$$ || $$1\,\!$$
 * style="text-align:right;" | $$\sin^2 \theta =\,\!$$ || $$1 - \frac{1}{16}$$
 * style="text-align:right;" | $$\sin^2 \theta =\,\!$$ || $$\frac{16}{16} - \frac{1}{16}$$
 * style="text-align:right;" | $$\sin^2 \theta =\,\!$$ || $$\frac{15}{16}$$
 * style="text-align:right;" | $$\sin \theta =\,\!$$ || $$\pm \sqrt{\frac{15}{16}}$$
 * style="text-align:right;" | $$\sin \theta =\,\!$$ || $$\pm \frac{\sqrt{15}}{4}$$
 * }
 * style="text-align:right;" | $$\sin^2 \theta =\,\!$$ || $$\frac{15}{16}$$
 * style="text-align:right;" | $$\sin \theta =\,\!$$ || $$\pm \sqrt{\frac{15}{16}}$$
 * style="text-align:right;" | $$\sin \theta =\,\!$$ || $$\pm \frac{\sqrt{15}}{4}$$
 * }
 * style="text-align:right;" | $$\sin \theta =\,\!$$ || $$\pm \frac{\sqrt{15}}{4}$$
 * }
 * }

Remember that it was given that θ is an angle in the first quadrant. Therefore, the sine value is positive, so sin θ = $$\tfrac{\sqrt{15}}{4}$$.

Use the identity cos2 θ + sin2 θ = 1 to show that cot2 θ + 1 = csc2 θ.

Solution:


 * {| cellpadding="3" style="background:transparent;"


 * style="text-align:right;" | $$\cos^2 \theta + \sin^2 \theta =\,\!$$ || $$1\,\!$$
 * style="padding-left:50px;" | Divide both sides by sin2 θ.
 * style="text-align:right;" | $$\frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta} =$$ || $$\frac{1}{\sin^2 \theta}$$ ||
 * style="text-align:right;" | $$\frac{\cos^2 \theta}{\sin^2 \theta} + \frac{\sin^2 \theta}{\sin^2 \theta} =$$ || $$\frac{1}{\sin^2 \theta}$$
 * style="padding-left:50px;" | $$\frac{\sin^2 \theta}{\sin^2 \theta} = 1$$
 * style="text-align:right;" | $$\frac{\cos^2 \theta}{\sin^2 \theta} + 1 =$$ || $$\frac{1}{\sin^2 \theta}$$ ||
 * style="text-align:right;" | $$\frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} + 1 =$$ || $$\frac{1}{\sin \theta} \cdot \frac{1}{\sin \theta}$$
 * style="padding-left:50px;" | Write the squared functions in terms of their factors.
 * style="text-align:right;" | $$\cot \theta \cdot \cot \theta + 1 =\,\!$$ || $$\csc \theta \cdot \csc \theta \,\!$$
 * style="padding-left:50px;" | Use the quotient and reciprocal identities.
 * style="text-align:right;" | $$\cot^2 \theta + 1 =\,\!$$ || $$\csc^2 \theta \,\!$$
 * style="padding-left:50px;" | Write the functions as squared functions.
 * }
 * style="text-align:right;" | $$\cot \theta \cdot \cot \theta + 1 =\,\!$$ || $$\csc \theta \cdot \csc \theta \,\!$$
 * style="padding-left:50px;" | Use the quotient and reciprocal identities.
 * style="text-align:right;" | $$\cot^2 \theta + 1 =\,\!$$ || $$\csc^2 \theta \,\!$$
 * style="padding-left:50px;" | Write the functions as squared functions.
 * }
 * }

Lesson Summary
In this lesson we have examined relationships between and among the trig functions. The reciprocal identities tell us the relationship between pairs of trig functions that are reciprocals of each other. The quotient identities tell us relationships among functions in threes: the tangent function is the quotient of the sine and cosine functions, and the cotangent function is the reciprocal of this quotient. The Pythagorean identities, which rely on the Pythagorean theorem, also tell us relationships among functions in threes. Each identity can be used to find values of trig functions, and as well as to prove other identities, which will be a focus of chapter 3. We can also use identities to determine the domain and range of functions, which will be useful in chapter 2, where we will graph the six trig functions.

Points to Consider

 * How do you know if an equation is an identity? [Hint: you could consider using a calculator and graphing a related function, or you could try to prove it mathematically.]
 * How can you verify the domain or range of a function?

Review Questions

 * 1) Use reciprocal identities to give the value of each expression.
 * (a) sec θ = 4, cos θ = ?
 * (b) sin θ = $$\tfrac{1}{3}$$, csc θ = ?
 * 1) In the lesson, the range of the cosecant function was given as: csc θ ≤ −1 or csc θ ≥ 1.
 * (a) Use a calculator to fill in the table below. Round values to 4 decimal places.
 * (b) Use the values in the table to explain in your own words what happens to the values of the cosecant function as the measure of the angle approaches 0 degrees.
 * (c) Explain what this tells you about the range of the cosecant function.
 * (d) Discuss how you might further explore values of the sine and cosecant to better understand the range of the cosecant function.
 * 1) In the lesson the domain of the secant function was given as: θϵ°, θ ≠ 90, 270, 450…  Explain why certain values are excluded from the domain.
 * 2) State the quadrant in which each angle lies, and state the sign of each expression.
 * (a) sin(80°)
 * (b) cos(200°)
 * (c) cot(325°)
 * (d) tan(110°)
 * 1) If cos θ = $$\tfrac{6}{10}$$ and sin θ = $$\tfrac{8}{10}$$, what is the value of tan θ?
 * 2) Use quotient identities to explain why the tangent and cotangent function have positive values for angles in the third quadrant.
 * 3) If sin θ = 0.4, what is the value of cos θ?  Assume that θ is an angle in the first quadrant.
 * 4) If cot θ = 2, what is the value of csc θ?  Assume that θ is an angle in the first quadrant.
 * 5) Show that 1 + tan2 θ = sec2 θ.
 * 6) Explain why it is necessary to state the quadrant in which the angle lies for problems such as #7.
 * 1) Explain why it is necessary to state the quadrant in which the angle lies for problems such as #7.

Review Answers

 * (a) $$\tfrac{1}{4}$$
 * (b) $$\tfrac{3}{1}$$ = 3
 * (a)
 * (b) As the angle gets smaller and smaller, the cosecant values get larger and larger.
 * (c) The range of the cosecant function does not have a maximum, like the sine function. The values get larger and larger.
 * (d) Answers will vary. For example, if we looked at values near 90 degrees, we would see the cosecant values get smaller and smaller, approaching 1.
 * 1) The values 90, 270, 450, etc., are excluded because they make the function undefined.
 * (a) Quadrant 1; positive
 * (b) Quadrant 3; negative
 * (c) Quadrant 4; negative
 * (d) Quadrant 2; negative
 * 1) $$\tfrac{8}{6}$$ = $$\tfrac{4}{3}$$
 * 2) The ratio of sine and cosine will be positive in the third quadrant because sine and cosine are both negative in the third quadrant.
 * 3) cos θ ≈ 0.92
 * 4) csc θ = &radic;5
 * 5) Using the Pythagorean identities results in a quadratic equation, which will have two solutions. Stating that the angle lies in a particular quadrant tells you which solution is the actual value of the expression. In #7, the angle is in the first quadrant, so both sine and cosine must be positive.
 * 1) The ratio of sine and cosine will be positive in the third quadrant because sine and cosine are both negative in the third quadrant.
 * 2) cos θ ≈ 0.92
 * 3) csc θ = &radic;5
 * 4) Using the Pythagorean identities results in a quadratic equation, which will have two solutions. Stating that the angle lies in a particular quadrant tells you which solution is the actual value of the expression. In #7, the angle is in the first quadrant, so both sine and cosine must be positive.
 * 1) Using the Pythagorean identities results in a quadratic equation, which will have two solutions. Stating that the angle lies in a particular quadrant tells you which solution is the actual value of the expression. In #7, the angle is in the first quadrant, so both sine and cosine must be positive.

Vocabulary

 * domain
 * The domain of a function is the set of all input (x) values for which the function is defined.


 * identity
 * An identity is an equation that is always true, as long as the variables and expressions involved are defined.


 * quotient
 * A quotient is the result of division. A fraction is one representation of a quotient.


 * range
 * The range of a function is the set of all output (y) values.


 * reciprocal
 * The reciprocal of a fraction is the fraction obtained by interchanging the numerator and denominator. That is, if you "flip over" a fraction, the result is the reciprocal.