Section 1.5 Relationships Between Trigonometric Functions

Supplemental Videos.

The main topics of this section are also presented in the following videos:

So far, we have defined the sine, cosine, and tangent functions. There are three more trigonometric functions, which we define in this section. We will also explore relationships between the six trigonometric functions.

Subsection 1.5.1 Reciprocal Trigonometric Functions

There are times when it can be useful to consider functions like \(1/\sin(\theta)\) or \(1/\cos(\theta)\text{.}\) This happens often enough that mathematicians have created special names for these types of functions. Generally, these functions are called reciprocal trigonometric functions, and they can be defined in terms of the sine, cosine, and tangent functions.

Reciprocal Trigonometric Functions.

Given an angle \(\theta\text{,}\) we define:

\begin{align*} \amp \text{the secant function as } \amp\amp \sec(\theta)=\frac{1}{\cos(\theta)}\\ \\ \amp \text{the cosecant function as } \amp\amp \csc(\theta)=\frac{1}{\sin(\theta)}\\ \\ \amp \text{the cotangent function as } \amp\amp \cot(\theta)=\frac{1}{\tan(\theta)} \end{align*}

Example 1.40.

Find \(\sec\left(\frac{5\pi}{6}\right)\text{,}\) \(\csc\left(\frac{5\pi}{6}\right)\text{,}\) and \(\cot\left(\frac{5\pi}{6}\right)\text{.}\)

Solution.

Since \(\frac{5\pi}{6}\) is a common angle on the unit circle, we have that

\begin{align*} \sec\left(\frac{5\pi}{6}\right)\amp=\frac{1}{\cos\left(\frac{5\pi}{6}\right)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\\ \\ \csc\left(\frac{5\pi}{6}\right)\amp=\frac{1}{\sin\left(\frac{5\pi}{6}\right)} = \frac{1}{\frac{1}{2}} = 2\\ \\ \cot\left(\frac{5\pi}{6}\right)\amp=\frac{1}{\tan\left(\frac{5\pi}{6}\right)} = \frac{1}{-\frac{1}{\sqrt{3}}} = -\sqrt{3} \end{align*}

Example 1.41.

Using the triangle shown, evaluate \(\sec(\theta)\text{,}\) \(\csc(\theta)\text{,}\) and \(\cot(\theta)\text{.}\)

triangle for reciprocal trig example

Solution.

To find values for reciprocal trigonometric functions, we must first find values for \(\cos(\theta)\text{,}\) \(\sin(\theta)\text{,}\) and \(\tan(\theta)\text{.}\) We must also find the length of the hypotenuse. Letting the variable \(c\) represent the length of the hypotenuse and using the Pythagorean Theorem, we get that

\begin{align*} c^2 \amp= 8^2 + 15^2 \\ c^2 \amp= 64 + 225 \\ c^2 \amp= 289 \\ c \amp= \pm \sqrt{289} \\ c \amp= 17 \end{align*}

triangle for reciprocal trig example with hypotenuse labeled

Now solving for \(\cos(\theta)\text{,}\) \(\sin(\theta)\text{,}\) and \(\tan(\theta)\text{,}\) we get that

\begin{align*} \cos(\theta) \amp= \frac{15}{17} \\ \\ \sin(\theta) \amp= \frac{8}{17} \\ \\ \tan(\theta) \amp= \frac{8}{15} \end{align*}

Finally, solving for the reciprocal trigonometric functions, we get

\begin{align*} \sec(\theta) \amp= \frac{1}{\cos(\theta)} = \frac{17}{15} \\ \\ \csc(\theta) \amp= \frac{1}{\sin(\theta)} = \frac{17}{8} \\ \\ \cot(\theta) \amp= \frac{1}{\tan(\theta)} = \frac{15}{8} \end{align*}

Subsection 1.5.2 Cofunctions

Trigonometric functions are related to each other in special ways. In the last part of this section, we explored how sine, cosine, and tangent are related to the reciprocal trig functions cosecant, secant, and cotangent. We now return to Example 1.31 from the previous Section 1.3 to illustrate a special relationship between sine and cosine.

Example 1.42.

Using the triangle shown below, find \(\cos(\alpha)\text{,}\) \(\sin(\alpha)\text{,}\) \(\cos(\beta)\text{,}\) and \(\sin(\beta)\text{.}\)

rotated right triangle example

Solution.

Before solving for the cosine and sine values of \(\alpha\) and \(\beta\text{,}\) we must first find the length of the hypotenuse. Using the Pythagorean theorem, we get that

\begin{align*} c^2 \amp= 3^2 + 4^2 \\ c^2 \amp= 9 + 16 \\ c^2 \amp= 25 \\ c \amp = 5 \end{align*}

rotated right triangle example with hypotenuse labeled

Now solving for the cosine and sine values of \(\alpha\) and \(\beta\text{,}\) we get that

\begin{equation*} \cos(\alpha) = \frac{\text{adjacent to } \alpha}{\text{hypotenuse}}=\frac{3}{5} \hspace{.5in} \cos(\beta) = \frac{\text{adjacent to } \beta}{\text{hypotenuse}}=\frac{4}{5} \end{equation*}

\begin{equation*} \sin(\alpha) = \frac{\text{opposite to } \alpha}{\text{hypotenuse}}=\frac{4}{5} \hspace{.5in} \sin(\beta) = \frac{\text{opposite to } \beta}{\text{hypotenuse}}=\frac{3}{5} \end{equation*}

You may have noticed in the previous example that \(\cos(\alpha) = \sin(\beta)\) and \(\cos(\beta) = \sin(\alpha)\text{.}\) If we think about Soh-Cah-Toa and the right triangle relationships, the side opposite of \(\alpha\) is the side that is adjacent to \(\beta\) so it would make sense that \(\cos(\alpha) = \sin(\beta)\text{.}\) Since the three angles in a triangle need to add to \(\pi\text{,}\) or 180 degrees, then the angles \(\alpha\) and \(\beta\) must add to \(\pi/2\text{,}\) or 90 degrees, so \(\alpha + \beta = \pi/2\) and we have that

\begin{equation*} \beta = \frac{\pi}{2} - \alpha \hspace{.25in} \text{ and } \hspace{.25in} \alpha = \frac{\pi}{2}-\beta \end{equation*}

Since \(\cos(\alpha) = \sin(\beta)\text{,}\) we get that

\begin{equation*} \cos(\alpha) = \sin\left(\frac{\pi}{2}-\alpha\right) \end{equation*}

Cofunctions.

Two functions are called cofunctions if they are equal on complementary angles (angles that add up to 90 degrees or \(\pi/2\) radians).

Sine and cosine are examples of cofunctions (hence the "co" in cosine). The cofunction identities for sine and cosine are

\begin{equation*} \cos(\theta) = \sin\left(\frac{\pi}{2}-\theta\right) \hspace{.25in} \text{ and } \hspace{.25in} \sin(\theta) = \cos\left(\frac{\pi}{2}-\theta\right) \end{equation*}

Secant and cosecant are another example of cofunctions as are tangent and cotangent, which gives us two more sets of identities.

\begin{equation*} \sec(\theta) = \csc\left(\frac{\pi}{2}-\theta\right) \hspace{.25in} \text{ and } \hspace{.25in} \csc(\theta) = \sec\left(\frac{\pi}{2}-\theta\right) \end{equation*}

\begin{equation*} \tan(\theta) = \cot\left(\frac{\pi}{2}-\theta\right) \hspace{.25in} \text{ and } \hspace{.25in} \cot(\theta) = \tan\left(\frac{\pi}{2}-\theta\right) \end{equation*}

Subsection 1.5.3 Even and Odd Properties of Trigonometric Functions

Recall that a function is called even if \(f(x)=f(-x)\) for all \(x\) in the domain of \(f\text{.}\) A function is called odd if \(f(x)=-f(-x)\) for all \(x\) in the domain of \(f\text{.}\) If you would like to review these concepts in further detail, please refer to Reflection and Even and Odd Functions II.5.3.

Now, suppose that \(\theta \) is an angle in the first quadrant. Then recall that the \((x,y)\) coordinates of the corresponding point on the unit circle satisfy the equations

\begin{equation*} \cos(\theta)=x \hspace{.25in} \text{and} \hspace{.25in} \sin(\theta)=y \end{equation*}

unit circle with \((x,y)\) and \(\theta\)

Then if we consider the angle \(-\theta \text{,}\) which is in the fourth quadrant, we can draw the following picture:

unit circle with (x,y) and theta

And use it to conclude that

\begin{equation*} \sin(-\theta) = -y = -\sin(\theta) \end{equation*}

and

\begin{equation*} \cos(-\theta) = x = \cos(\theta). \end{equation*}

As an exercise, you can check that these equations are also true for angles in the second, third, and fourth quadrants. Therefore, we can conclude that sine is an odd function and that cosine is an even function.

We can use the definition of tangent, and the even and odd properties of cosine and sine, respectively, to show that tangent is an odd function:

\begin{equation*} \begin{aligned} \tan(-\theta) \amp= \frac{\sin(-\theta)}{\cos(-\theta)} \amp\amp\text{Definition of tangent}\\ \amp= \frac{-\sin(\theta)}{\cos(\theta)} \amp\amp\text{Since sine is odd and cosine is even}\\ \amp= -\left(\frac{\sin(\theta)}{\cos(\theta)}\right) \\ \amp =-\tan(\theta) \end{aligned} \end{equation*}

Our findings are summarized below.

Cosine is an even function, i.e.

\begin{equation*} \cos(-\theta)=\cos(\theta) \text{ for all angles }\theta \end{equation*}

Sine is an odd function, i.e.

\begin{equation*} \sin(-\theta)=-\sin(\theta) \text{ for all angles }\theta \end{equation*}

Tangent is an odd function, i.e.

\begin{equation*} \tan(-\theta)=-\tan(\theta) \text{ for all angles }\theta \end{equation*}

Example 1.43.

Find \(\sin(-\theta+4\pi)\) given that \(\sin(\theta)=\frac{1}{3}\text{.}\)

Solution.

Recall that sine is \(2\pi\) periodic. This means that the value of the sine function is same same whenever the input differs by \(2\pi\text{;}\) in other words, if two angles are coterminal, then they have the same output in the sine function. Therefore

\begin{equation*} \sin(-\theta+4\pi)=\sin(-\theta+2\pi)=\sin(-\theta) \end{equation*}

Since sine is a odd function, \(\sin(-\theta)=-\sin(\theta)\text{,}\) so

\begin{equation*} \sin(-\theta+4\pi)=\sin(-\theta)=-\sin(\theta)=-\frac{1}{3} \end{equation*}

Subsection 1.5.4 The Pythagorean Identity

The relationship between right triangles and trigonometric functions of angles on the unit circle can also be used to derive a new identity. Consider the same right triangle we used above.

unit circle with right triangle

By using the Pythagorean Theorem and the definitions of cosine and sine, we can establish a new identity. We start by relating the sides of the right triangle shown above using the Pythagorean Theorem.

\begin{align*} x^2+y^2 \amp= r^2 \amp\amp \text{Substituting in } r=1 \\ \\ x^2+y^2 \amp= 1^2 \amp\amp \text{Substituting in } x=\cos(\theta) \text{ and } y=\sin(\theta) \\ \\ (\cos(\theta))^2+(\sin(\theta))^2 \amp= 1^2 \amp\amp \text{Simplifying and using shorthand notation } \\ \\ \cos^2(\theta) + \sin^2(\theta) \amp= 1 \end{align*}

This result is known as the Pythagorean Identity .

Note 1.44.

As shown above, \(\cos^2(\theta)\) and \(\sin^2(\theta)\) are commonly used shorthand notations for \((\cos(\theta))^2\) and \((\sin(\theta))^2\text{.}\) Be aware that some calculators and computers may not understand the shorthand notation.

The Pythagorean Identity.

For any angle \(\theta\text{,}\)

\begin{equation*} \cos^2(\theta)+\sin^2(\theta)=1 \end{equation*}

The Pythagorean Identity can help us to find a cosine value of an angle if we know the sine value of that angle or vice versa. However, solving the equation will yield two possible values, so we will need to utilize additional knowledge of the angle to help us find the desired value.

Example 1.45.

If \(\displaystyle \sin(\theta)=\frac{3}{7}\) and \(\theta\) is in the second quadrant, find \(\cos(\theta)\text{.}\)

Solution.

Substituting the known value for sine into the Pythagorean identity, we have

\begin{align*} \cos^2(\theta) + \left(\frac{3}{7}\right)^2 \amp= 1 \amp\amp \text{Simplifying}\\ \\ \cos^2(\theta) + \frac{9}{49} \amp= 1 \amp\amp \text{Subtracting } \frac{9}{49} \text{ from both sides}\\ \\ \cos^2(\theta) \amp= \frac{40}{49} \amp\amp \text{Taking the square root of both sides }\\ \\ \cos(\theta) \amp= \pm \sqrt{\frac{40}{49}} \amp\amp \text{Simplifying} \\ \\ \amp= \pm \frac{\sqrt{40}}{7} \\ \\ \amp= \pm \frac{2\sqrt{10}}{7} \end{align*}

Since \(\theta\) is in the second quadrant, we know the \(x\) value of the corresponding point on the unit circle is negative. Therefore, the cosine value should also be negative. Using this additional information, we can conclude that

\begin{equation*} \cos(\theta) = -\frac{2\sqrt{10}}{7} \end{equation*}

Exercises 1.5.5 Exercises

1. Calculating Exact Values.

Use the diagram below to find the exact values of each of the following trigonometric functions, where \(a = 8\) and \(b=3\text{.}\)

\(\sin(\theta)\) =

\(\cos(\theta)\) =

\(\tan(\theta)\) =

\(\csc(\theta)\) =

\(\sec(\theta)\) =

\(\cot(\theta)\) =

Hint.

How can we find the hypotenuse of the right triangle? After this, recall the acronym "SOHCAHTOA" and the definitions of the cofunctions.

2. Using the Unit Circle.

[(a)]

Use your unit circle to fill in the tables below:

\(\theta\)	\(5\pi/6\)	\(2\pi/3\)	\(3\pi/4\)
\(\sin(\theta)\)
\(\sin(-\theta)\)

Based on the table above, which of the following statements about the sine function could be true? (Select all that apply.)

The equation \(\sin(\theta) = \sin(-\theta)\) is true for all angles \(\theta\text{.}\)
The equation \(\sin(\theta) = -\sin(-\theta)\) is true for all angles \(\theta\text{.}\)
Sine is an odd function.
Sine is an even function.
None of the above statements are true.

Using the definitions of reciprocal functions and your work in the above problems, which of the following statements about the cosecant function could be true? (Select all that apply.)

The equation \(\csc(\theta) = \csc(-\theta)\) is true for all angles \(\theta\text{.}\)
The equation \(\csc(\theta) = -\csc(-\theta)\) is true for all angles \(\theta\text{.}\)
Cosecant is an odd function.
Cosecant is an even function.
None of the above statements are true.

Hint.

Recall how a negative angle differs from a positive angle (in relation to the unit circle). In the coordinate pair \((x,y)\text{,}\) which coordinate is the sine function associated with on the unit circle? How is the cosecant function related to the sine function?

3. Graphs of Tangent and Cofunctions.

Consider the following graphs:


A	B	C	D

(a) Which (if any) of the graphs A-D represents a graph of the function \(f(\theta)=\sin(\theta)\text{?}\)

Select
A
B
C
D
None of the graphs above

(b) Which (if any) of the graphs A-D represents a graph of the function \(g(\theta)=\cos(\frac{\pi}{2}-\theta)\text{?}\)

Select
A
B
C
D
None of the graphs above

(c) Based on your answers to (a) and (b), what can you say about the two functions \(f(\theta)=\sin(\theta)\) and \(g(\theta)=\cos(\frac{\pi}{2}-\theta)\text{?}\)

\(\cos(\frac{\pi}{2}-\theta)\) is a horizontal shift of \(\sin(\theta)\text{.}\)
\(\cos(\frac{\pi}{2}-\theta)\) is a horizontal reflection of \(\sin(\theta)\text{.}\)
The two functions are the same.
There is no relationship between the two functions.

(d) Based on your answers to parts (a) through (c), what can you say about the two functions \(h(\theta) = \csc(\theta)\) and \(j(\theta) = \sec(\frac{\pi}{2}-\theta)\text{?}\)

\(\sec(\frac{\pi}{2}-\theta)\) is a horizontal shift of \(\csc(\theta)\text{.}\)
\(\sec(\frac{\pi}{2}-\theta)\) is a horizontal reflection of \(\csc(\theta)\text{.}\)
The two functions are the same.
There is no relationship between the two functions.

Hint.

a) When \(\theta=0\text{,}\) what is the value of \(\sin(\theta)\text{?}\) As the angle increases, what happens to the sine function (think about what happens when the angle increases on the unit circle).

b) When \(\theta =0\text{,}\) what is the value of \(\cos(\frac{\pi}{2}-\theta)\text{?}\) As \(\theta\) increases, what happens to this function? Note that we are subtracting \(\theta\) from \(\frac{\pi}{2}\text{.}\)

c)-d) Based on the information found in the previous parts, how do these functions compare?

4. Simplifying Sine and Cosine.

This problem is similar to Problem 6 on your 1.3 worksheet.

Simplify the expression as much as possible.

\(\cos^2(6 t) + \sin^2(6 t)\) =

5. The Pythagorean Theorem and Sine and Cosine.

(a) Using your calculator, fill in the table below. What do you notice?

\(A\)	0	\(\frac{\pi}{3}\)	\(\frac{5\pi}{4}\)	\(2\pi\)
\(\sin^2(A)+\cos^2(A)\)
\(\sin^2(A+5)+\cos^2(A+5)\)
\(\sin^2(2 A)+\cos^2(2 A)\)

(b) Using the Pythagorean theorem, relate the side lengths of the triangle shown in the diagram above.

\(\displaystyle \sin^2(x)+\cos^2(y)=1^2\)
\(\displaystyle x^2+y^2=c^2\)
\(\displaystyle x^2+y^2=1\)
\(\displaystyle a^2+b^2=c^2\)
\(\displaystyle x^2+y^2=\sin^2(A)\)
None of the above

(c)

By writing \(x\) and \(y\) in terms of trigonometric functions of \(A\text{,}\) rewrite your equation from part (b) in terms of the angle \(A\text{.}\) This result is called the Pythagorean identity.

\(\displaystyle x^2+y^2=\sin^2(x)+\cos^2(y)\)
\(\displaystyle \sin^2(A)+\cos^2(A)=1\)
\(\displaystyle \sin^2(x)+\cos^2(y)=1\)
\(\displaystyle a^2+b^2=c^2\)
\(\displaystyle \sin(x)+\cos(y)=\tan(A)\)
None of the above

(d) Suppose we are given that \(\cos(A)=\frac{1}{2}\) and \(A\) is in the first quadrant. Use your work from part (c) to find the exact value of \(\sin(A)\text{.}\)

\(\sin(A)=\)

(e) Suppose we are given that \(\sin(A)=-\frac{1}{7}\) and \(A\) is in the third quadrant. Use your work from part (c) to find the exact value of \(\cos(A)\text{.}\)

\(\cos(A)=\)

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