8.2: Simplifying Trigonometric Expressions with Identities

Skills to Develop

• Verify the fundamental trigonometric identities.
• Simplify trigonometric expressions using algebra and the identities.

In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.

Figure \(\PageIndex{1}\): International passports and travel documents

In this section, we will examine how to use the fundamental trigonometric identities, to verify new identities and to simplify trigonometric expressions.  You are asked to solve problems involving trigonometric identities in WeBWorK in the assignments titled "Chapter 7.1" and "Chapter 7.2."

Verifying Trigonometric Identities

Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.  We have already seen and proved the fundamental trigonometric identities in Sections 6.3 and 6.4.  They are summarized below.

SUMMARIZING TRIGONOMETRIC IDENTITIES

The Pythagorean identities are based on the properties of a right triangle.  (See Sections 6.3 and 6.4.)

\[{\cos}^2 \theta+{\sin}^2 \theta=1 \nonumber\]

\[1+{\cot}^2 \theta={\csc}^2 \theta \nonumber\]

\[1+{\tan}^2 \theta={\sec}^2 \theta \nonumber\]

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.  (See Section 6.4.)

\[\tan(−\theta)=−\tan \theta \nonumber\]

\[\cot(−\theta)=−\cot \theta \nonumber\]

\[\sin(−\theta)=−\sin \theta \nonumber\]

\[\csc(−\theta)=−\csc \theta \nonumber\]

\[\cos(−\theta)=\cos \theta \nonumber\]

\[\sec(−\theta)=\sec \theta \nonumber\]

The reciprocal identities define reciprocals of the trigonometric functions.  (See Section 6.4.)

\[\sin \theta=\dfrac{1}{\csc \theta} \nonumber\]

\[\cos \theta=\dfrac{1}{\sec \theta} \nonumber\]

\[\tan \theta=\dfrac{1}{\cot \theta} \nonumber\]

\[\csc \theta=\dfrac{1}{\sin \theta} \nonumber\]

\[\sec \theta=\dfrac{1}{\cos \theta} \nonumber\]

\[\cot \theta=\dfrac{1}{\tan \theta} \nonumber\]

The quotient identities define the relationship among the trigonometric functions.  (See Section 6.4.)

\[\tan \theta=\dfrac{\sin \theta}{\cos \theta} \nonumber\]

\[\cot \theta=\dfrac{\cos \theta}{\sin \theta} \nonumber\]

Verifying other Trigonometric Identities

Because the six trigonometric functions are so interrelated with one another, there are an endless number of identities that can be created.  By an identity, we mean two different trigonometric expressions that are equal to one another, no matter what the input value.  When we are given a trigonometric identity, we are usually asked to verify it.  This means that we can believe that the identity is true, but we have to show how it is true, mathematically.  To verify trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result.

Given a trigonometric identity, verify that it is true

1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
2. Look for opportunities to factor expressions, square a binomial, or add fractions.
3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
4. If these steps do not yield the desired result, try converting all terms to sines and cosines.

Example \(\PageIndex{1}\): Verifying a Trigonometric Identity

Verify \(\tan \theta \cos \theta=\sin \theta\).

Solution

We will start on the left side, as it is the more complicated side:

\[ \begin{align*} \tan \theta \cos \theta &=\left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta \\[5pt] &= \left(\dfrac{\sin \theta}{\cos \theta}\right)\left(\dfrac{\cos \theta}{1}\right)\\[5pt] &=\sin \theta. \end{align*}\]

Analysis

This identity was fairly simple to verify, as it only required writing \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\), and then using algebra to simplify.

\(\PageIndex{1}\)

Verify the identity \(\csc \theta \cos \theta \tan \theta=1\).

Answer

\[ \begin{align*} \csc \theta \cos \theta \tan \theta &=\left(\dfrac{1}{\sin \theta}\right)\cos \theta\left(\dfrac{\sin \theta}{\cos \theta}\right) \\[5pt] & =\dfrac{\cos \theta}{\sin \theta}\left(\dfrac{\sin \theta}{\cos \theta}\right) \\[5pt] & =\dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[5pt] &=1 \end{align*}\]

Example \(\PageIndex{2}\): Verifying an Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

\((1+\sin x)[1+\sin(−x)]={\cos}^2 x\)

Solution

Working on the left side of the equation, we have

\[ \begin{align*} (1+\sin x)[1+\sin(−x)]&=(1+\sin x)(1-\sin x), \quad \text{since } \sin(-x)= -\sin x \\[5pt] &=1-\sin^2x \quad \text{(difference of squares)} \\[5pt] &= \cos^2x, \quad \text{since } \sin^2 \theta + \cos^2 \theta = 1. \end{align*}\]

Example \(\PageIndex{3}\): Verifying a Trigonometric Identity Involving \({\sec}^2 \theta\)

Verify the identity \(\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta}={\sin}^2 \theta\)

Solution

As the left side is more complicated, let’s begin there.

\[\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{{\sec}^2 \theta}{{\sec}^2 \theta}-\dfrac{1}{{\sec}^2 \theta}\\[5pt] &= 1-{\cos}^2 \theta\\[5pt] &= {\sin}^2 \theta \end{align*}\]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

\[\begin{align*}
\dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{({\tan}^2 \theta +1)-1}{{\sec}^2 \theta},\qquad \text{ since }
{\sec}^2 \theta= {\tan}^2 \theta +1\\[5pt]
&= \dfrac{{\tan}^2 \theta}{{\sec}^2 \theta}\\[5pt]
&= {\tan}^2 \theta\left (\dfrac{1}{{\sec}^2 \theta}\right )\\[5pt]
&= {\tan}^2 \theta \left ({\cos}^2 \theta\right ),\qquad \text{ since }
{\cos}^2 \theta= \dfrac{1}{{\sec}^2 \theta}\\[5pt]
&= \left (\dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\right )\cos^2 \theta,\qquad \text{ since }
{\tan}^2 \theta= \dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\\[5pt]
&= {\sin}^2 \theta
\end{align*}\]

Analysis

In the first method, we split the fraction, putting both terms in the numerator over a common denominator. In the second method, we used the identity \({\sec}^2 \theta={\tan}^2 \theta+1\) and continued to simplify. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

\(\PageIndex{2}\)

Show that \(\dfrac{\cot \theta}{\csc \theta}=\cos \theta\).

Answer

\[\begin{align*} \dfrac{\cot \theta}{\csc \theta}&= \dfrac{\tfrac{\cos \theta}{\sin \theta}}{\tfrac{1}{\sin \theta}}\\[4pt] &= \dfrac{\cos \theta}{\sin \theta}\cdot \dfrac{\sin \theta}{1}\\[4pt] &= \cos \theta \end{align*}\]

Example \(\PageIndex{4}\): Creating an Identity

Create an identity for the expression \(2 \tan \theta \sec \theta\) by rewriting strictly in terms of sine.

Solution

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

\[\begin{align*} 2 \tan \theta \sec \theta&= 2\left (\dfrac{\sin \theta}{\cos \theta}\right )\left(\dfrac{1}{\cos \theta}\right )\\ &= \dfrac{2\sin \theta}{{\cos}^2 \theta}\\ &= \dfrac{2\sin \theta}{1-{\sin}^2 \theta}\qquad \text{Substitute } 1-{\sin}^2 \theta \text{ for } {\cos}^2 \theta \end{align*}\]

Thus,

\(2 \tan \theta \sec \theta=\dfrac{2 \sin \theta}{1−{\sin}^2 \theta}\)

Example \(\PageIndex{5}\): Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

\(\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)}=\cos \theta−\sin \theta\)

Solution

Let’s start with the left side and simplify:

\[\begin{align*} \dfrac{\sin^2(-\theta)-\cos^2(-\theta)}{\sin(-\theta)-\cos(-\theta)} &= \dfrac{[\sin(-\theta]^2-[\cos(-\theta)]^2}{\sin(-\theta)-\cos(-\theta)} \\[5pt] &= \dfrac{(-\sin \theta)^2 - (\cos \theta)^2}{-\sin \theta - \cos \theta}, \quad \text{ since } \sin(-x) = -\sin x \text{ and } \cos(-x)=\cos x\\[5pt] &= \dfrac{(\sin \theta)^2-(\cos \theta)^2}{-\sin \theta -\cos \theta}\\[5pt] &= \dfrac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{-(\sin \theta+\cos \theta)} \qquad \text{(difference of squares in the numerator)}\\[5pt] &= \cos \theta-\sin \theta \end{align*}\]

\(\PageIndex{3}\)

Verify the identity \(\dfrac{{\sin}^2 \theta−1}{\tan \theta \sin \theta−\tan \theta}=\dfrac{\sin \theta+1}{\tan \theta}\).

Answer

\[\begin{align*} \dfrac{{\sin}^2 \theta-1}{\tan \theta \sin \theta-\tan \theta}&= \dfrac{(\sin \theta +1)(\sin \theta -1)}{\tan \theta(\sin \theta -1)}\\[4pt] &= \dfrac{\sin \theta+1}{\tan \theta} \end{align*}\]

Example \(\PageIndex{6}\): Verifying an Identity Involving Cosines and Cotangents

Verify the identity: \((1−{\cos}^2 x)(1+{\cot}^2 x)=1\).

Solution:

\[\begin{align*} (1-{\cos}^2 x)(1+{\cot}^2 x)&= (1-{\cos}^2 x)\left(1+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right)\\[4pt] &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x}{{\sin}^2 x}+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right )\qquad \text{Rewrite using a common denominator}\\[4pt] &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x +{\cos}^2 x}{{\sin}^2 x}\right)\\[4pt] &= ({\sin}^2 x)\left (\dfrac{1}{{\sin}^2 x}\right )\\[4pt] &= 1 \end{align*}\]

 

Example \(\PageIndex{7}\): Simplify an expression by Rewriting

Simplify the expression by rewriting:

\({\csc}^2 \theta−{\cot}^2 \theta.\)

Solution:

We will use Ratio Identities and a Pythagorean Identity:

\[\begin{align*} {\csc}^2 \theta−{\cot}^2 \theta&= \frac{1}{\sin^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}\\[2pt] &= \frac{1-\cos^2\theta}{\sin^2\theta}\\[2pt] &=\frac{\sin^2\theta}{\sin^2\theta}\\&= 1 \end{align*}\]

\(\PageIndex{4}\)

Use algebraic techniques to verify the identity: \(\dfrac{\cos \theta}{1+\sin \theta}=\dfrac{1−\sin \theta}{\cos \theta}\).

(Hint: Work on the left side. Multiply the numerator and denominator by \(1−\sin \theta\).)

Answer

\[\begin{align*} \dfrac{\cos \theta}{1+\sin \theta}\left(\dfrac{1-\sin \theta}{1-\sin \theta}\right)&= \dfrac{\cos \theta (1-\sin \theta)}{1-{\sin}^2 \theta}\\ &= \dfrac{\cos \theta (1-\sin \theta)}{{\cos}^2 \theta}\\ &= \dfrac{1-\sin \theta}{\cos \theta} \end{align*}\]

 

Media

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

• Fundamental Trigonometric Identities
• Verifying Trigonometric Identities

Key Equations

Pythagorean identities	\({\cos}^2 \theta+{\sin}^2 \theta=1\) \(1+{\cot}^2 \theta={\csc}^2 \theta\) \(1+{\tan}^2 \theta={\sec}^2 \theta\)
Even-odd identities	\(\tan(−\theta)=-\tan \theta\) \(\cot(-\theta)=-\cot \theta\) \(\sin(-\theta)=-\sin \theta\) \(\csc(-\theta)=-\csc \theta\) \(\cos(-\theta)=\cos \theta\) \(\sec(-\theta)=\sec \theta\)
Reciprocal identities	\(\sin \theta=\dfrac{1}{\csc \theta}\) \(\cos \theta=\dfrac{1}{\sec \theta}\) \(\tan \theta=\dfrac{1}{\cot \theta}\) \(\csc \theta=\dfrac{1}{\sin \theta}\) \(\sec \theta=\dfrac{1}{\cos \theta}\) \(\cot \theta=\dfrac{1}{\tan \theta}\)
Quotient identities	\(\tan \theta=\dfrac{\sin \theta}{\cos \theta}\) \(\cot \theta=\dfrac{\cos \theta}{\sin \theta}\)

Key Concepts

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Simplifying one side of the equation to equal the other side is a method for verifying an identity. See Example \(\PageIndex{1}\) and Example \(\PageIndex{2}\).
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example \(\PageIndex{3}\).
• We can create an identity from a given expression. See Example \(\PageIndex{4}\).
• Verifying an identity may involve algebra with the fundamental identities. See Example \(\PageIndex{5}\) and Example \(\PageIndex{6}\).

Contributors

• Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.