Learning Objectives
In this section, you will:
- 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.
In this section, we will review some trigonometric identities that we have already seen in earlier sections, create some new ones and show how we can use identities, along with basic tools of algebra, to simplify trigonometric expressions.
Some Fundamental Trigonometric Identities
We have previously discussed the set of even-odd identities. 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 and determine whether the identity is odd or even. (See Table 1).
Table 1
| Even-Odd Identities |
| [latex]\begin{array}{l}\mathrm{tan}\left(-\theta \right)=-\mathrm{tan}\text{ }\left(\theta\right)\hfill \\ \mathrm{cot}\left(-\theta \right)=-\mathrm{cot}\text{ }\left(\theta\right) \hfill \end{array}[/latex] |
[latex]\begin{array}{l}\mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\text{ }\left(\theta\right) \hfill \\ \mathrm{csc}\left(-\theta \right)=-\mathrm{csc}\text{ }\left(\theta\right) \hfill \end{array}[/latex] |
[latex]\begin{array}{l}\mathrm{cos}\left(-\theta \right)=\mathrm{cos}\text{ }\left(\theta\right) \hfill \\ \mathrm{sec}\left(-\theta \right)=\mathrm{sec}\text{ }\left(\theta\right) \hfill \end{array}[/latex] |
The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. See Table 2.
Table 2
| Reciprocal Identities |
| [latex]\mathrm{sin}\text{ }\left(\theta\right) =\frac{1}{\mathrm{csc}\text{ }\left(\theta\right)}[/latex] |
[latex]\mathrm{csc}\text{ }\left(\theta\right) =\frac{1}{\mathrm{sin}\text{ }\left(\theta\right)}[/latex] |
| [latex]\mathrm{cos}\text{ }\left(\theta\right) =\frac{1}{\mathrm{sec}\text{ }\left(\theta\right)}[/latex] |
[latex]\mathrm{sec}\text{ }\left(\theta\right) =\frac{1}{\mathrm{cos}\text{ }\left(\theta\right)}[/latex] |
| [latex]\mathrm{tan}\text{ }\left(\theta\right) =\frac{1}{\mathrm{cot}\text{ }\left(\theta\right) }[/latex] |
[latex]\mathrm{cot}\text{ }\left(\theta\right) =\frac{1}{\mathrm{tan}\text{ }\left(\theta\right)}[/latex] |
Another set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See Table 3.
Table 3
| Quotient Identities |
| [latex]\mathrm{tan}\left(\theta\right)=\frac{\mathrm{sin}\left(\theta\right) }{\mathrm{cos}\left(\theta\right)}[/latex] |
[latex]\mathrm{cot}\left(\theta\right)=\frac{\mathrm{cos}\left(\theta\right) }{\mathrm{sin}\left(\theta\right) }[/latex] |
Alternate Forms of the Pythagorean Identity
We can use these fundamental identities to derive alternative forms of the Pythagorean Identity,
[latex]{\mathrm{cos}}^{2}\left(\theta\right)+{\mathrm{sin}}^{2}\left(\theta\right)=1.[/latex]
You should recall that the identity shown above is a direct result of our definition of the sine and cosine functions in terms of the coordinates of points on the unit circle. We can derive two more identities using the methods shown below.
The identity [latex]1+\mathrm{cot}^{2}\left(\theta\right) =\mathrm{csc}^{2}\left(\theta\right)[/latex] is found by dividing each term of the first identity by [latex]\mathrm{sin}^{2}\left(\theta\right)[/latex], and then rewriting each part of the equation using the identities we have already discussed in earlier sections.
[latex]\begin{align*}\frac{\mathrm{sin}^{2}\left(\theta \right)}{\mathrm{sin}^{2}\left(\theta\right)}+\frac{\mathrm{cos}^{2}\left(\theta\right)}{\mathrm{sin}^{2}\left(\theta\right)}& =\frac{1}{\mathrm{sin}^{2}\left(\theta\right)}\end{align*}[/latex]
We can then use our earlier quotient and reciprocal identities to rewrite the expression in this equation as shown below.
[latex]\begin{align*}1+\mathrm{cot}^{2}\left(\theta\right)& =\mathrm{csc}^{2}\left(\theta\right)\end{align*}[/latex]
Similarly, [latex]1+\mathrm{tan}^{2}\left(\theta\right) =\mathrm{sec}^{2}\left(\theta\right)[/latex] can be obtained by dividing all terms in the first identity by [latex]\mathrm{cos}^{2}\left(\theta\right)[/latex], and then rewriting each part of the equation using the quotient and reciprocal identities .
[latex]\begin{align*}\frac{\mathrm{sin}^{2}\left(\theta \right)}{\mathrm{cos}^{2}\left(\theta\right)}+\frac{\mathrm{cos}^{2}\left(\theta\right)}{\mathrm{cos}^{2}\left(\theta\right)}& =\frac{1}{\mathrm{cos}^{2}\left(\theta\right)}\end{align*}[/latex]
[latex]\begin{align*}\mathrm{tan}^{2}\left(\theta\right)+1& =\mathrm{sec}^{2}\left(\theta\right)\end{align*}[/latex]
We now have the three Pythagorean identities shown in Table 4.
Table 4
| Pythagorean Identities |
| [latex]\mathrm{sin}^{2}\left(\theta\right) +\mathrm{cos}^{2}\left(\theta\right) =1[/latex] |
[latex]1+{\mathrm{cot}}^{2}\left(\theta\right) ={\mathrm{csc}}^{2}\left(\theta\right)[/latex] |
[latex]1+{\mathrm{tan}}^{2}\left(\theta\right) ={\mathrm{sec}}^{2}\left(\theta\right)[/latex] |
Example 1: Graphing the Equations of an Identity
Graph both sides of the identity [latex]\mathrm{cot}\left(\theta\right) =\frac{1}{\mathrm{tan}\left(\theta\right)}.[/latex] In other words, on the graphing calculator, graph [latex]y=\mathrm{cot}\left(\theta\right)[/latex] and [latex]y=\frac{1}{\mathrm{tan}\left(\theta\right)}.[/latex]
How To:
Given a trigonometric identity, verify that it is true.
- 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.
- Look for opportunities to factor expressions, square a binomial, or add fractions.
- Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
- If these steps do not yield the desired result, try converting all terms to sines and cosines.
- Note the values not in the domain of the expression on the left and right as the identity does not hold for those values.
Example 2: Verifying a Trigonometric Identity
Verify [latex]\mathrm{tan}\left(\theta\right) \mathrm{cos}\left(\theta\right)=\mathrm{sin}\left(\theta\right) .[/latex]
Show Solution
We will start on the left side, as it is the more complicated side:
[latex]\begin{align*}\mathrm{tan}\left(\theta\right)\mathrm{cos}\left(\theta\right)&=\left(\frac{\mathrm{sin}\text{ }\left(\theta\right)}{\mathrm{cos}\left(\theta\right)}\right)\mathrm{cos}\left(\theta\right)\\&=\mathrm{sin}\text{ }\left(\theta\right)\end{align*}[/latex]
Analysis
This identity was fairly simple to verify, as it only required writing [latex]\mathrm{tan}\left(\theta\right)[/latex] in terms of [latex]\mathrm{sin}\left(\theta\right)[/latex] and [latex]\mathrm{cos}\left(\theta\right) .[/latex] When determining identities we need to also consider the domains of the expressions on the left and right sides of the equation. The identity is only valid where both expressions are defined. For this problem, the domain of [latex]\mathrm{tan}\left(\theta\right)\mathrm{cos}\left(\theta\right)[/latex] is all real numbers except [latex]\frac{\pi}{2}\pm n\pi[/latex] where [latex]n[/latex] is an integer. The expression [latex]\frac{\pi}{2}\pm n\pi[/latex] means that we will have breaks in the domain starting at [latex]\frac{\pi}{2}[/latex] and then every time we add or subtract [latex]\pi[/latex] from there. This means the identity will not be valid at these points. We also need to consider the left hand side of the equation. However, since the domain of [latex]\mathrm{sin}\left(\theta\right)[/latex] is all real numbers there are no additional places that the identity is not valid.
Try it #1
Verify the identity [latex]\mathrm{csc}\left(\theta\right)\text{ } \mathrm{cos}\left(\theta\right)\text{ } \mathrm{tan}\left(\theta\right) =1.[/latex]
Show Solution
[latex]\begin{align*}\mathrm{csc}\left(\theta\right)\mathrm{cos}\left(\theta\right) \mathrm{tan}\left(\theta\right) &=\left(\frac{1}{\mathrm{sin}\left(\theta\right) }\right)\mathrm{cos}\left(\theta\right) \left(\frac{\mathrm{sin}\left(\theta\right) }{\mathrm{cos}\left(\theta\right) }\right) \\ &=\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{sin}\left(\theta\right) }\left(\frac{\mathrm{sin}\left(\theta\right) }{\mathrm{cos}\left(\theta\right) }\right) \\ &=\frac{\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) }{\mathrm{sin}\left(\theta\right) \mathrm{cos}\left(\theta\right)} \\ &=1\end{align*}[/latex]
This identity is not valid where [latex]\mathrm{csc}\left(\theta\right)[/latex] or [latex]\mathrm{tan}\left(\theta\right)[/latex] are not defined. This means that it is not valid at [latex]... -\pi, \frac{-\pi}{2}, 0, \frac{\pi}{2}, \pi, ...[/latex].
We can capture these values with the expression [latex]\frac{\pi}{2}\pm\frac{n\pi}{2}[/latex] where [latex]n[/latex] is an integer.
Example 3: Verifying the Equivalency Using the Even-Odd Identities
Verify the following equivalency using the even-odd identities:
[latex]\left(1+\mathrm{sin}\text{ }\left(x\right)\right)\left(1+\mathrm{sin}\left(-x\right)\right)={\mathrm{cos}}^{2}\left(x\right).[/latex]
Show Solution
Working on the left side of the equation, we have
[latex]\begin{align*}\left(1+\mathrm{sin}\left(x\right)\right)\left(1+\mathrm{sin}\left(-x\right)\right)&=\left(1+\mathrm{sin}\left(x\right)\right)\left(1-\mathrm{sin}\left(x\right)\right)&& \text{Since sin}\left(-x\right)=-\mathrm{sin}\left(x\right).\\ &=1-\mathrm{sin}^{2}\left(x\right)&&\text{Difference of squares}.\\ &=\mathrm{cos}^{2}\left(x\right)&& \mathrm{cos}^{2}\left(x\right)=1-\mathrm{sin}^{2}\left(x\right)\text{ from Pythagorean Identity}.\end{align*}[/latex]
This identity is valid for all real numbers since [latex]\mathrm{sin}\left(\theta\right)[/latex] and [latex]\mathrm{cos}\left(\theta\right)[/latex] have a domain of all real numbers.
Example 4: Verifying a Trigonometric Identity Involving sec2(θ)
Verify the identity [latex]\frac{{\mathrm{sec}}^{2}\left(\theta\right) -1}{{\mathrm{sec}}^{2}\left(\theta\right) }={\mathrm{sin}}^{2}\left(\theta\right).[/latex]
Show Solution
As the left side is more complicated, let’s begin there.
[latex]\begin{align*}\frac{\mathrm{sec}^{2}\left(\theta\right) -1}{\mathrm{sec}^{2}\left(\theta\right) }&=\frac{\left(\mathrm{tan}^{2}\left(\theta\right) +1\right)-1}{\mathrm{sec}^{2}\left(\theta\right)}&&\text{Because }\mathrm{sec}^{2}\left(\theta\right) =\mathrm{tan}^{2}\left(\theta\right) +1. \\ &=\frac{\mathrm{tan}^{2}\left(\theta\right)}{\mathrm{sec}^{2}\left(\theta\right)}\\ &=\mathrm{tan}^{2}\left(\theta\right) \left(\frac{1}{\mathrm{sec}^{2}\left(\theta\right)} \right)\\ &=\mathrm{tan}^{2}\left(\theta\right)\mathrm{cos}^{2}\left(\theta\right)&&\text{Because }\mathrm{cos}^{2}\left(\theta\right) =\frac{1}{\mathrm{sec}^{2}\left(\theta\right)}.\\ &=\frac{\mathrm{sin}^{2}\left(\theta\right)} {\mathrm{cos}^{2}\left(\theta\right)}\mathrm{cos}^{2}\left(\theta\right) && \text{Because }\mathrm{tan}^{2}\left(\theta\right) =\frac{\mathrm{sin}^{2}\left(\theta\right)}{\mathrm{cos}^{2}\left(\theta\right)}.\\ &=\mathrm{sin}^{2}\left(\theta\right)\end{align*}[/latex]
There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.
[latex]\begin{align*}\frac{\mathrm{sec}^{2}\left(\theta\right) -1}{\mathrm{sec}^{2}\left(\theta\right) }&=\frac{\mathrm{sec}^{2}\left(\theta\right) }{\mathrm{sec}^{2}\left(\theta\right) }-\frac{1}{\mathrm{sec}^{2}\left(\theta\right) }\\&=1-\mathrm{cos}^{2}\left(\theta\right) \\ &=\mathrm{sin}^{2}\left(\theta\right)\end{align*}[/latex]
This identity is valid on the domain of [latex]\mathrm{sec}\left(\theta\right).[/latex] This means that we need to exclude values where [latex]\mathrm{cos}\left(\theta\right)=0.[/latex]
This would give us all real numbers except [latex]\frac{\pi}{2}\pm n\pi[/latex] where [latex]n[/latex] is an integer.
Analysis
In the first method, we used the identity [latex]\mathrm{sec}^{2}\left(\theta\right) = \mathrm{tan}^{2}\left(\theta\right) +1[/latex] and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. 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.
Try it #2
Show that [latex]\frac{\mathrm{cot}\text{ }\left(\theta\right) }{\mathrm{csc}\left(\theta\right)}=\mathrm{cos}\left(\theta\right) .[/latex]
Show Solution
[latex]\begin{align*}\frac{\mathrm{cot}\left(\theta\right)}{\mathrm{csc}\left(\theta\right)}&=\frac{\frac{\mathrm{cos}\left(\theta\right) }{\mathrm{sin}\left(\theta\right) }}{\frac{1}{\mathrm{sin}\left(\theta\right)}} && \text{This is a complex fraction.}\\ &=\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{sin}\left(\theta\right) }\cdot \frac{\mathrm{sin}\left(\theta\right)}{ 1}&& \text{Multiple the numerator by the reciprocal of the denominator.}\\ &=\mathrm{cos}\left(\theta\right)\end{align*}[/latex]
This identity is valid on the domain of [latex]\mathrm{cot}\left(\theta\right)[/latex] and [latex]\mathrm{csc}\left(\theta\right).[/latex] This means that the identity is not valid for the values [latex]... -2\pi, -\pi, 0, \pi, 2\pi, ...[/latex]
This would give us all real numbers except [latex]\pi\pm n\pi[/latex] where [latex]n[/latex] is an integer.
Example 5: Creating and Verifying an Identity
Create an identity for the expression [latex]2\text{ }\mathrm{tan}\left(\theta\right) \mathrm{sec}\left(\theta\right)[/latex] by rewriting strictly in terms of sine.
Show Solution
There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:
[latex]\begin{align*}2\text{ }\mathrm{tan}\left(\theta\right)\mathrm{sec}\left(\theta\right)&=2\left(\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{cos}\left(\theta\right)}\right)\left(\frac{1}{\mathrm{cos}\left(\theta\right)}\right)\\ &=\frac{2\text{ }\mathrm{sin}\left(\theta\right)}{\mathrm{cos}^{2}\left(\theta\right)}\\ &=\frac{2\text{ }\mathrm{sin}\left(\theta\right)}{1-\mathrm{sin}^{2}\left(\theta\right)}&&\text{Substitute }1-\mathrm{sin}^{2}\left(\theta\right)\text{ for }\mathrm{cos}^{2}\left(\theta\right).\\ 2\mathrm{tan}\left(\theta\right)\mathrm{sec}\left(\theta\right)&=\frac{2\text{ }\mathrm{sin}\left(\theta\right)}{1-\mathrm{sin}^{2}\left(\theta\right)}\end{align*}[/latex]
This identity holds on values in both of the domains of [latex]\mathrm{tan}\left(\theta\right)[/latex] and [latex]\mathrm{sec}\left(\theta\right).[/latex] This means it holds for all real numbers except [latex]... -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}...[/latex]
This would give us all real numbers except [latex]\frac{\pi}{2}\pm n\pi[/latex] where [latex]n[/latex] is an integer.
The expression on the left also needs to be considered for restrictions but the denominator is zero exactly when the left side of the expression is undefined so no more values need to be excluded.
Example 6: Verifying an Identity Using Algebra and Even/Odd Identities
Verify the identity:
[latex]\frac{\mathrm{sin}^{2}\left(-\theta \right)-\mathrm{cos}^{2}\left(-\theta \right)}{\mathrm{sin}\left(-\theta \right)-\mathrm{cos}\left(-\theta \right)}=\mathrm{cos}\left(\theta \right) -\mathrm{sin}\left(\theta \right)[/latex]
Show Solution
Let’s start with the left side and simplify:
[latex]\begin{align*}\frac{\mathrm{sin}^{2}\left(-\theta\right)-\mathrm{cos}^{2}\left(-\theta \right)}{\mathrm{sin}\left(-\theta\right)-\mathrm{cos}\left(-\theta \right)}&=\frac{\left(\mathrm{sin}\left(-\theta \right)-\mathrm{cos}\left(-\theta \right)\right)\left(\mathrm{sin}\left(-\theta \right)+\mathrm{cos}\left(-\theta \right)\right)}{\mathrm{sin}\left(-\theta \right)-\mathrm{cos}\left(-\theta \right)}&& \text{Difference of squares.}\\&=\frac{\left(-\mathrm{sin}\left(\theta \right)-\mathrm{cos}\left(\theta \right)\right)\left(-\mathrm{sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)\right)}{-\mathrm{sin}\left(\theta \right)-\mathrm{cos}\left(\theta \right)}&&\text{Odd-Even Identities.}\\&=\mathrm{-sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)&&\text{Cancel the common factor.}\\&=\mathrm{cos}\left(\theta \right)\mathrm{-sin}\left(\theta \right)&&\text{Reorder terms.}\end{align*}[/latex][latex]\\[/latex]
This identity will be valid when [latex]\mathrm{sin}\left(-\theta \right)-\mathrm{cos}\left(-\theta \right)\ne 0.[/latex] Solving this type of equation is discussed in section 3.8 Solving Trigonometric Equations.
Try it #3
Verify the identity [latex]\frac{\mathrm{sin}^{2}\left(\theta \right) -1}{\mathrm{tan}\left(\theta \right)\mathrm{sin}\left(\theta\right) -\mathrm{tan}\left(\theta\right)}=\frac{\mathrm{sin}\left(\theta\right) +1}{\mathrm{tan}\left(\theta \right)}.[/latex]
Show Solution
[latex]\begin{align*}\frac{\mathrm{sin}^{2}\left(\theta \right) -1}{\mathrm{tan}\left(\theta \right) \mathrm{sin}\left(\theta \right) -\mathrm{tan}\left(\theta \right)}&=\frac{\left(\mathrm{sin}\left(\theta \right) +1\right)\left(\mathrm{sin}\left(\theta \right) -1\right)}{\mathrm{tan}\left(\theta \right)\left(\mathrm{sin}\left(\theta \right) -1\right)}\\ &=\frac{\mathrm{sin}\left(\theta \right)+1}{\mathrm{tan}\left(\theta \right)}\end{align*}[/latex]
This equation is valid on the domain of tangent and where we don’t get a divide by zero. Determining when the denominator in the expression on the left requires equation solving discussed in section 3.8 Solving Trigonometric Equations.
Example 7: Verifying an Identity Involving Cosines and Cotangents
Verify the identity: [latex]\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)\left(1+{\mathrm{cot}}^{2}\left(x\right)\right)=1.[/latex]
Show Solution
We will work on the left side of the equation.
[latex]\begin{align*}\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)&\left(1+\mathrm{cot}^{2}\left(x\right)\right)\\&=\left(1-\mathrm{cos}^{2}\left(x\right)\right)\left(1+\frac{\mathrm{cos}^{2}\left(x\right)}{\mathrm{sin}^{2}\left(x\right)}\right)&&\text{Rewrite }\mathrm{cot}^{2}\left(x\right). \\ &=\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)\left(\frac{{\mathrm{sin}}^{2}\left(x\right)}{{\mathrm{sin}}^{2}x}+\frac{{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}\left(x\right)}\right) &&\text{Find the common denominator.}\\ &=\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)\left(\frac{\mathrm{sin}^{2}\left(x\right)+\mathrm{cos}^{2}\left(x\right)}{\mathrm{sin}^{2}\left(x\right)}\right) \\ &=\left({\mathrm{sin}}^{2}\left(x\right)\right)\left(\frac{\mathrm{sin}^{2}\left(x\right)+\mathrm{cos}^{2}\left(x\right)}{\mathrm{sin}^{2}\left(x\right)}\right)&&\text{Because }1-\mathrm{cos}^{2}\left(x\right) =\mathrm{sin}^{2}\left(x\right).\\&=\left({\mathrm{sin}}^{2}\left(x\right)\right)\left(\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\right)&& \text{Because } \mathrm{sin}^{2}\left(x\right)+\mathrm{cos}^{2}\left(x\right)=1.\\ &=1\end{align*}[/latex]
[latex]\\[/latex]
This identity is valid on the domain of cotangent. This would give us all real numbers except [latex]\pi\pm n\pi[/latex] where [latex]n[/latex] is an integer.
An alternate way to verify this identity is to begin by multiplying out the expression on the left and simplifying.
[latex]\begin{align*}\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)\left(1+\mathrm{cot}^{2}\left(x\right)\right)&=1-\mathrm{cos}^{2}\left(x\right)+\mathrm{cot}^{2}\left(x\right)-\mathrm{cot}^{2}\left(x\right)\mathrm{cos}^{2}\left(x\right)\\&=1-\mathrm{cos}^{2}\left(x\right)+\mathrm{cot}^{2}\left(x\right)\left(1-\mathrm{cos}^{2}\left(x\right)\right)\\&=1-\mathrm{cos}^{2}\left(x\right)+\frac{\mathrm{cos}^{2}\left(x\right)}{\mathrm{sin}^{2}\left(x\right)}\mathrm{sin}^{2}\left(x\right)\\&=1-\mathrm{cos}^{2}\left(x\right)+\mathrm{cos}^{2}\left(x\right)\\&=1\end{align*}[/latex]
Using Algebra to Simplify Trigonometric Expressions
We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions when we are solving equations. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.
An example is the difference of squares formula, [latex]{a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right),[/latex] which is widely used in many areas other than mathematics, such as engineering, architecture, and physics
For example, the expression [latex]\mathrm{sin}^2\left(x\right)-1[/latex] resembles the difference of squares [latex]x^2-1.[/latex] Recognizing that [latex]x^2-1[/latex] can be factored as [latex]\left(x+1\right)\left(x-1\right)[/latex] helps us quickly recognize that [latex]\mathrm{sin}^2\left(x\right)-1[/latex] can be factored as [latex]\left(\mathrm{sin}\left(x\right)+1\right)\left(\mathrm{sin}\left(x\right)-1\right).[/latex]
We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric expressions and equations easier to work with.
Example 8: Writing the Trigonometric Expression as an Algebraic Expression
Write the following trigonometric expression as an algebraic expression: [latex]2\mathrm{cos}^{2}\left(\theta\right) +\mathrm{cos}\left(\theta\right) -1.[/latex]
Show Solution
Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c.[/latex] Letting [latex]\mathrm{cos}\left(\theta\right) =x,[/latex] we can rewrite the expression as follows:
[latex]2{x}^{2}+x-1[/latex]
This expression can be factored as [latex]\left(2x-1\right)\left(x+1\right).[/latex] If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for [latex]x.[/latex] At this point, we would replace [latex]x[/latex] with [latex]\mathrm{cos}\left(\theta\right)[/latex] and solve for [latex]\theta .[/latex]
Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares
Rewrite the trigonometric expression: [latex]4\mathrm{cos}^{2}\left(\theta\right) -1.[/latex]
Show Solution
Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,
[latex]\begin{align*}4\text{ }\mathrm{cos}^{2}\left(\theta\right) -1&=\left(2 \mathrm{cos}\left(\theta\right)-1 \right)\left(2\text{ }\mathrm{cos}\left(\theta\right) +1\right)\end{align*}[/latex]
Analysis
If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let [latex]\mathrm{cos}\left(\theta\right) =x,[/latex] rewrite the expression as [latex]4{x}^{2}-1,[/latex] and factor [latex]\left(2x-1\right)\left(2x+1\right).[/latex] Then replace [latex]x[/latex] with [latex]\mathrm{cos}\left(\theta\right)[/latex] and solve for the angle.
Try it #4
Rewrite the trigonometric expression: [latex]25-9\text{ }\mathrm{sin}^{2}\left(\theta\right) .[/latex]
Show Solution
This is a difference of squares formula: [latex]25-9\text{ }\mathrm{sin}^{2}\left(\theta\right)=\left(5-3\text{ }\mathrm{sin}\left(\theta\right) \right)\left(5+3\text{ }\mathrm{sin}\left(\theta\right)\right).[/latex]
Example 10: Simplify by Rewriting and Using Substitution
Simplify the expression by rewriting and using identities:
[latex]\mathrm{csc}^{2}\left(\theta\right) -\mathrm{cot}^{2}\left(\theta\right)[/latex]
Show Solution
We can start with the Pythagorean identity. We know: [latex]1+\mathrm{cot}^{2}\left(\theta\right)=\mathrm{csc}^{2}\left(\theta\right)[/latex]
[latex]\begin{align*} \mathrm{csc}^{2}\left(\theta\right)-\mathrm{cot}^{2}\left(\theta\right)&=1+\mathrm{cot}^{2}\left(\theta\right)-\mathrm{cot}^{2}\left(\theta\right)&&\text{Substitute expression in for }\mathrm{csc}^{2}\left(\theta\right).\\ &=1\end{align*}[/latex]
This identity is valid on the domains of cotangent and cosecant. This would give us all real numbers except [latex]\pi\pm n\pi[/latex] where [latex]n[/latex] is an integer.
Try it #5
Use algebraic techniques to verify the identity: [latex]\frac{\mathrm{cos}\left(\theta\right) }{1+\mathrm{sin}\left(\theta\right) }=\frac{1-\mathrm{sin}\left(\theta\right) }{\mathrm{cos}\left(\theta\right)}.[/latex]
(Hint: Multiply the numerator and denominator on the left side by [latex]1-\mathrm{sin}\left(\theta\right) .)[/latex]
Show Solution
[latex]\begin{align*}\frac{\mathrm{cos}\left(\theta\right)}{1+\mathrm{sin}\left(\theta\right)}\left(\frac{1-\mathrm{sin}\left(\theta\right)}{1-\mathrm{sin}\left(\theta\right)}\right)&=\frac{\mathrm{cos}\left(\theta\right)\left(1-\mathrm{sin}\left(\theta\right)\right)}{1-\mathrm{sin}^{2}\left(\theta\right)}\\ &=\frac{\mathrm{cos}\left(\theta\right)\left(1-\mathrm{sin}\left(\theta\right)\right)}{\mathrm{cos}^{2}\left(\theta\right))}&&\text{We know}\text{ } 1-\mathrm{sin}^{2}\left(\theta\right)=\mathrm{cos}^{2}\left(\theta\right).\\ &=\frac{1-\mathrm{sin}\left(\theta\right)}{\mathrm{cos}\left(\theta\right)}&&\text{Cancel a factor of}\text{ }\mathrm{cos}\left(\theta\right).\end{align*}[/latex]
Key Equations
| Pythagorean identities |
[latex]\begin{align*}{\mathrm{sin}}^{2}\left(\theta\right) +{\mathrm{cos}}^{2}\left(\theta\right)&=1\\ 1+{\mathrm{cot}}^{2}\left(\theta\right) &={\mathrm{csc}}^{2}\left(\theta\right)\\ 1+{\mathrm{tan}}^{2}\left(\theta\right) &={\mathrm{sec}}^{2}\left(\theta\right) \end{align*}[/latex] |
| Even-odd identities |
[latex]\begin{align*}\mathrm{tan}\left(-\theta \right)&=-\mathrm{tan}\left(\theta\right) \\ \mathrm{cot}\left(-\theta \right)&=-\mathrm{cot}\left(\theta\right) \\ \mathrm{sin}\left(-\theta \right)&=-\mathrm{sin}\left(\theta\right)\\ \mathrm{csc}\left(-\theta \right)&=-\mathrm{csc}\left(\theta\right) \\ \mathrm{cos}\left(-\theta \right)&=\mathrm{cos}\left(\theta\right) \\ \mathrm{sec}\left(-\theta \right)&=\mathrm{sec}\left(\theta\right) \end{align*}[/latex] |
| Reciprocal identities |
[latex]\begin{align*}\mathrm{sin}\left(\theta\right)&=\frac{1}{\mathrm{csc}\left(\theta\right) }\\ \mathrm{cos}\left(\theta\right)&=\frac{1}{\mathrm{sec}\left(\theta\right)}\\ \mathrm{tan}\left(\theta\right) &=\frac{1}{\mathrm{cot}\left(\theta\right) }\\ \mathrm{csc}\left(\theta\right)&=\frac{1}{\mathrm{sin}\left(\theta\right) }\\ \mathrm{sec}\left(\theta\right) &=\frac{1}{\mathrm{cos}\left(\theta\right)}\\ \mathrm{cot}\left(\theta\right)&=\frac{1}{\mathrm{tan}\left(\theta\right)}\end{align*}[/latex] |
| Quotient identities |
[latex]\begin{align*} \mathrm{tan}\left(\theta\right) &=\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{cos}\left(\theta\right) }\\ \mathrm{cot}\left(\theta\right) &=\frac{\mathrm{cos}\left(\theta\right) }{\mathrm{sin}\left(\theta\right) }\end{align*}[/latex] |
| Shift Identities |
[latex]\begin{align*}\mathrm{sin}\left(\theta\right)&=\mathrm{cos}\left(\theta-\frac{\pi}{2}\right)\\ \mathrm{cos}\left(\theta\right)&=\mathrm{sin}\left(\theta-\frac{\pi}{2}\right)\end{align*}[/latex] |
Key Concepts
- There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
- Graphing both sides of an identity will verify it.
- Simplifying one side of the equation to equal the other side is another method for verifying an identity.
- 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.
- We can create an identity by simplifying an expression and then verifying it.
- Verifying an identity may involve algebra with the fundamental identities.
- Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics.
Glossary
- even-odd identities
- set of equations involving trigonometric functions such that if [latex]f\left(-x\right)=-f\left(x\right),[/latex]the identity is odd, and if [latex]f\left(-x\right)=f\left(x\right),[/latex]the identity is even
- Pythagorean identities
- set of equations involving trigonometric functions based on the right triangle properties
- quotient identities
- pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine
- reciprocal identities
- set of equations involving the reciprocals of basic trigonometric definitions
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