Solving Trigonometric Equations With Identities
Verify the fundamental 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. To verify the 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. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities. We will begin with the Pythagorean identities, which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.| Pythagorean Identities | ||
|---|---|---|
| [latex]{\sin }^{2}\theta +{\cos }^{2}\theta =1\\[/latex] | [latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta \\[/latex] | [latex]1+{\tan }^{2}\theta ={\sec }^{2}\theta \\[/latex] |
[latex]\begin{array}{lllll}1+{\cot }^{2}\theta =\left(1+\frac{{\cos }^{2}\theta }{{\sin }^{2}\theta }\right)\hfill & \hfill & \hfill & \hfill & \text{Rewrite the left side}.\hfill \\ \text{ }=\left(\frac{{\sin }^{2}\theta }{{\sin }^{2}\theta }\right)+\left(\frac{{\cos }^{2}\theta }{{\sin }^{2}\theta }\right)\hfill & \hfill & \hfill & \hfill & \text{Write both terms with the common denominator}.\hfill \\ \text{ }=\frac{{\sin }^{2}\theta +{\cos }^{2}\theta }{{\sin }^{2}\theta }\hfill & \hfill & \hfill & \hfill & \hfill \\ \text{ }=\frac{1}{{\sin }^{2}\theta }\hfill & \hfill & \hfill & \hfill & \hfill \\ \text{ }={\csc }^{2}\theta \hfill & \hfill & \hfill & \hfill & \hfill \end{array}\\[/latex]
Similarly, [latex]1+{\tan }^{2}\theta ={\sec }^{2}\theta \\[/latex] can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives
[latex]\begin{array}{llll}1+{\tan }^{2}\theta =1+{\left(\frac{\sin \theta }{\cos \theta }\right)}^{2}\hfill & \hfill & \hfill & \text{Rewrite left side}.\hfill \\ ={\left(\frac{\cos \theta }{\cos \theta }\right)}^{2}+{\left(\frac{\sin \theta }{\cos \theta }\right)}^{2}\hfill & \hfill & \hfill & \text{Write both terms with the common denominator}.\hfill \\ =\frac{{\cos }^{2}\theta +{\sin }^{2}\theta }{{\cos }^{2}\theta }\hfill & \hfill & \hfill & \hfill \\ =\frac{1}{{\cos }^{2}\theta }\hfill & \hfill & \hfill & \hfill \\ ={\sec }^{2}\theta \hfill & \hfill & \hfill & \hfill \end{array}\\[/latex]
The next set of fundamental identities is 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.
| Even-Odd Identities | ||
|---|---|---|
| [latex]\begin{array}{l}\tan \left(-\theta \right)=-\tan \theta \hfill \\ \cot \left(-\theta \right)=-\cot \theta \hfill \end{array}\\[/latex] | [latex]\begin{array}{l}\sin \left(-\theta \right)=-\sin \theta \hfill \\ \csc \left(-\theta \right)=-\csc \theta \hfill \end{array}\\[/latex] | [latex]\begin{array}{l}\cos \left(-\theta \right)=\cos \theta \hfill \\ \sec \left(-\theta \right)=\sec \theta \hfill \end{array}\\[/latex] |
[latex]\begin{array}{l}\sin \left(\frac{\pi }{2}\right)=1\hfill \\ \text{and}\hfill \\ \sin \left(-\frac{\pi }{2}\right)=-\sin \left(\frac{\pi }{2}\right)\hfill \\ =-1\hfill \end{array}\\[/latex]
This is shown in Figure 2.
Figure 2. Graph of [latex]y=\sin \theta \\[/latex]
Recall that an even function is one in which[latex]f\left(-x\right)=f\left(x\right)\text{ for all }x\text{ in the domain of }f\\[/latex]
The graph of an even function is symmetric about the y-axis. The cosine function is an even function because [latex]\cos \left(-\theta \right)=\cos \theta \\[/latex].
For example, consider corresponding inputs [latex]\frac{\pi }{4}\\[/latex] and [latex]-\frac{\pi }{4}\\[/latex]. The output of [latex]\cos \left(\frac{\pi }{4}\right)\\[/latex] is the same as the output of [latex]\cos \left(-\frac{\pi }{4}\right)\\[/latex]. Thus,
[latex]\begin{array}{l}\cos \left(-\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)\hfill \\ \text{ }\approx 0.707\hfill \end{array}\\[/latex]
See Figure 3.
Figure 3. Graph of [latex]y=\cos \theta \\[/latex]
For all [latex]\theta \\[/latex] in the domain of the sine and cosine functions, respectively, we can state the following:- Since [latex]\sin \left(-\theta \right)=-\sin \theta \\[/latex], sine is an odd function.
- Since, [latex]\cos \left(-\theta \right)=\cos \theta \\[/latex], cosine is an even function.
| Reciprocal Identities | |
|---|---|
| [latex]\sin \theta =\frac{1}{\csc \theta }\\[/latex] | [latex]\csc \theta =\frac{1}{\sin \theta }\\[/latex] |
| [latex]\cos \theta =\frac{1}{\sec \theta }\\[/latex] | [latex]\sec \theta =\frac{1}{\cos \theta }\\[/latex] |
| [latex]\tan \theta =\frac{1}{\cot \theta }\\[/latex] | [latex]\cot \theta =\frac{1}{\tan \theta }\\[/latex] |
| Quotient Identities | |
|---|---|
| [latex]\tan \theta =\frac{\sin \theta }{\cos \theta }\\[/latex] | [latex]\cot \theta =\frac{\cos \theta }{\sin \theta }\\[/latex] |
A General Note: Summarizing Trigonometric Identities
The Pythagorean identities are based on the properties of a right triangle.[latex]{\cos }^{2}\theta +{\sin }^{2}\theta =1\\[/latex]
[latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta \\[/latex]
[latex]1+{\tan }^{2}\theta ={\sec }^{2}\theta \\[/latex]
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.
[latex]\tan \left(-\theta \right)=-\tan \theta \\[/latex]
[latex]\cot \left(-\theta \right)=-\cot \theta \\[/latex]
[latex]\sin \left(-\theta \right)=-\sin \theta \\[/latex]
[latex]\csc \left(-\theta \right)=-\csc \theta \\[/latex]
[latex]\cos \left(-\theta \right)=\cos \theta \\[/latex]
[latex]\sec \left(-\theta \right)=\sec \theta \\[/latex]
The reciprocal identities define reciprocals of the trigonometric functions.
[latex]\sin \theta =\frac{1}{\csc \theta }\\[/latex]
[latex]\cos \theta =\frac{1}{\sec \theta }\\[/latex]
[latex]\tan \theta =\frac{1}{\cot \theta }\\[/latex]
[latex]\csc \theta =\frac{1}{\sin \theta }\\[/latex]
[latex]\sec \theta =\frac{1}{\cos \theta }\\[/latex]
[latex]\cot \theta =\frac{1}{\tan \theta }\\[/latex]
The quotient identities define the relationship among the trigonometric functions.
[latex]\tan \theta =\frac{\sin \theta }{\cos \theta }\\[/latex]
[latex]\cot \theta =\frac{\cos \theta }{\sin \theta }\\[/latex]
Example 1: Graphing the Equations of an Identity
Graph both sides of the identity [latex]\cot \theta =\frac{1}{\tan \theta }\\[/latex]. In other words, on the graphing calculator, graph [latex]y=\cot \theta \\[/latex] and [latex]y=\frac{1}{\tan \theta }\\[/latex].Solution
Analysis of the Solution
We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.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.
Example 2: Verifying a Trigonometric Identity
Verify [latex]\tan \theta \cos \theta =\sin \theta \\[/latex].Solution
We will start on the left side, as it is the more complicated side:[latex]\begin{array}{l}\tan \theta \cos \theta =\left(\frac{\sin \theta }{\cos \theta }\right)\cos \theta \hfill \\ \text{ }=\left(\frac{\sin \theta }{\cancel{\cos \theta }}\right)\cancel{\cos \theta }\hfill \\ \text{ }=\sin \theta \hfill \end{array}\\[/latex]
Analysis of the Solution
This identity was fairly simple to verify, as it only required writing [latex]\tan \theta \\[/latex] in terms of [latex]\sin \theta \\[/latex] and [latex]\cos \theta \\[/latex].Try It 1
Verify the identity [latex]\csc \theta \cos \theta \tan \theta =1\\[/latex]. SolutionExample 3: Verifying the Equivalency Using the Even-Odd Identities
Verify the following equivalency using the even-odd identities:[latex]\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]={\cos }^{2}x\\[/latex]
Solution
Working on the left side of the equation, we have[latex]\begin{array}{llll}\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]=\left(1+\sin x\right)\left(1-\sin x\right)\hfill & \hfill & \hfill & \text{Since sin(-}x\text{)=}-\sin x\hfill \\ \text{ }=1-{\sin }^{2}x\hfill & \hfill & \hfill & \text{Difference of squares}\hfill \\ \text{ }={\cos }^{2}x\hfill & \hfill & \hfill & {\text{cos}}^{2}x=1-{\sin }^{2}x\hfill \end{array}\\[/latex]
Example 4: Verifying a Trigonometric Identity Involving sec2θ
Verify the identity [latex]\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }={\sin }^{2}\theta \\[/latex]Solution
As the left side is more complicated, let’s begin there.[latex]\begin{array}{llll}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }=\frac{\left({\tan }^{2}\theta +1\right)-1}{{\sec }^{2}\theta }\hfill & \hfill & \hfill & {\text{sec}}^{2}\theta ={\tan }^{2}\theta +1\hfill \\ \text{ }=\frac{{\tan }^{2}\theta }{{\sec }^{2}\theta }\hfill & \hfill & \hfill & \hfill \\ \text{ }={\tan }^{2}\theta \left(\frac{1}{{\sec }^{2}\theta }\right)\hfill & \hfill & \hfill & \hfill \\ \text{ }={\tan }^{2}\theta \left({\cos }^{2}\theta \right)\hfill & \hfill & \hfill & {\cos }^{2}\theta =\frac{1}{{\sec }^{2}\theta }\hfill \\\text{ }=\left(\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta }\right)\left({\cos }^{2}\theta \right)\hfill & \hfill & \begin{array}{cc}\begin{array}{cc}& \end{array}& \end{array}\hfill & {\text{tan}}^{2}\theta =\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta }\hfill\\ \text{ }=\left(\frac{{\sin }^{2}\theta }{\cancel{{\cos }^{2}\theta}}\right)\left(\cancel{{\cos }^{2}\theta} \right)\hfill & \hfill & \hfill & \hfill\\ \text{ }={\sin }^{2}\theta \hfill & \hfill & \hfill & \hfill \end{array}\\[/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{array}{l}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }=\frac{{\sec }^{2}\theta }{{\sec }^{2}\theta }-\frac{1}{{\sec }^{2}\theta }\hfill \\ \text{ }=1-{\cos }^{2}\theta \hfill \\ \text{ }={\sin }^{2}\theta \hfill \end{array}\\[/latex]
Analysis
In the first method, we used the identity [latex]{\sec }^{2}\theta ={\tan }^{2}\theta +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{\cot \theta }{\csc \theta }=\cos \theta \\[/latex]. SolutionExample 5: Creating and Verifying an Identity
Create an identity for the expression [latex]2\tan \theta \sec \theta \\[/latex] 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:[latex]\begin{array}{ll}2\tan \theta \sec \theta =2\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{1}{\cos \theta }\right)\hfill & \hfill \\ \text{ }=\frac{2\sin \theta }{{\cos }^{2}\theta }\hfill & \hfill \\ \text{ }=\frac{2\sin \theta }{1-{\sin }^{2}\theta }\hfill & \text{Substitute }1-{\sin }^{2}\theta \text{ for }{\cos }^{2}\theta \hfill \end{array}\\[/latex]
Thus,
[latex]2\tan \theta \sec \theta =\frac{2\sin \theta }{1-{\sin }^{2}\theta }\\[/latex]
Example 6: Verifying an Identity Using Algebra and Even/Odd Identities
Verify the identity:[latex]\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}=\cos \theta -\sin \theta \\[/latex]
Solution
Let’s start with the left side and simplify:[latex]\begin{array}{llll}\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}=\frac{{\left[\sin \left(-\theta \right)\right]}^{2}-{\left[\cos \left(-\theta \right)\right]}^{2}}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}\hfill & \hfill & \hfill & \hfill \\ \text{ }=\frac{{\left(-\sin \theta \right)}^{2}-{\left(\cos \theta \right)}^{2}}{-\sin \theta -\cos \theta }\hfill & \hfill & \hfill & \sin \left(-x\right)=-\sin x\text{and}\cos \left(-x\right)=\cos x\hfill \\ \text{ }=\frac{{\left(\sin \theta \right)}^{2}-{\left(\cos \theta \right)}^{2}}{-\sin \theta -\cos \theta }\hfill & \hfill & \hfill & \text{Difference of squares}\hfill \\ \text{ }=\frac{\left(\sin \theta -\cos \theta \right)\left(\sin \theta +\cos \theta \right)}{-\left(\sin \theta +\cos \theta \right)}\hfill & \hfill & \hfill & \hfill \\ \text{ }=\frac{\left(\sin \theta -\cos \theta \right)\left(\cancel{\sin \theta +\cos \theta }\right)}{-\left(\cancel{\sin \theta +\cos \theta }\right)}\hfill & \hfill & \hfill & \hfill \\ \text{ }=\cos \theta -\sin \theta \hfill & \hfill & \hfill & \hfill \end{array}\\[/latex]
Try It 3
Verify the identity [latex]\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }=\frac{\sin \theta +1}{\tan \theta }\\[/latex]. SolutionExample 7: Verifying an Identity Involving Cosines and Cotangents
Verify the identity: [latex]\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)=1\\[/latex].Solution
We will work on the left side of the equation.[latex]\begin{array}{ll}\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)=\left(1-{\cos }^{2}x\right)\left(1+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right)\hfill & \hfill \\ \text{ }=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) \hfill & \text{Find the common denominator}.\hfill \\ \text{ }=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\right)\hfill & \hfill \\ \text{ }=\left({\sin }^{2}x\right)\left(\frac{1}{{\sin }^{2}x}\right)\hfill & \hfill \\ \text{ }=1\hfill & \hfill \end{array}\\[/latex]
Simplify trigonometric expressions using algebra and the identities
We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. 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. For example, the equation [latex]\left(\sin x+1\right)\left(\sin x - 1\right)=0[/latex] resembles the equation [latex]\left(x+1\right)\left(x - 1\right)=0[/latex], which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations. Another 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. 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 equations easier to understand and solve.Example 8: Writing the Trigonometric Expression as an Algebraic Expression
Write the following trigonometric expression as an algebraic expression: [latex]2{\cos }^{2}\theta +\cos \theta -1[/latex].Solution
Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c[/latex]. Letting [latex]\cos \theta =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]\cos \theta [/latex] and solve for [latex]\theta [/latex].
Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares
Rewrite the trigonometric expression: [latex]4{\cos }^{2}\theta -1[/latex].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{array}{l}4{\cos }^{2}\theta -1={\left(2\cos \theta \right)}^{2}-1\hfill \\ \text{ }=\left(2\cos \theta -1\right)\left(2\cos \theta +1\right)\hfill \end{array}[/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]\cos \theta =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]\cos \theta [/latex] and solve for the angle.Try It 4
Rewrite the trigonometric expression: [latex]25 - 9{\sin }^{2}\theta [/latex]. SolutionExample 10: Simplify by Rewriting and Using Substitution
Simplify the expression by rewriting and using identities:[latex]{\csc }^{2}\theta -{\cot }^{2}\theta [/latex]
Solution
We can start with the Pythagorean identity.[latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta [/latex]
Now we can simplify by substituting [latex]1+{\cot }^{2}\theta [/latex] for [latex]{\csc }^{2}\theta [/latex]. We have
[latex]\begin{array}{l}{\csc }^{2}\theta -{\cot }^{2}\theta =1+{\cot }^{2}\theta -{\cot }^{2}\theta \hfill \\ \text{ }=1\hfill \end{array}[/latex]
Try It 5
Use algebraic techniques to verify the identity: [latex]\frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}[/latex]. (Hint: Multiply the numerator and denominator on the left side by [latex]1-\sin\theta[/latex]). SolutionKey Equations
| Pythagorean identities | [latex]\begin{array}{l}{\sin }^{2}\theta +{\cos }^{2}\theta =1\\ 1+{\cot }^{2}\theta ={\csc }^{2}\theta \\ 1+{\tan }^{2}\theta ={\sec }^{2}\theta \end{array}[/latex] |
| Even-odd identities | [latex]\begin{array}{l}\tan \left(-\theta \right)=-\tan \theta \\ \cot \left(-\theta \right)=-\cot \theta \\ \sin \left(-\theta \right)=-\sin \theta \\ \csc \left(-\theta \right)=-\csc \theta \\ \cos \left(-\theta \right)=\cos \theta \\ \sec \left(-\theta \right)=\sec \theta \end{array}[/latex] |
| Reciprocal identities | [latex]\begin{array}{l}\sin \theta =\frac{1}{\csc \theta }\\ \cos \theta =\frac{1}{\sec \theta }\\ \tan \theta =\frac{1}{\cot \theta }\\ \csc \theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ \cot \theta =\frac{1}{\tan \theta }\end{array}[/latex] |
| Quotient identities | [latex]\begin{array}{l}\hfill \\ \tan \theta =\frac{\sin \theta }{\cos \theta }\hfill \\ \cot \theta =\frac{\cos \theta }{\sin \theta }\hfill \end{array}[/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
- set of equations involving trigonometric functions based on the right triangle properties
- 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
- set of equations involving the reciprocals of basic trigonometric definitions
Section Exercises
1. We know [latex]g\left(x\right)=\cos x[/latex] is an even function, and [latex]f\left(x\right)=\sin x[/latex] and [latex]h\left(x\right)=\tan x[/latex] are odd functions. What about [latex]G\left(x\right)={\cos }^{2}x,F\left(x\right)={\sin }^{2}x[/latex], and [latex]H\left(x\right)={\tan }^{2}x?[/latex] Are they even, odd, or neither? Why? 2. Examine the graph of [latex]f\left(x\right)=\sec x[/latex] on the interval [latex]\left[-\pi ,\pi \right][/latex]. How can we tell whether the function is even or odd by only observing the graph of [latex]f\left(x\right)=\sec x?[/latex] 3. After examining the reciprocal identity for [latex]\sec t[/latex], explain why the function is undefined at certain points. 4. All of the Pythagorean identities are related. Describe how to manipulate the equations to get from [latex]{\sin }^{2}t+{\cos }^{2}t=1[/latex] to the other forms. For the following exercises, use the fundamental identities to fully simplify the expression. 5. [latex]\sin x\cos x\sec x[/latex] 6. [latex]\sin \left(-x\right)\cos \left(-x\right)\csc \left(-x\right)[/latex] 7. [latex]\tan x\sin x+\sec x{\cos }^{2}x[/latex] 8. [latex]\csc x+\cos x\cot \left(-x\right)[/latex] 9. [latex]\frac{\cot t+\tan t}{\sec \left(-t\right)}[/latex] 10. [latex]3{\sin }^{3}t\csc t+{\cos }^{2}t+2\cos \left(-t\right)\cos t[/latex] 11. [latex]-\tan \left(-x\right)\cot \left(-x\right)[/latex] 12. [latex]\frac{-\sin \left(-x\right)\cos x\sec x\csc x\tan x}{\cot x}[/latex] 13. [latex]\frac{1+{\tan }^{2}\theta }{{\csc }^{2}\theta }+{\sin }^{2}\theta +\frac{1}{{\sec }^{2}\theta }[/latex] 14. [latex]\left(\frac{\tan x}{{\csc }^{2}x}+\frac{\tan x}{{\sec }^{2}x}\right)\left(\frac{1+\tan x}{1+\cot x}\right)-\frac{1}{{\cos }^{2}x}[/latex] 15. [latex]\frac{1-{\cos }^{2}x}{{\tan }^{2}x}+2{\sin }^{2}x[/latex] For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression. 16. [latex]\frac{\tan x+\cot x}{\csc x};\cos x[/latex] 17. [latex]\frac{\sec x+\csc x}{1+\tan x};\sin x[/latex] 18. [latex]\frac{\cos x}{1+\sin x}+\tan x;\cos x[/latex] 19. [latex]\frac{1}{\sin x\cos x}-\cot x;\cot x[/latex] 20. [latex]\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x};\csc x[/latex] 21. [latex]\left(\sec x+\csc x\right)\left(\sin x+\cos x\right)-2-\cot x;\tan x[/latex] 22. [latex]\frac{1}{\csc x-\sin x};\sec x\text{ and }\tan x[/latex] 23. [latex]\frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x};\sec x\text{ and }\tan x[/latex] 24. [latex]\tan x;\sec x[/latex] 25. [latex]\sec x;\cot x[/latex] 26. [latex]\sec x;\sin x[/latex] 27. [latex]\cot x;\sin x[/latex] 28. [latex]\cot x;\csc x[/latex] For the following exercises, verify the identity. 29. [latex]\cos x-{\cos }^{3}x=\cos x{\sin }^{2}x[/latex] 30. [latex]\cos x\left(\tan x-\sec \left(-x\right)\right)=\sin x - 1[/latex] 31. [latex]\frac{1+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}=1+2{\tan }^{2}x[/latex] 32. [latex]{\left(\sin x+\cos x\right)}^{2}=1+2\sin x\cos x[/latex] 33. [latex]{\cos }^{2}x-{\tan }^{2}x=2-{\sin }^{2}x-{\sec }^{2}x[/latex] For the following exercises, prove or disprove the identity. 34. [latex]\frac{1}{1+\cos x}-\frac{1}{1-\cos \left(-x\right)}=-2\cot x\csc x[/latex] 35. [latex]{\csc }^{2}x\left(1+{\sin }^{2}x\right)={\cot }^{2}x[/latex] 36. [latex]\left(\frac{{\sec }^{2}\left(-x\right)-{\tan }^{2}x}{\tan x}\right)\left(\frac{2+2\tan x}{2+2\cot x}\right)-2{\sin }^{2}x=\cos 2x[/latex] 37. [latex]\frac{\tan x}{\sec x}\sin \left(-x\right)={\cos }^{2}x[/latex] 38. [latex]\frac{\sec \left(-x\right)}{\tan x+\cot x}=-\sin \left(-x\right)[/latex] 39. [latex]\frac{1+\sin x}{\cos x}=\frac{\cos x}{1+\sin \left(-x\right)}[/latex] For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 40. [latex]\frac{{\cos }^{2}\theta -{\sin }^{2}\theta }{1-{\tan }^{2}\theta }={\sin }^{2}\theta [/latex] 41. [latex]3{\sin }^{2}\theta +4{\cos }^{2}\theta =3+{\cos }^{2}\theta [/latex] 42. [latex]\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }={\sec }^{2}\theta [/latex]Licenses & Attributions
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- Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
