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Study Guides > Precalculus II

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
sin2θ+cos2θ=1{\sin }^{2}\theta +{\cos }^{2}\theta =1\\ 1+cot2θ=csc2θ1+{\cot }^{2}\theta ={\csc }^{2}\theta \\ 1+tan2θ=sec2θ1+{\tan }^{2}\theta ={\sec }^{2}\theta \\
The second and third identities can be obtained by manipulating the first. The identity 1+cot2θ=csc2θ1+{\cot }^{2}\theta ={\csc }^{2}\theta \\ is found by rewriting the left side of the equation in terms of sine and cosine. Prove: 1+cot2θ=csc2θ1+{\cot }^{2}\theta ={\csc }^{2}\theta \\
1+cot2θ=(1+cos2θsin2θ)Rewrite the left side. =(sin2θsin2θ)+(cos2θsin2θ)Write both terms with the common denominator. =sin2θ+cos2θsin2θ =1sin2θ =csc2θ\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}\\
Similarly, 1+tan2θ=sec2θ1+{\tan }^{2}\theta ={\sec }^{2}\theta \\ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives
1+tan2θ=1+(sinθcosθ)2Rewrite left side.=(cosθcosθ)2+(sinθcosθ)2Write both terms with the common denominator.=cos2θ+sin2θcos2θ=1cos2θ=sec2θ\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}\\
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
tan(θ)=tanθcot(θ)=cotθ\begin{array}{l}\tan \left(-\theta \right)=-\tan \theta \hfill \\ \cot \left(-\theta \right)=-\cot \theta \hfill \end{array}\\ sin(θ)=sinθcsc(θ)=cscθ\begin{array}{l}\sin \left(-\theta \right)=-\sin \theta \hfill \\ \csc \left(-\theta \right)=-\csc \theta \hfill \end{array}\\ cos(θ)=cosθsec(θ)=secθ\begin{array}{l}\cos \left(-\theta \right)=\cos \theta \hfill \\ \sec \left(-\theta \right)=\sec \theta \hfill \end{array}\\
Recall that an odd function is one in which f(x)=f(x)f\left(-x\right)= -f\left(x\right)\\ for all xx\\ in the domain of ff\\. The sine function is an odd function because sin(θ)=sinθ\sin \left(-\theta \right)=-\sin \theta \\. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of π2\frac{\pi }{2}\\ and π2-\frac{\pi }{2}\\. The output of sin(π2)\sin \left(\frac{\pi }{2}\right)\\ is opposite the output of sin(π2)\sin \left(-\frac{\pi }{2}\right)\\. Thus,
sin(π2)=1andsin(π2)=sin(π2)=1\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}\\
This is shown in Figure 2. Graph of y=sin(theta) from -2pi to 2pi, showing in particular that it is symmetric about the origin. Points given are (pi/2, 1) and (-pi/2, -1).

Figure 2. Graph of y=sinθy=\sin \theta \\

  Recall that an even function is one in which
f(x)=f(x) for all x in the domain of ff\left(-x\right)=f\left(x\right)\text{ for all }x\text{ in the domain of }f\\
The graph of an even function is symmetric about the y-axis. The cosine function is an even function because cos(θ)=cosθ\cos \left(-\theta \right)=\cos \theta \\. For example, consider corresponding inputs π4\frac{\pi }{4}\\ and π4-\frac{\pi }{4}\\. The output of cos(π4)\cos \left(\frac{\pi }{4}\right)\\ is the same as the output of cos(π4)\cos \left(-\frac{\pi }{4}\right)\\. Thus,
cos(π4)=cos(π4) 0.707\begin{array}{l}\cos \left(-\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)\hfill \\ \text{ }\approx 0.707\hfill \end{array}\\
See Figure 3.Graph of y=cos(theta) from -2pi to 2pi, showing in particular that it is symmetric about the y-axis. Points given are (-pi/4, .707) and (pi/4, .707).

Figure 3. Graph of y=cosθy=\cos \theta \\

For all θ\theta \\ in the domain of the sine and cosine functions, respectively, we can state the following:
  • Since sin(θ)=sinθ\sin \left(-\theta \right)=-\sin \theta \\, sine is an odd function.
  • Since, cos(θ)=cosθ\cos \left(-\theta \right)=\cos \theta \\, cosine is an even function.
The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, tan(θ)=tanθ\tan \left(-\theta \right)=\mathrm{-tan}\theta \\. We can interpret the tangent of a negative angle as tan(θ)=sin(θ)cos(θ)=sinθcosθ=tanθ\tan \left(-\theta \right)=\frac{\sin \left(-\theta \right)}{\cos \left(-\theta \right)}=\frac{-\sin \theta }{\cos \theta }=-\tan \theta \\. Tangent is therefore an odd function, which means that tan(θ)=tan(θ)\tan \left(-\theta \right)=-\tan \left(\theta \right)\\ for all θ\theta \\ in the domain of the tangent function. The cotangent identity, cot(θ)=cotθ\cot \left(-\theta \right)=-\cot \theta \\, also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as cot(θ)=cos(θ)sin(θ)=cosθsinθ=cotθ\cot \left(-\theta \right)=\frac{\cos \left(-\theta \right)}{\sin \left(-\theta \right)}=\frac{\cos \theta }{-\sin \theta }=-\cot \theta \\. Cotangent is therefore an odd function, which means that cot(θ)=cot(θ)\cot \left(-\theta \right)=-\cot \left(\theta \right)\\ for all θ\theta \\ in the domain of the cotangent function. The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as csc(θ)=1sin(θ)=1sinθ=cscθ\csc \left(-\theta \right)=\frac{1}{\sin \left(-\theta \right)}=\frac{1}{-\sin \theta }=-\csc \theta \\. The cosecant function is therefore odd. Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as sec(θ)=1cos(θ)=1cosθ=secθ\sec \left(-\theta \right)=\frac{1}{\cos \left(-\theta \right)}=\frac{1}{\cos \theta }=\sec \theta \\. The secant function is therefore even. To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities. 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.
Reciprocal Identities
sinθ=1cscθ\sin \theta =\frac{1}{\csc \theta }\\ cscθ=1sinθ\csc \theta =\frac{1}{\sin \theta }\\
cosθ=1secθ\cos \theta =\frac{1}{\sec \theta }\\ secθ=1cosθ\sec \theta =\frac{1}{\cos \theta }\\
tanθ=1cotθ\tan \theta =\frac{1}{\cot \theta }\\ cotθ=1tanθ\cot \theta =\frac{1}{\tan \theta }\\
The final 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.
Quotient Identities
tanθ=sinθcosθ\tan \theta =\frac{\sin \theta }{\cos \theta }\\ cotθ=cosθsinθ\cot \theta =\frac{\cos \theta }{\sin \theta }\\
The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

A General Note: Summarizing Trigonometric Identities

The Pythagorean identities are based on the properties of a right triangle.
cos2θ+sin2θ=1{\cos }^{2}\theta +{\sin }^{2}\theta =1\\
1+cot2θ=csc2θ1+{\cot }^{2}\theta ={\csc }^{2}\theta \\
1+tan2θ=sec2θ1+{\tan }^{2}\theta ={\sec }^{2}\theta \\
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.
tan(θ)=tanθ\tan \left(-\theta \right)=-\tan \theta \\
cot(θ)=cotθ\cot \left(-\theta \right)=-\cot \theta \\
sin(θ)=sinθ\sin \left(-\theta \right)=-\sin \theta \\
csc(θ)=cscθ\csc \left(-\theta \right)=-\csc \theta \\
cos(θ)=cosθ\cos \left(-\theta \right)=\cos \theta \\
sec(θ)=secθ\sec \left(-\theta \right)=\sec \theta \\
The reciprocal identities define reciprocals of the trigonometric functions.
sinθ=1cscθ\sin \theta =\frac{1}{\csc \theta }\\
cosθ=1secθ\cos \theta =\frac{1}{\sec \theta }\\
tanθ=1cotθ\tan \theta =\frac{1}{\cot \theta }\\
cscθ=1sinθ\csc \theta =\frac{1}{\sin \theta }\\
secθ=1cosθ\sec \theta =\frac{1}{\cos \theta }\\
cotθ=1tanθ\cot \theta =\frac{1}{\tan \theta }\\
The quotient identities define the relationship among the trigonometric functions.
tanθ=sinθcosθ\tan \theta =\frac{\sin \theta }{\cos \theta }\\
cotθ=cosθsinθ\cot \theta =\frac{\cos \theta }{\sin \theta }\\

Example 1: Graphing the Equations of an Identity

Graph both sides of the identity cotθ=1tanθ\cot \theta =\frac{1}{\tan \theta }\\. In other words, on the graphing calculator, graph y=cotθy=\cot \theta \\ and y=1tanθy=\frac{1}{\tan \theta }\\.

Solution Graph of y = cot(theta) and y=1/tan(theta) from -2pi to 2pi. They are the same!

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.

  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 2: Verifying a Trigonometric Identity

Verify tanθcosθ=sinθ\tan \theta \cos \theta =\sin \theta \\.

Solution

We will start on the left side, as it is the more complicated side:
tanθcosθ=(sinθcosθ)cosθ =(sinθcosθ)cosθ =sinθ\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}\\

Analysis of the Solution

This identity was fairly simple to verify, as it only required writing tanθ\tan \theta \\ in terms of sinθ\sin \theta \\ and cosθ\cos \theta \\.

Try It 1

Verify the identity cscθcosθtanθ=1\csc \theta \cos \theta \tan \theta =1\\. Solution

Example 3: Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:
(1+sinx)[1+sin(x)]=cos2x\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]={\cos }^{2}x\\

Solution

Working on the left side of the equation, we have
(1+sinx)[1+sin(x)]=(1+sinx)(1sinx)Since sin(-x)=sinx =1sin2xDifference of squares =cos2xcos2x=1sin2x\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}\\

Example 4: Verifying a Trigonometric Identity Involving sec2θ

Verify the identity sec2θ1sec2θ=sin2θ\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }={\sin }^{2}\theta \\

Solution

As the left side is more complicated, let’s begin there.
sec2θ1sec2θ=(tan2θ+1)1sec2θsec2θ=tan2θ+1 =tan2θsec2θ =tan2θ(1sec2θ) =tan2θ(cos2θ)cos2θ=1sec2θ =(sin2θcos2θ)(cos2θ)tan2θ=sin2θcos2θ =(sin2θcos2θ)(cos2θ) =sin2θ\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}\\
There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.
sec2θ1sec2θ=sec2θsec2θ1sec2θ =1cos2θ =sin2θ\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}\\

Analysis

In the first method, we used the identity sec2θ=tan2θ+1{\sec }^{2}\theta ={\tan }^{2}\theta +1\\ 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 cotθcscθ=cosθ\frac{\cot \theta }{\csc \theta }=\cos \theta \\. Solution

Example 5: Creating and Verifying an Identity

Create an identity for the expression 2tanθsecθ2\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:
2tanθsecθ=2(sinθcosθ)(1cosθ) =2sinθcos2θ =2sinθ1sin2θSubstitute 1sin2θ for cos2θ\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}\\
Thus,
2tanθsecθ=2sinθ1sin2θ2\tan \theta \sec \theta =\frac{2\sin \theta }{1-{\sin }^{2}\theta }\\

Example 6: Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:
sin2(θ)cos2(θ)sin(θ)cos(θ)=cosθsinθ\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}=\cos \theta -\sin \theta \\

Solution

Let’s start with the left side and simplify:
sin2(θ)cos2(θ)sin(θ)cos(θ)=[sin(θ)]2[cos(θ)]2sin(θ)cos(θ) =(sinθ)2(cosθ)2sinθcosθsin(x)=sinxandcos(x)=cosx =(sinθ)2(cosθ)2sinθcosθDifference of squares =(sinθcosθ)(sinθ+cosθ)(sinθ+cosθ) =(sinθcosθ)(sinθ+cosθ)(sinθ+cosθ) =cosθsinθ\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}\\

Try It 3

Verify the identity sin2θ1tanθsinθtanθ=sinθ+1tanθ\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }=\frac{\sin \theta +1}{\tan \theta }\\. Solution

Example 7: Verifying an Identity Involving Cosines and Cotangents

Verify the identity: (1cos2x)(1+cot2x)=1\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)=1\\.

Solution

We will work on the left side of the equation.
(1cos2x)(1+cot2x)=(1cos2x)(1+cos2xsin2x) =(1cos2x)(sin2xsin2x+cos2xsin2x)Find the common denominator. =(1cos2x)(sin2x+cos2xsin2x) =(sin2x)(1sin2x) =1\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}\\

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 (sinx+1)(sinx1)=0\left(\sin x+1\right)\left(\sin x - 1\right)=0 resembles the equation (x+1)(x1)=0\left(x+1\right)\left(x - 1\right)=0, 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, a2b2=(ab)(a+b){a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right), 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: 2cos2θ+cosθ12{\cos }^{2}\theta +\cos \theta -1.

Solution

Notice that the pattern displayed has the same form as a standard quadratic expression, ax2+bx+ca{x}^{2}+bx+c. Letting cosθ=x\cos \theta =x, we can rewrite the expression as follows:
2x2+x12{x}^{2}+x - 1
This expression can be factored as (2x+1)(x1)\left(2x+1\right)\left(x - 1\right). 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 xx. At this point, we would replace xx with cosθ\cos \theta and solve for θ\theta .

Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression: 4cos2θ14{\cos }^{2}\theta -1.

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,
4cos2θ1=(2cosθ)21 =(2cosθ1)(2cosθ+1)\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}

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 cosθ=x\cos \theta =x, rewrite the expression as 4x214{x}^{2}-1, and factor (2x1)(2x+1)\left(2x - 1\right)\left(2x+1\right). Then replace xx with cosθ\cos \theta and solve for the angle.

Try It 4

Rewrite the trigonometric expression: 259sin2θ25 - 9{\sin }^{2}\theta . Solution

Example 10: Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:
csc2θcot2θ{\csc }^{2}\theta -{\cot }^{2}\theta

Solution

We can start with the Pythagorean identity.
1+cot2θ=csc2θ1+{\cot }^{2}\theta ={\csc }^{2}\theta
Now we can simplify by substituting 1+cot2θ1+{\cot }^{2}\theta for csc2θ{\csc }^{2}\theta . We have
csc2θcot2θ=1+cot2θcot2θ =1\begin{array}{l}{\csc }^{2}\theta -{\cot }^{2}\theta =1+{\cot }^{2}\theta -{\cot }^{2}\theta \hfill \\ \text{ }=1\hfill \end{array}

Try It 5

Use algebraic techniques to verify the identity: cosθ1+sinθ=1sinθcosθ\frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}. (Hint: Multiply the numerator and denominator on the left side by 1sinθ1-\sin\theta). Solution

Key Equations

Pythagorean identities sin2θ+cos2θ=11+cot2θ=csc2θ1+tan2θ=sec2θ\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}
Even-odd identities tan(θ)=tanθcot(θ)=cotθsin(θ)=sinθcsc(θ)=cscθcos(θ)=cosθsec(θ)=secθ\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}
Reciprocal identities sinθ=1cscθcosθ=1secθtanθ=1cotθcscθ=1sinθsecθ=1cosθcotθ=1tanθ\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}
Quotient identities tanθ=sinθcosθcotθ=cosθsinθ\begin{array}{l}\hfill \\ \tan \theta =\frac{\sin \theta }{\cos \theta }\hfill \\ \cot \theta =\frac{\cos \theta }{\sin \theta }\hfill \end{array}

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 f(x)=f(x)f\left(-x\right)=-f\left(x\right), the identity is odd, and if f(x)=f(x)f\left(-x\right)=f\left(x\right), 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

Section Exercises

1. We know g(x)=cosxg\left(x\right)=\cos x is an even function, and f(x)=sinxf\left(x\right)=\sin x and h(x)=tanxh\left(x\right)=\tan x are odd functions. What about G(x)=cos2x,F(x)=sin2xG\left(x\right)={\cos }^{2}x,F\left(x\right)={\sin }^{2}x, and H(x)=tan2x?H\left(x\right)={\tan }^{2}x? Are they even, odd, or neither? Why? 2. Examine the graph of f(x)=secxf\left(x\right)=\sec x on the interval [π,π]\left[-\pi ,\pi \right]. How can we tell whether the function is even or odd by only observing the graph of f(x)=secx?f\left(x\right)=\sec x? 3. After examining the reciprocal identity for sect\sec t, 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 sin2t+cos2t=1{\sin }^{2}t+{\cos }^{2}t=1 to the other forms. For the following exercises, use the fundamental identities to fully simplify the expression. 5. sinxcosxsecx\sin x\cos x\sec x 6. sin(x)cos(x)csc(x)\sin \left(-x\right)\cos \left(-x\right)\csc \left(-x\right) 7. tanxsinx+secxcos2x\tan x\sin x+\sec x{\cos }^{2}x 8. cscx+cosxcot(x)\csc x+\cos x\cot \left(-x\right) 9. cott+tantsec(t)\frac{\cot t+\tan t}{\sec \left(-t\right)} 10. 3sin3tcsct+cos2t+2cos(t)cost3{\sin }^{3}t\csc t+{\cos }^{2}t+2\cos \left(-t\right)\cos t 11. tan(x)cot(x)-\tan \left(-x\right)\cot \left(-x\right) 12. sin(x)cosxsecxcscxtanxcotx\frac{-\sin \left(-x\right)\cos x\sec x\csc x\tan x}{\cot x} 13. 1+tan2θcsc2θ+sin2θ+1sec2θ\frac{1+{\tan }^{2}\theta }{{\csc }^{2}\theta }+{\sin }^{2}\theta +\frac{1}{{\sec }^{2}\theta } 14. (tanxcsc2x+tanxsec2x)(1+tanx1+cotx)1cos2x\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} 15. 1cos2xtan2x+2sin2x\frac{1-{\cos }^{2}x}{{\tan }^{2}x}+2{\sin }^{2}x For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression. 16. tanx+cotxcscx;cosx\frac{\tan x+\cot x}{\csc x};\cos x 17. secx+cscx1+tanx;sinx\frac{\sec x+\csc x}{1+\tan x};\sin x 18. cosx1+sinx+tanx;cosx\frac{\cos x}{1+\sin x}+\tan x;\cos x 19. 1sinxcosxcotx;cotx\frac{1}{\sin x\cos x}-\cot x;\cot x 20. 11cosxcosx1+cosx;cscx\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x};\csc x 21. (secx+cscx)(sinx+cosx)2cotx;tanx\left(\sec x+\csc x\right)\left(\sin x+\cos x\right)-2-\cot x;\tan x 22. 1cscxsinx;secx and tanx\frac{1}{\csc x-\sin x};\sec x\text{ and }\tan x 23. 1sinx1+sinx1+sinx1sinx;secx and tanx\frac{1-\sin x}{1+\sin x}-\frac{1+\sin x}{1-\sin x};\sec x\text{ and }\tan x 24. tanx;secx\tan x;\sec x 25. secx;cotx\sec x;\cot x 26. secx;sinx\sec x;\sin x 27. cotx;sinx\cot x;\sin x 28. cotx;cscx\cot x;\csc x For the following exercises, verify the identity. 29. cosxcos3x=cosxsin2x\cos x-{\cos }^{3}x=\cos x{\sin }^{2}x 30. cosx(tanxsec(x))=sinx1\cos x\left(\tan x-\sec \left(-x\right)\right)=\sin x - 1 31. 1+sin2xcos2x=1cos2x+sin2xcos2x=1+2tan2x\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 32. (sinx+cosx)2=1+2sinxcosx{\left(\sin x+\cos x\right)}^{2}=1+2\sin x\cos x 33. cos2xtan2x=2sin2xsec2x{\cos }^{2}x-{\tan }^{2}x=2-{\sin }^{2}x-{\sec }^{2}x For the following exercises, prove or disprove the identity. 34. 11+cosx11cos(x)=2cotxcscx\frac{1}{1+\cos x}-\frac{1}{1-\cos \left(-x\right)}=-2\cot x\csc x 35. csc2x(1+sin2x)=cot2x{\csc }^{2}x\left(1+{\sin }^{2}x\right)={\cot }^{2}x 36. (sec2(x)tan2xtanx)(2+2tanx2+2cotx)2sin2x=cos2x\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 37. tanxsecxsin(x)=cos2x\frac{\tan x}{\sec x}\sin \left(-x\right)={\cos }^{2}x 38. sec(x)tanx+cotx=sin(x)\frac{\sec \left(-x\right)}{\tan x+\cot x}=-\sin \left(-x\right) 39. 1+sinxcosx=cosx1+sin(x)\frac{1+\sin x}{\cos x}=\frac{\cos x}{1+\sin \left(-x\right)} For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 40. cos2θsin2θ1tan2θ=sin2θ\frac{{\cos }^{2}\theta -{\sin }^{2}\theta }{1-{\tan }^{2}\theta }={\sin }^{2}\theta 41. 3sin2θ+4cos2θ=3+cos2θ3{\sin }^{2}\theta +4{\cos }^{2}\theta =3+{\cos }^{2}\theta 42. secθ+tanθcotθ+cosθ=sec2θ\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }={\sec }^{2}\theta

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