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Simplifying and Verifying Trigonometric Identities

Simplifying and Verifying Trigonometric Identities

By the end of this section, you will be able to:

  • 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 begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

Verifying 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, shown below, which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identities, but now we will also use additional identities.

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}

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

Rewrite the left side.1+cot2θ=1+cos2θsin2θWrite both terms with the common denominator.=sin2θsin2θ+cos2θsin2θ=sin2θ+cos2θsin2θ=1sin2θ=csc2θ \begin{array}{lrcl} \text{Rewrite the left side.} & 1+\cot^2\theta &=& 1+\tfrac{\cos^2\theta}{\sin^2\theta} \\[4pt] \text{Write both terms with the common denominator.} & &=& \tfrac{\sin^2\theta}{\sin^2\theta}+\tfrac{\cos^2\theta}{\sin^2\theta} \\[4pt] & &=& \tfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta} \\[4pt] & &=& \tfrac{1}{\sin^2\theta} \\[4pt] & &=& \csc^2\theta \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

Rewrite left side.1+tan2θ=1+(sinθcosθ)2Write both terms with the common denominator.=(cosθcosθ)2+(sinθcosθ)2=cos2θ+sin2θcos2θ=1cos2θ=sec2θ \begin{array}{lrcl} \text{Rewrite left side.} & 1+\tan^2\theta &=& 1+\left(\tfrac{\sin\theta}{\cos\theta}\right)^2 \\[4pt] \text{Write both terms with the common denominator.} & &=& \left(\tfrac{\cos\theta}{\cos\theta}\right)^2+\left(\tfrac{\sin\theta}{\cos\theta}\right)^2 \\[4pt] & &=& \tfrac{\cos^2\theta+\sin^2\theta}{\cos^2\theta} \\[4pt] & &=& \tfrac{1}{\cos^2\theta} \\[4pt] & &=& \sec^2\theta \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, shown below.

Even-Odd Identities.

tan(θ)=tanθcot(θ)=cotθsin(θ)=sinθcsc(θ)=cscθcos(θ)=cosθsec(θ)=secθ \begin{array}{l} \tan(-\theta)=-\tan\theta \\ \cot(-\theta)=-\cot\theta \\ \sin(-\theta)=-\sin\theta \\ \csc(-\theta)=-\csc\theta \\ \cos(-\theta)=\cos\theta \\ \sec(-\theta)=\sec\theta \end{array}

Recall that an odd function is one in which f(x)=f(x)f(-x)=-f(x) for all xx in the domain of ff. The sine function is an odd function because sin(θ)=sinθ\sin(-\theta)=-\sin\theta. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of π2\tfrac{\pi}{2} and π2-\tfrac{\pi}{2}. The output of sin(π2)\sin\left(\tfrac{\pi}{2}\right) is opposite the output of sin(π2)\sin\left(-\tfrac{\pi}{2}\right). Thus,

sin(π2)=1andsin(π2)=sin(π2)=1 \begin{array}{l} \sin\left(\tfrac{\pi}{2}\right)=1 \\ \text{and} \\ \sin\left(-\tfrac{\pi}{2}\right)=-\sin\left(\tfrac{\pi}{2}\right) \\ =-1 \end{array}

This is shown in the figure below.

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(-x)=f(x)\ \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(-\theta)=\cos\theta. For example, consider corresponding inputs π4\tfrac{\pi}{4} and π4-\tfrac{\pi}{4}. The output of cos(π4)\cos\left(\tfrac{\pi}{4}\right) is the same as the output of cos(π4)\cos\left(-\tfrac{\pi}{4}\right). Thus,

cos(π4)=cos(π4)0.707 \begin{array}{l} \cos\left(-\tfrac{\pi}{4}\right)=\cos\left(\tfrac{\pi}{4}\right) \\ \approx0.707 \end{array}

See the figure below.

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(-\theta)=-\sin\theta, sine is an odd function.
  • Since, cos(θ)=cosθ\cos(-\theta)=\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(-\theta)=-\tan\theta. We can interpret the tangent of a negative angle as tan(θ)=sin(θ)cos(θ)=sinθcosθ=tanθ\tan(-\theta)=\tfrac{\sin(-\theta)}{\cos(-\theta)}=\tfrac{-\sin\theta}{\cos\theta}=-\tan\theta. Tangent is therefore an odd function, which means that tan(θ)=tan(θ)\tan(-\theta)=-\tan(\theta) for all θ\theta in the domain of the tangent function.

The cotangent identity, cot(θ)=cotθ\cot(-\theta)=-\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(-\theta)=\tfrac{\cos(-\theta)}{\sin(-\theta)}=\tfrac{\cos\theta}{-\sin\theta}=-\cot\theta. Cotangent is therefore an odd function, which means that cot(θ)=cot(θ)\cot(-\theta)=-\cot(\theta) 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(-\theta)=\tfrac{1}{\sin(-\theta)}=\tfrac{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(-\theta)=\tfrac{1}{\cos(-\theta)}=\tfrac{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, shown below.

Reciprocal Identities.

sinθ=1cscθcscθ=1sinθcosθ=1secθsecθ=1cosθtanθ=1cotθcotθ=1tanθ \begin{array}{l} \sin\theta=\tfrac{1}{\csc\theta} \\ \csc\theta=\tfrac{1}{\sin\theta} \\ \cos\theta=\tfrac{1}{\sec\theta} \\ \sec\theta=\tfrac{1}{\cos\theta} \\ \tan\theta=\tfrac{1}{\cot\theta} \\ \cot\theta=\tfrac{1}{\tan\theta} \end{array}

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, shown below.

Quotient Identities.

tanθ=sinθcosθcotθ=cosθsinθ \begin{array}{l} \tan\theta=\tfrac{\sin\theta}{\cos\theta} \\ \cot\theta=\tfrac{\cos\theta}{\sin\theta} \end{array}

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

Summarizing Trigonometric Identities.

The Pythagorean Identities are based on the properties of a right triangle.

cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta=11+cot2θ=csc2θ1+\cot^2\theta=\csc^2\theta1+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(-\theta)=-\tan\thetacot(θ)=cotθ\cot(-\theta)=-\cot\thetasin(θ)=sinθ\sin(-\theta)=-\sin\thetacsc(θ)=cscθ\csc(-\theta)=-\csc\thetacos(θ)=cosθ\cos(-\theta)=\cos\thetasec(θ)=secθ\sec(-\theta)=\sec\theta

The reciprocal identities define reciprocals of the trigonometric functions.

sinθ=1cscθ\sin\theta=\tfrac{1}{\csc\theta}cosθ=1secθ\cos\theta=\tfrac{1}{\sec\theta}tanθ=1cotθ\tan\theta=\tfrac{1}{\cot\theta}cscθ=1sinθ\csc\theta=\tfrac{1}{\sin\theta}secθ=1cosθ\sec\theta=\tfrac{1}{\cos\theta}cotθ=1tanθ\cot\theta=\tfrac{1}{\tan\theta}

The quotient identities define the relationship among the trigonometric functions.

tanθ=sinθcosθ\tan\theta=\tfrac{\sin\theta}{\cos\theta}cotθ=cosθsinθ\cot\theta=\tfrac{\cos\theta}{\sin\theta}

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

Solution. See the figure below.

Analysis. 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 confirm an identity verified with analytical means. If both expressions give the same graph, then they are most likely 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. 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}{lrcl} & \tan\theta\cos\theta &=& \left(\tfrac{\sin\theta}{\cos\theta}\right)\cos\theta \\[4pt] & &=& \left(\tfrac{\sin\theta}{\cancel{\cos\theta}}\right)\cancel{\cos\theta} \\[4pt] & &=& \sin\theta \end{array}

Analysis. 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.

Evaluate each of the following, as in the example above.

Simplifycscθcosθtanθ\csc\theta\cos\theta\tan\thetato a single number.

Example. Verify the following equivalency using the even-odd identities:

(1+sinx)[1+sin(x)]=cos2x(1+\sin x)[1+\sin(-x)]=\cos^2x

Solution. Working on the left side of the equation, we have

Since sin(x)=sinx.(1+sinx)[1+sin(x)]=(1+sinx)(1sinx)Difference of squares=1sin2xcos2x=1sin2x=cos2x \begin{array}{lrcl} \text{Since }\sin(-x)=-\sin x. & (1+\sin x)[1+\sin(-x)] &=& (1+\sin x)(1-\sin x) \\[4pt] \text{Difference of squares} & &=& 1-\sin^2x \\[4pt] \cos^2x=1-\sin^2x & &=& \cos^2x \end{array}

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

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

sec2θ=tan2θ+1sec2θ1sec2θ=(tan2θ+1)1sec2θ=tan2θsec2θ=tan2θ(1sec2θ)cos2θ=1sec2θ=tan2θ(cos2θ)tan2θ=sin2θcos2θ=(sin2θcos2θ)(cos2θ)=(sin2θcos2θ)cos2θ=sin2θ \begin{array}{lrcl} \sec^2\theta=\tan^2\theta+1 & \tfrac{\sec^2\theta-1}{\sec^2\theta} &=& \tfrac{(\tan^2\theta+1)-1}{\sec^2\theta} \\[4pt] & &=& \tfrac{\tan^2\theta}{\sec^2\theta} \\[4pt] & &=& \tan^2\theta\left(\tfrac{1}{\sec^2\theta}\right) \\[4pt] \cos^2\theta=\tfrac{1}{\sec^2\theta} & &=& \tan^2\theta(\cos^2\theta) \\[4pt] \tan^2\theta=\tfrac{\sin^2\theta}{\cos^2\theta} & &=& \left(\tfrac{\sin^2\theta}{\cos^2\theta}\right)(\cos^2\theta) \\[4pt] & &=& \left(\tfrac{\sin^2\theta}{\cancel{\cos^2\theta}}\right)\cancel{\cos^2\theta} \\[4pt] & &=& \sin^2\theta \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}{lrcl} & \tfrac{\sec^2\theta-1}{\sec^2\theta} &=& \tfrac{\sec^2\theta}{\sec^2\theta}-\tfrac{1}{\sec^2\theta} \\[4pt] & &=& 1-\cos^2\theta \\[4pt] & &=& \sin^2\theta \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.

Simplifycotθcscθ\tfrac{\cot\theta}{\csc\theta}to a single trigonometric function.

Example. 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)=2sinθcos2θSubstitute 1sin2θ for cos2θ.=2sinθ1sin2θ \begin{array}{lrcl} & 2\tan\theta\sec\theta &=& 2\left(\tfrac{\sin\theta}{\cos\theta}\right)\left(\tfrac{1}{\cos\theta}\right) \\[4pt] & &=& \tfrac{2\sin\theta}{\cos^2\theta} \\[4pt] \text{Substitute }1-\sin^2\theta\text{ for }\cos^2\theta. & &=& \tfrac{2\sin\theta}{1-\sin^2\theta} \end{array}

Thus,

2tanθsecθ=2sinθ1sin2θ2\tan\theta\sec\theta=\tfrac{2\sin\theta}{1-\sin^2\theta}

Example. Verify the identity:

sin2(θ)cos2(θ)sin(θ)cos(θ)=cosθsinθ\tfrac{\sin^2(-\theta)-\cos^2(-\theta)}{\sin(-\theta)-\cos(-\theta)}=\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)sin(x)=sinx and cos(x)=cosx=(sinθ)2(cosθ)2sinθcosθDifference of squares=(sinθ)2(cosθ)2sinθcosθ=(sinθcosθ)(sinθ+cosθ)(sinθ+cosθ)=(sinθcosθ)(sinθ+cosθ)(sinθ+cosθ)=cosθsinθ \begin{array}{lrcl} & \tfrac{\sin^2(-\theta)-\cos^2(-\theta)}{\sin(-\theta)-\cos(-\theta)} &=& \tfrac{[\sin(-\theta)]^2-[\cos(-\theta)]^2}{\sin(-\theta)-\cos(-\theta)} \\[4pt] \sin(-x)=-\sin x\text{ and }\cos(-x)=\cos x & &=& \tfrac{(-\sin\theta)^2-(\cos\theta)^2}{-\sin\theta-\cos\theta} \\[4pt] \text{Difference of squares} & &=& \tfrac{(\sin\theta)^2-(\cos\theta)^2}{-\sin\theta-\cos\theta} \\[4pt] & &=& \tfrac{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}{-(\sin\theta+\cos\theta)} \\[4pt] & &=& \tfrac{(\sin\theta-\cos\theta)\cancel{(\sin\theta+\cos\theta)}}{-\cancel{(\sin\theta+\cos\theta)}} \\[4pt] & &=& \cos\theta-\sin\theta \end{array}

Which expression completes the identity’s verification below?

Which expression is the simplified form ofsin2θ1tanθsinθtanθ\tfrac{\sin^2\theta-1}{\tan\theta\sin\theta-\tan\theta}?

Example. Verify the identity: (1cos2x)(1+cot2x)=1(1-\cos^2x)(1+\cot^2x)=1.

Solution. We will work on the left side of the equation.

(1cos2x)(1+cot2x)=(1cos2x)(1+cos2xsin2x)Find the common denominator.=(1cos2x)(sin2xsin2x+cos2xsin2x)=(1cos2x)(sin2x+cos2xsin2x)=(sin2x)(1sin2x)=1 \begin{array}{lrcl} & (1-\cos^2x)(1+\cot^2x) &=& (1-\cos^2x)\left(1+\tfrac{\cos^2x}{\sin^2x}\right) \\[4pt] \text{Find the common denominator.} & &=& (1-\cos^2x)\left(\tfrac{\sin^2x}{\sin^2x}+\tfrac{\cos^2x}{\sin^2x}\right) \\[4pt] & &=& (1-\cos^2x)\left(\tfrac{\sin^2x+\cos^2x}{\sin^2x}\right) \\[4pt] & &=& (\sin^2x)\left(\tfrac{1}{\sin^2x}\right) \\[4pt] & &=& 1 \end{array}

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 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(\sin x+1)(\sin x-1)=0 resembles the equation (x+1)(x1)=0(x+1)(x-1)=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=(a-b)(a+b), 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. 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+cax^2+bx+c. Letting cosθ=x\cos\theta=x, we can rewrite the expression as follows:

2x2+x12x^2+x-1

This expression can be factored as (2x1)(x+1)(2x-1)(x+1). 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. 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 11 is 11. This is the difference of squares. Thus,

4cos2θ1=(2cosθ)21=(2cosθ1)(2cosθ+1) \begin{array}{lrcl} & 4\cos^2\theta-1 &=& (2\cos\theta)^2-1 \\[4pt] & &=& (2\cos\theta-1)(2\cos\theta+1) \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 4x214x^2-1, and factor (2x1)(2x+1)(2x-1)(2x+1). Then replace xx with cosθ\cos\theta and solve for the angle.

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

Example. 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}{lrcl} & \csc^2\theta-\cot^2\theta &=& 1+\cot^2\theta-\cot^2\theta \\[4pt] & &=& 1 \end{array}

Use algebraic techniques to verify the identity: which expression equalscosθ1+sinθ\tfrac{\cos\theta}{1+\sin\theta}? (Hint: Multiply the numerator and denominator on the left side by1sinθ1-\sin\theta.)

Key equations

Pythagorean Identitiessin2θ+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 identitiestan(θ)=tanθcot(θ)=cotθsin(θ)=sinθcsc(θ)=cscθcos(θ)=cosθsec(θ)=secθ\begin{array}{l} \tan(-\theta)=-\tan\theta \\ \cot(-\theta)=-\cot\theta \\ \sin(-\theta)=-\sin\theta \\ \csc(-\theta)=-\csc\theta \\ \cos(-\theta)=\cos\theta \\ \sec(-\theta)=\sec\theta \end{array}
Reciprocal identitiessinθ=1cscθcosθ=1secθtanθ=1cotθcscθ=1sinθsecθ=1cosθcotθ=1tanθ\begin{array}{l} \sin\theta=\tfrac{1}{\csc\theta} \\ \cos\theta=\tfrac{1}{\sec\theta} \\ \tan\theta=\tfrac{1}{\cot\theta} \\ \csc\theta=\tfrac{1}{\sin\theta} \\ \sec\theta=\tfrac{1}{\cos\theta} \\ \cot\theta=\tfrac{1}{\tan\theta} \end{array}
Quotient identitiestanθ=sinθcosθcotθ=cosθsinθ\begin{array}{l} \tan\theta=\tfrac{\sin\theta}{\cos\theta} \\ \cot\theta=\tfrac{\cos\theta}{\sin\theta} \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. See Example 1.
  • Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example 2 and Example 3.
  • 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 4.
  • We can create an identity by simplifying an expression and then verifying it. See Example 5.
  • Verifying an identity may involve algebra with the fundamental identities. See Example 6 and Example 7.
  • 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. See Example 8, Example 9, and Example 10.

Key terms

even-odd identities — set of equations involving trigonometric functions such that if f(x)=f(x)f(-x)=-f(x), the identity is odd, and if f(x)=f(x)f(-x)=f(x), 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.

Practice

Verify the fundamental trigonometric identities

Determine whether the following statement is true or false:csc2x(1+sin2x)=cot2x\csc^2x(1+\sin^2x)=\cot^2x.

Determine whether the following statement is true or false:tanxsecxsin(x)=cos2x\tfrac{\tan x}{\sec x}\sin(-x)=\cos^2x.

Determine whether the following statement is true or false:3sin2θ+4cos2θ=3+cos2θ3\sin^2\theta+4\cos^2\theta=3+\cos^2\theta.

Simplify trigonometric expressions using algebra and the identities

Use the fundamental identities to fully simplify the expression:sinxcosxsecx\sin x\cos x\sec x.

Use the fundamental identities to fully simplify the expression:tanxsinx+secxcos2x\tan x\sin x+\sec x\cos^2x.

Simplify the first trigonometric expression by writing the simplified form in terms of the second:secx+cscx1+tanx\tfrac{\sec x+\csc x}{1+\tan x}, in terms ofsinx\sin x.


This section is adapted from Precalculus 2e, Section 7.1: Simplifying and Verifying Trigonometric Identities by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the decorative photograph of international passports (Figure 1), which illustrates only the section’s opening analogy; recreated the three instructional figures as accessible spec-first SVGs built from exact coordinates — the graph of y=sinθy=\sin\theta and the graph of y=cosθy=\cos\theta, each from 2π-2\pi to 2π2\pi with their marked symmetry points, and the overlaid graphs of y=cotθy=\cot\theta and y=1tanθy=\tfrac{1}{\tan\theta} confirming the identity by coinciding as one curve, with dashed vertical asymptotes at every multiple of π\pi. Converted the four boxed identity tables (Pythagorean, even-odd, reciprocal, quotient) and the closing “Summarizing Trigonometric Identities” recap into the book’s callout convention. Every retained Try It became a real interactive component: the three whose left side simplifies to a single value, a single trigonometric function, or a factorable difference of squares became fillin components (evaluated-trig, single-trig-function, and factored respectively); the two verify-type Try Its whose simplified side still holds more than one trigonometric function (so no answerForm token can separate the printed subject from the key) became multiplechoice questions on the identity’s load-bearing simplification step instead, per this chapter’s proof-item policy. Struck-through cancellation steps shown in the printed derivations (Examples 2, 4, and 6) are rendered with \cancel{}. Omitted the “Access these online resources” media links. Adapted three true/false “prove or disprove” exercises (themselves ungradeable as free response) into multiplechoice True/False questions for the “verify the fundamental trigonometric identities” objective, and three “fully simplify”/“simplify in terms of” exercises into single-trig-function fillin questions for the “simplify trigonometric expressions” objective, into a closing Practice block, one group per objective.