7.1: Verifying Trigonometric Identities
 Page ID
 34929
Learning Objectives
 Verify the fundamental trigonometric identities.
 Simplify trigonometric expressions using algebra and the identities.
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.
Basic Trigonometric Identities
Identities enable us to simplify complicated expressions. We can use algebraic techniques 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. 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.
We will begin with reviewing the fundamental identities already introduced in a previous section: the Pythagorean Identities, the EvenOdd (or Negative Angle) Identities, the Reciprocal Identities, and the Quotient Identitie
Definitions: Basic TRIGONOMETRIC IDENTITIES
The Pythagorean Identities are based on the properties of a right triangle.
\({\sin}^2 \theta+{\cos}^2 \theta=1\)  \(1+{\cot}^2 \theta={\csc}^2 \theta\)  \(1+{\tan}^2 \theta={\sec}^2 \theta\) 
The EvenOdd (or Negative Angle) Identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.
\(\tan(−\theta)=−\tan \theta\)  \(\sin(−\theta)=−\sin \theta\)  \(\cos(−\theta)=\cos \theta\) 
\(\cot(−\theta)=−\cot \theta\)  \(\csc(−\theta)=−\csc \theta\)  \(\sec(−\theta)=\sec \theta\) 
The Reciprocal Identities define reciprocals of the trigonometric functions.
\(\sin \theta=\dfrac{1}{\csc \theta}\)  \(\cos \theta = \dfrac{1}{\sec \theta}\)  \(\tan \theta=\dfrac{1}{\cot \theta}\) 
\(\csc \theta=\dfrac{1}{\sin \theta}\)  \(\sec \theta=\dfrac{1}{\cos \theta}\)  \(\cot \theta=\dfrac{1}{\tan \theta}\) 
The Quotient Identities define the relationship among the trigonometric functions.
\(\tan \theta=\dfrac{\sin \theta}{\cos \theta}\)  \(\cot \theta=\dfrac{\cos \theta}{\sin \theta}\) 
\(\PageIndex{1}\) Summary of Basic Trigonometric Identities
These relationships are called identities. Identities are statements that are true for all values of the input on which they are defined. Identities are usually something that can be derived from definitions and relationships we already know. The Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine. We will discuss the role of identities more after an example.
Simplifying a Trigonometric Expression
Just as we often need to simplify algebraic expressions, it is often also necessary or helpful to simplify trigonometric expressions. To do so, we utilize the definitions and identities we have established.
Example \(\PageIndex{2}\)
Simplify \(\dfrac{\sec \left(\theta \right)}{\tan \left(\theta \right)}\).
Solution
\( \begin{array} {lll }
\dfrac{ \;\; \sec \left(\theta \right)\;\;}{\tan \left(\theta \right)} &= \dfrac{\;\; \dfrac{1}{\cos(\theta)} \;\;}{ \dfrac{\sin(\theta)}{\cos(\theta)} } &\text{Simplify: rewrite both functions in terms of sine and cosine } \\
&= \dfrac{1}{\cos \left(\theta \right)} \cdot \dfrac{\cos \left(\theta \right)}{\sin \left(\theta \right)}&\text{Divide fractions: invert and multiply } \\
&= \dfrac{1}{\sin \left(\theta \right)} &\text{Cancel the cosines } \\
&= \csc \left(\theta \right) &\text{Simplify: use Reciprocal Identity } \\
\end{array} \)
By showing that \(\dfrac{\sec \left(\theta \right)}{\tan \left(\theta \right)}\) can be simplified to \(\csc \left(\theta \right)\), we have, in fact, established a new identity: \(\dfrac{\sec \left(\theta \right)}{\tan \left(\theta \right)} =\csc \left(\theta \right)\).
Try It \(\PageIndex{2}\): Simplify by Rewriting and Using Substitution
Simplify the expression by rewriting and using identities: \({\csc}^2 \theta−{\cot}^2 \theta\)
 Answer

\[\begin{align*}
{\csc}^2 \theta−{\cot}^2 \theta &= (1+{\cot}^2 \theta) {\cot}^2 \theta & \text{Use the Pythagorean Identity: } 1+{\cot}^2 \theta = {\csc}^2 \\
&= 1+{\cot}^2 \theta{\cot}^2 \theta\\
&= 1 \end{align*}\]
Verifying A Trigonometric Identity
Occasionally a question may ask you to “prove the identity” or “establish the identity.” This is the same idea as when an algebra book asks a question like “show that \((x1)^{2} =x^{2} 2x+1\).” In this type of question, we must show the algebraic manipulations that demonstrate that the left and right side of the equation are in fact equal. You can think of a “prove the identity” problem as a simplification problem where you know the answer: you know what the end goal of the simplification should be, and just need to show the steps to get there.
To prove an identity, in most cases you will start with the expression on one side of the identity and manipulate it using algebra and trigonometric identities until you have simplified it to the expression on the other side of the equation. Do not treat the identity like an equation to solve – it isn’t! The proof is establishing the two expressions are equal, so we must take care to work with one side at a time rather than applying an operation simultaneously to both sides of the equation.
Example \(\PageIndex{3}\)
Prove the identity \(\dfrac{1+\cot (\alpha )}{\csc (\alpha )} =\sin (\alpha )+\cos (\alpha )\).
Solution
Since the left side seems a bit more complicated, we will start there and simplify the expression until we obtain the right side. We can use the right side as a guide for what might be good steps to make. In this case, the left side involves a fraction while the right side doesn’t, which suggests we should look to see if the fraction can be reduced.
Additionally, since the right side involves sine and cosine and the left does not, it suggests that rewriting the cotangent and cosecant using sine and cosine might be a good idea.
\( \begin{array} {lll }
\dfrac{1+\cot (\alpha )}{\csc (\alpha )} &= \sin (\alpha )+\cos (\alpha ) & \text{Rewrite the cotangent and cosecant} \\[2pt]
= \dfrac{\; 1+\dfrac{\cos (\alpha )\; \;}{\sin (\alpha )} }{\dfrac{1}{\sin (\alpha )} } & & \text{Divide fractions: invert and multiply} \\[2pt]
= \left(1+\dfrac{\cos (\alpha )}{\sin (\alpha )} \right) \cdot \dfrac{\sin (\alpha )}{1} & & \text{Distribute} \\[2pt]
= 1\cdot \dfrac{\sin (\alpha )}{1} +\dfrac{\cos (\alpha )}{\sin (\alpha )} \cdot \dfrac{\sin (\alpha )}{1} & & \text{Simplify fractions} \\[2pt]
= \sin (\alpha )+\cos (\alpha ) \;\;\color{Cerulean}{✓}& &\text{Establish the identity} \\
\end{array} \)
Notice that in the second step, we could have combined the 1 and \(\dfrac{\cos (\alpha )}{\sin (\alpha )}\) before inverting and multiplying. It is very common when proving or simplifying identities for there to be more than one way to obtain the same result.
We can also utilize identities we have previously learned, like the Pythagorean Identity, while simplifying or proving identities.
Confirming a Trigonometric Identity Graphically
One way to quickly confirm whether or not an identity is valid, is to graph the expression on each side of the equal sign. If the resulting gtaphs are identical, then the equation is an identity.
Example \(\PageIndex{4}\): Graphing the Equations of an Identity
Graph both sides of the identity \(\cot \theta=\dfrac{1}{\tan \theta}\). In other words, on the graphing calculator, graph \(y=\cot \theta\) and \(y=\dfrac{1}{\tan \theta}\). Solution See Figure \(\PageIndex{4}\). 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 prove any identity. If both expressions give the same graph, then they must be identities. 
Verifying Trig Identities
How to: Given a trigonometric identity, verify that it is true.
 Work on one side of the equation. Do NOTNOTNOT use Properties of Equality like adding/subtracting/multiplying/or dividing the same expression to both sides of the equal sign. Work ONLY on one side of the equal sign.
 It is usually better to start with the more complex side, as it is easier to simplify than to build.
 Look for opportunities to
 multiply expressions out and combine like terms or
 Factor expressions and cancel,r
 Split apart fractions or
 rewrite fractions with a common denominator and combine fractions into a single fraction,
 Simplify two term denominators by using a Pythagorean substitution or
 simplify two term denominators by multiplying numerator and denominator by the conjugate of the binomial denominator.
 Observe which functions are in the final expression, and look for opportunities to use identities and/or employ substitutions that would make both sides of both sides of the equal sign have
 the number of terms
 the same functions in the numerators and/or denominators.
 If these steps do not yield the desired result, try converting all terms to sines and cosines.
Example \(\PageIndex{5}\): Creating and Verifying an Identity
Create an identity for the expression \(2 \tan \theta \sec \theta\) by rewriting strictly in terms of sine.
Solution
There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:
\[\begin{align*} 2 \tan \theta \sec \theta
&= 2\left (\dfrac{\sin \theta}{\cos \theta}\right )\left(\dfrac{1}{\cos \theta}\right )\\
&= \dfrac{2\sin \theta}{{\cos}^2 \theta}\\
&= \dfrac{2\sin \theta}{1{\sin}^2 \theta} &&\text{Substitute } 1{\sin}^2 \theta \text{ for } {\cos}^2 \theta
\end{align*}\]
Thus, \(2 \tan \theta \sec \theta=\dfrac{2 \sin \theta}{1−{\sin}^2 \theta}\)
Example \(\PageIndex{6}\): Verify a Trigonometric Identity  Cancel
Verify \(\tan \theta \cos \theta=\sin \theta\).
Solution
We will start on the left side, as it is the more complicated side:
\( \begin{array} {lll }
\tan \theta \cos \theta &= \sin \theta &\text{Use the Quotient Identity: } \dfrac{\sin \theta}{\cos \theta} \\[2pt]
= \left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta & &\text{Rewrite as product of fractions} \\[2pt]
= \dfrac{\sin \theta}{\cos \theta} \cdot \dfrac{\cos \theta }{1} & &\text{Cancel} \\[2pt]
= \sin \left(\theta \right) \;\;\color{Cerulean}{✓}& &\text{Establish the identity} \\
\end{array} \)
Analysis
This identity was fairly simple to verify, as it only required writing \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).
Try It \(\PageIndex{6}\)
Verify the identity \(\csc \theta \cos \theta \tan \theta=1\).
 Answer

\( \begin{array} {lll }
\csc \theta \cos \theta \tan \theta &= 1 & \\[2pt]
= \left(\dfrac{1}{\sin \theta}\right) \cos \theta \left(\dfrac{\sin \theta}{\cos \theta}\right) & & \\[4pt]
= \dfrac{1}{\sin \theta} \cdot \dfrac{\cos \theta}{1} \cdot \dfrac{\sin \theta}{\cos \theta} & & \\[4pt]
= \dfrac{\cos \theta \sin \theta}{ \sin \theta \cos \theta} & & \\[2pt]
= 1 \;\;\color{Cerulean}{✓}& & \\
\end{array} \)
Example \(\PageIndex{7}\): Verify a Trigonometric Identity  Multiply Out
Verify the following equivalency:
\((1+\sin x)(1+\sin(−x))={\cos}^2 x\)
Solution
Working on the left side of the equation, we have
\( \begin{array} {lll }
(1+\sin x)(1+\sin(−x)) &= \cos^2 x &\text{Use the Even/Odd Identity: } \sin(x) = \sin x \\[2pt]
= (1+\sin x)(1\sin x) & &\text{Multiply out} \\[2pt]
= 1  \sin x + \sin x \sin^2 x & &\text{Combine like terms} \\[2pt]
= 1  \sin^2 x & &\text{Use the Pythagorean Identity: } \cos ^{2} x+\sin ^{2} x=1 \\[2pt]
= \cos^2 x \;\;\color{Cerulean}{✓} & &\text{Establish the identity} \\
\end{array} \)
Example \(\PageIndex{8}\): Verify a Trigonometric Identity  Factor
Establish the identity \(\dfrac{\cos ^{2} \left(\theta \right)}{1+\sin \left(\theta \right)} =1\sin \left(\theta \right)\).
Solution
Since the left side of the identity is more complicated, it makes sense to start there. To simplify this, we will have to reduce the fraction, which would require the numerator to have a factor in common with the denominator. Additionally, we notice that the right side only involves sine. Both of these suggest that we need to convert the cosine into something involving sine.
Recall the Pythagorean Identity told us \(\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\). By moving one of the trig functions to the other side, we can establish:
\[\sin ^{2} (\theta )=1\cos ^{2} (\theta )\quad \text{ and }\quad \cos ^{2} (\theta )=1\sin ^{2} (\theta )\nonumber\]
Utilizing this, we now can establish the identity.
\( \begin{array} {lll }
\dfrac{\cos ^{2} \left(\theta \right)}{1+\sin \left(\theta \right)} &= 1\sin \left(\theta \right) &\text{Use the Pythagorean Identity: } \cos ^{2} (\theta )+\sin ^{2} (\theta )=1 \\[2pt]
= \dfrac{1\sin ^{2} \left(\theta \right)}{1+\sin \left(\theta \right)} & &\text{Factor the numerator} \\[2pt]
= \dfrac{\left(1\sin \left(\theta \right)\right)\left(1+\sin \left(\theta \right)\right)}{1+\sin \left(\theta \right)} & &\text{Cancel like factors } \\[2pt]
= 1\sin \left(\theta \right) \;\;\color{Cerulean}{✓}& &\text{Establish the identity} \\
\end{array} \)
Example \(\PageIndex{9}\): Verify a Trigonometric Identity  Factor
Verify the identity:
\(\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)}=\cos \theta−\sin \theta\)
Solution
Let’s start with the left side and simplify:
\( \begin{array} {lll }
\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)} &= \cos \theta−\sin \theta &\text{ } \\[4pt]
= \dfrac{{[\sin(\theta)]}^2{[\cos(\theta)]}^2}{\sin(\theta)\cos(\theta)} & &\text{Use Even\Odd Identities: } \sin(x) = \sin\space x\text { and } \cos(x)=\cos \space x \\[2pt]
= \dfrac{{(\sin \theta)}^2{(\cos \theta)}^2}{\sin \theta \cos \theta} & &\text{Factor a difference of squares; Factor out a common negative. } \\[2pt]
= \dfrac{(\sin \theta\cos \theta)(\sin \theta+\cos \theta)}{(\sin \theta+\cos \theta)} & &\text{Cancel} \\
= \dfrac{(\sin \theta\cos \theta)}{1} & &\text{ } \\
= \cos \theta\sin \theta \;\;\color{Cerulean}{✓} & &\text{Establish the identity} \\
\end{array} \)
Try It \(\PageIndex{10}\): Factor
Verify the identity \(\dfrac{{\sin}^2 \theta−1}{\tan \theta \sin \theta−\tan \theta}=\dfrac{\sin \theta+1}{\tan \theta}\).
 Answer

\( \dfrac{{\sin}^2 \theta1}{\tan \theta \sin \theta\tan \theta}= \dfrac{(\sin \theta +1)(\sin \theta 1)}{\tan \theta(\sin \theta 1)}= \dfrac{\sin \theta+1}{\tan \theta} \)
Example \(\PageIndex{11}\): Verify a Trigonometric Identity  Add fractions
Verify the identity: \((1−{\cos}^2 x)(1+{\cot}^2 x)=1\).
Solution
\( \begin{array} {lll }
(1{\cos}^2 x)(1+{\cot}^2 x) &= 1 &\text{Use Quotient Identity: } \cot x = \dfrac{\cos x}{\sin x} \\
= (1{\cos}^2 x)\left(1+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right) && \text{Find the common denominator} \\
= (1{\cos}^2 x)\left(\dfrac{{\sin}^2 x}{{\sin}^2 x}+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right ) \\
= (1{\cos}^2 x)\left(\dfrac{{\sin}^2 x +{\cos}^2 x}{{\sin}^2 x}\right) &&\text{Use the Pythagorean Identity: } \cos ^{2} x )+\sin ^{2} x )=1 \\
= ({\sin}^2 x)\left (\dfrac{1}{{\sin}^2 x}\right ) && \text{Cancel} \\
= 1 \end{array}\)
Example \(\PageIndex{12}\): Verify a Trigonometric Identity  Split or Shear Fractions
Verify the identity \(\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta}={\sin}^2 \theta\)
Solution
As the left side is more complicated, let’s begin there.
\( \begin{array} {lll }
\dfrac{{\sec}^2 \theta−1}{\sec^2 \theta} &= {\sin}^2 \theta &\text{Use the Pythagorean Identity: } \tan ^2 (\theta )+1=\sec ^{2} (\theta ) \\[2pt]
= \dfrac{ (\tan ^2 (\theta )+1)−1}{\sec^2 \theta} & &\text{Simplify} \\[2pt]
= \dfrac{{\tan}^2 \theta}{{\sec}^2 \theta} & &\text{"Shear" the fraction} \\[2pt]
= \dfrac{\tan^2 \theta}{1} \cdot \dfrac{1}{\sec^2 \theta} & &\text{Use Reciprocal Identity: } \dfrac{1}{\sec \theta}=\cos\theta \\[2pt]
= \tan^2 \theta \cdot \cos^2 \theta & &\text{Use Quotient Identity: } \tan \theta = \dfrac{\sin \theta}{\cos\theta} \\[2pt]
= \dfrac{\sin^2 \theta}{\cos^2 \theta}\cdot \cos^2 \theta & &\text{Cancel} \\[2pt]
= {\sin}^2 \theta \;\;\color{Cerulean}{✓} & &\text{Establish the identity} \\
\end{array} \)
There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.
\( \begin{array} {lll }
\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta} &= {\sin}^2 \theta &\text{Split the fraction apart} \\[4pt]
= \dfrac{{\sec}^2 \theta}{{\sec}^2 \theta}\dfrac{1}{{\sec}^2 \theta} & &\text{Simplify. Use Reciprocal Identity: } \dfrac{1}{\sec \theta}=\cos\theta \\[2pt]
= 1{\cos}^2 \theta & &\text{Use the Pythagorean Identity: } \cos ^{2} (\theta )+\sin ^{2} (\theta )=1 \\[2pt]
= {\sin}^2 \theta \;\;\color{Cerulean}{✓} & &\text{Establish the identity} \\
\end{array} \)
Analysis
In the first method, we used the identity \({\sec}^2 \theta={\tan}^2 \theta+1\) 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 \(\PageIndex{13}\): Complex Fraction
Show that \(\dfrac{\cot \theta}{\csc \theta}=\cos \theta\).
 Answer

\( \dfrac{\cot \theta}{\csc \theta} = \dfrac{\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}} = \dfrac{\cos \theta}{\sin \theta}\cdot \dfrac{\sin \theta}{1} = \cos \theta \)
Example \(\PageIndex{14}\): Verify a Trigonometric Identity  2 term denominator
Use algebraic techniques to verify the identity: \(\dfrac{\cos \theta}{1+\sin \theta}=\dfrac{1−\sin \theta}{\cos \theta}\).
(Hint: Multiply the numerator and denominator on the left side by \(1−\sin \theta\), the conjugate of the denominator.)
Solution
\(\begin{array} {lll}
\dfrac{\cos \theta}{1+\sin \theta} &=\dfrac{1−\sin \theta}{\cos \theta} & \text{Multiply by a \(\color{Cerulean}{"1"}\) composed of the conjugate of the denominator}\\
=\dfrac{\cos \theta}{1+\sin \theta} \color{Cerulean}{ \left(\dfrac{1\sin \theta}{1\sin \theta}\right) }&&\\
= \dfrac{\cos \theta (1\sin \theta)}{1{\sin}^2 \theta} &&\text{Use the Pythagorean Identity: } \cos ^{2} \theta +\sin ^{2} \theta =1 \\
= \dfrac{\cos \theta (1\sin \theta)}{{\cos}^2 \theta}&&\\
= \dfrac{1\sin \theta}{\cos \theta} \;\;\color{Cerulean}{✓} & &\text{Establish the identity}
\end{array}\)
Use Identities to Solve Trigonometric Equations
When solving some trigonometric equations, it becomes necessary to first rewrite the equation using trigonometric identities. One of the most common is the Pythagorean Identity, \(\sin ^{2} (\theta )+\cos ^{2} (\theta )=1\) which allows you to rewrite \(\sin ^{2} (\theta )\) in terms of \(\cos ^{2} (\theta )\) or vice versa. These identities become very useful whenever an equation involves a combination of sine and cosine functions.
Example \(\PageIndex{15}\): Use Pythagorean Identities
Solve \(2\sin ^{2} (t)\cos (t)=1\) for all solutions with \(0\le t<2\pi\).
Solution
Since this equation has a mix of sine and cosine functions, it becomes more complicated to solve. It is usually easier to work with an equation involving only one trig function. This is where we can use the Pythagorean Identity.
\[2\sin ^{2} (t)\cos (t)=1\nonumber\]Using \(\sin ^{2} (\theta )=1\cos ^{2} (\theta )\)
\[2\left(1\cos ^{2} (t)\right)\cos (t)=1\nonumber\]Distributing the 2
\[22\cos ^{2} (t)\cos (t)=1\nonumber\]
Since this is now quadratic in cosine, we rearrange the equation so one side is zero and factor.
\[2\cos ^{2} (t)\cos (t)+1=0\nonumber\]Multiply by 1 to simplify the factoring
\[2\cos ^{2} (t)+\cos (t)1=0\nonumber\]Factor
\[\left(2\cos (t)1\right)\left(\cos (t)+1\right)=0\nonumber\]
This product will be zero if either factor is zero, so we can break this into two separate cases and solve each independently.
\[2\cos (t)1=0\text{ or }\cos (t)+1=0\nonumber\]
\[\cos (t)=\dfrac{1}{2}\text{ or }\cos (t)=1\nonumber\]
\[t=\dfrac{\pi }{3}\text{ or }t=\dfrac{5\pi }{3}\text{ or }t=\pi\nonumber\]
Try It \(\PageIndex{15}\)
Solve \(2\sin ^{2} (t)=3\cos (t)\) for all solutions with \(0\le t<2\pi\).
 Answer

\[2\left(1\cos ^{2} (t)\right)=3\cos (t)\nonumber\]
\[2\cos ^{2} (t)+3\cos (t)2=0\nonumber\]
\[\left(2\cos (t)1\right)\left(\cos (t)+2\right)=0\nonumber\]
\(\cos (t)+2=0\) has no solutions \(2\cos (t)1=0\) at \(t=\dfrac{\pi }{3} ,\dfrac{5\pi }{3}\)
In addition to the Pythagorean Identity, it is often necessary to rewrite the tangent, secant, cosecant, and cotangent as part of solving an equation.
Example \(\PageIndex{16}\): Use Quotient Identity
Solve \(\tan (x)=3\sin (x)\) for all solutions with \(0\le x<2\pi\).
Solution
With a combination of tangent and sine, we might try rewriting tangent
\[\tan (x)=3\sin (x)\nonumber\]
\[\dfrac{\sin (x)}{\cos (x)} =3\sin (x)\nonumber\]Multiplying both sides by cosine
\[\sin (x)=3\sin (x)\cos (x) \nonumber\]
At this point, you may be tempted to divide both sides of the equation by sin(\(x\)).
Resist the urge. When we divide both sides of an equation by a quantity, we are assuming the quantity is never zero. In this case, when sin(\(x\)) = 0 the equation is satisfied, so we’d lose those solutions if we divided by the sine.
To avoid this problem, we can rearrange the equation so that one side is zero (You technically can divide by sin(x), as long as you separately consider the case where sin(x) = 0. Since it is easy to forget this step, the factoring approach used in the example is recommended.).
\[\sin (x)3\sin (x)\cos (x)=0\nonumber\]Factoring out sin(\(x\)) from both parts
\[\sin (x)\left(13\cos (x)\right)=0 \nonumber\]
From here, we can see we get solutions when \(\sin (x)=0\) or \(13\cos (x)=0\).
Using our knowledge of the special angles of the unit circle,
\[\sin (x)=0\text{ when }x = 0\text{ or }x = \pi\nonumber\]
For the second equation, we will need the inverse cosine.
\[13\cos (x)=0\nonumber\]
\[\cos (x)=\dfrac{1}{3}\nonumber\]Using our calculator or technology
\[x=\cos ^{1} \left(\dfrac{1}{3} \right)\approx 1.231\nonumber\]Using symmetry to find a second solution
\[x=2\pi 1.231=5.052 \nonumber\]
We have four solutions on \(0 \le x<2\pi\):
\[x = 0, 1.231, \quad\pi , 5.052\nonumber\]
Example \(\PageIndex{17}\): Use Reciprocal Identities
Solve \(\sec (\theta )=2\cos (\theta )\) to find the first four positive solutions.
 Answer

\[\dfrac{1}{\cos (\theta )} =2\cos (\theta )\nonumber\]
\[\dfrac{1}{2} =\cos ^{2} (\theta )\nonumber\]
\[\cos (\theta )=\pm \sqrt{\dfrac{1}{2} } =\pm \dfrac{\sqrt{2} }{2}\nonumber\]
\[\theta =\dfrac{\pi }{4} ,\dfrac{3\pi }{4} ,\dfrac{5\pi }{4} ,\dfrac{7\pi }{4}\nonumber\]
Example \(\PageIndex{18}\): Use Reciprocal Identities
Solve \(\dfrac{4}{\sec ^{2} (\theta )} +3\cos \left(\theta \right)=2\cot \left(\theta \right)\tan \left(\theta \right)\) for all solutions with \(0\le \theta <2\pi\).
Solution
\[\dfrac{4}{\sec ^{2} (\theta )} +3\cos \left(\theta \right)=2\cot \left(\theta \right)\tan \left(\theta \right)\nonumber\] Using the reciprocal identities
\[4\cos ^{2} (\theta )+3\cos (\theta )=2\dfrac{1}{\tan (\theta )} \tan (\theta )\nonumber\] Simplifying
\[4\cos ^{2} \left(\theta \right)+3\cos \left(\theta \right)=2\nonumber\] Subtracting 2 from each side
\[4\cos ^{2} \left(\theta \right)+3\cos \left(\theta \right)2=0\nonumber\]
This does not appear to factor nicely so we use the quadratic formula, remembering that we are solving for cos( \(\theta\)).
\[\cos (\theta )=\dfrac{3\pm \sqrt{3^{2} 4(4)(2)} }{2(4)} =\dfrac{3\pm \sqrt{41} }{8}\nonumber\]
Using the negative square root first,
\[\cos (\theta )=\dfrac{3\sqrt{41} }{8} =1.175\nonumber\]
This has no solutions, since the cosine can’t be less than 1.
Using the positive square root,
\[\cos (\theta )=\dfrac{3+\sqrt{41} }{8} =0.425\nonumber\]
\[\theta =\cos ^{1} \left(0.425\right)=1.131\nonumber\] By symmetry, a second solution can be found
\[\theta =2\pi 1.131=5.152\nonumber\]
Important Topics of This Section
 Review of Trig Identities
 Solving Trig Equations
 By Factoring
 Using the Quadratic Formula
 Utilizing Trig Identities to simplify
Media
Access these online resources for additional instruction and practice with the fundamental trigonometric identities.
 Fundamental Trigonometric Identities
 Verifying Trigonometric Identities
Key Equations
Pythagorean identities 
\({\cos}^2 \theta+{\sin}^2 \theta=1\) \(1+{\cot}^2 \theta={\csc}^2 \theta\) \(1+{\tan}^2 \theta={\sec}^2 \theta\) 
Evenodd identities 
\(\tan(−\theta)=\tan \theta\) \(\cot(\theta)=\cot \theta\) \(\sin(\theta)=\sin \theta\) \(\csc(\theta)=\csc \theta\) \(\cos(\theta)=\cos \theta\) \(\sec(\theta)=\sec \theta\) 
Reciprocal identities 
\(\sin \theta=\dfrac{1}{\csc \theta}\) \(\cos \theta=\dfrac{1}{\sec \theta}\) \(\tan \theta=\dfrac{1}{\cot \theta}\) \(\csc \theta=\dfrac{1}{\sin \theta}\) \(\sec \theta=\dfrac{1}{\cos \theta}\) \(\cot \theta=\dfrac{1}{\tan \theta}\) 
Quotient identities 
\(\tan \theta=\dfrac{\sin \theta}{\cos \theta}\) \(\cot \theta=\dfrac{\cos \theta}{\sin \theta}\) 
Key Concepts
 There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
 Graphing both sides of an identity will verify it. See Example \(\PageIndex{1}\).
 Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example \(\PageIndex{2}\) and Example \(\PageIndex{3}\).
 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 \(\PageIndex{4}\).
 We can create an identity and then verify it. See Example \(\PageIndex{5}\).
 Verifying an identity may involve algebra with the fundamental identities. See Example \(\PageIndex{6}\) and Example \(\PageIndex{7}\).
 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 \(\PageIndex{8}\), Example \(\PageIndex{9}\), and Example \(\PageIndex{10}\).
Contributors and Attributions
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.