7.1: Solving Trigonometric Equations with Identities
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 (Table \(\PageIndex{1}\)), 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.
| \({\sin}^2 \theta+{\cos}^2 \theta=1\) | \(1+{\cot}^2 \theta={\csc}^2 \theta\) | \(1+{\tan}^2 \theta={\sec}^2 \theta\) |
The second and third identities can be obtained by manipulating the first. The identity \(1+{\cot}^2 \theta={\csc}^2 \theta\) is found by rewriting the left side of the equation in terms of sine and cosine.
Prove: \(1+{\cot}^2 \theta={\csc}^2 \theta\)
\[\begin{align*} 1+{\cot}^2 \theta&= (1+\dfrac{{\cos}^2}{{\sin}^2})\qquad \text{Rewrite the left side}\\ &= \left(\dfrac{{\sin}^2}{{\sin}^2}\right)+\left (\dfrac{{\cos}^2}{{\sin}^2}\right)\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\sin}^2+{\cos}^2}{{\sin}^2}\\ &= \dfrac{1}{{\sin}^2}\\ &= {\csc}^2 \end{align*}\]
Similarly,\(1+{\tan}^2 \theta={\sec}^2 \theta\)can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives
\[\begin{align*} 1+{\tan}^2 \theta&= 1+{\left(\dfrac{\sin \theta}{\cos \theta}\right )}^2\qquad \text{Rewrite left side}\\ &= {\left (\dfrac{\cos \theta}{\cos \theta}\right )}^2+{\left (\dfrac{\sin \theta}{\cos \theta}\right)}^2\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\cos}^2 \theta+{\sin}^2 \theta}{{\cos}^2 \theta}\\ &= \dfrac{1}{{\cos}^2 \theta}\\ &= {\sec}^2 \theta \end{align*}\]
Recall that we determined which trigonometric functions are odd and which are even. 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 (Table \(\PageIndex{2}\)).
| \(\tan(−\theta)=−\tan \theta\) | \(\sin(−\theta)=−\sin \theta\) | \(\cos(−\theta)=\cos \theta\) |
| \(\cot(−\theta)=−\cot \theta\) | \(\csc(−\theta)=−\csc \theta\) | \(\sec(−\theta)=\sec \theta\) |
Recall that an odd function is one in which \(f(−x)= −f(x)\) for all \(x\) in the domain off. f. The sine function is an odd function because \(\sin(−\theta)=−\sin \theta\). The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of \(\dfrac{\pi}{2}\) and \(−\dfrac{\pi}{2}\). The output of \(\sin\left (\dfrac{\pi}{2}\right )\) is opposite the output of \(\sin \left (−\dfrac{\pi}{2}\right )\). Thus,
\[\begin{align*} \sin\left (\dfrac{\pi}{2}\right)&=1 \\[4pt] \sin\left (-\dfrac{\pi}{2}\right) &=-\sin\left (\dfrac{\pi}{2}\right) \\[4pt] &=-1 \end{align*}\]
This is shown in Figure \(\PageIndex{2}\).
Recall that an even function is one in which
\(f(−x)=f(x)\) for all \(x\) 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 \(\cos(−\theta)=\cos \theta\). For example, consider corresponding inputs \(\dfrac{\pi}{4}\) and \(−\dfrac{\pi}{4}\). The output of \(\cos\left (\dfrac{\pi}{4}\right)\) is the same as the output of \(\cos\left (−\dfrac{\pi}{4}\right)\). Thus,
\[\begin{align*} \cos\left (−\dfrac{\pi}{4}\right ) &=\cos\left (\dfrac{\pi}{4}\right) \\[4pt] &≈0.707 \end{align*}\]
See Figure \(\PageIndex{3}\).
For all \(\theta\) in the domain of the sine and cosine functions, respectively, we can state the following:
- Since \(\sin(−\theta)=−\sin \theta\),sine is an odd function.
- Since \(\cos(−\theta)=\cos \theta\),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,\(\tan(−\theta)=−\tan \theta\). We can interpret the tangent of a negative angle as
\[\tan (−\theta)=\dfrac{\sin (−\theta)}{\cos (−\theta)}=\dfrac{−\sin \theta}{\cos \theta}=−\tan \theta. \nonumber\]
Tangent is therefore an odd function, which means that \(\tan(−\theta)=−\tan(\theta)\) for all \(\theta\) in the domain of the tangent function.
The cotangent identity, \(\cot(−\theta)=−\cot \theta\),also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as
\[\cot(−\theta)=\dfrac{\cos(−\theta)}{\sin(−\theta)}=\dfrac{\cos \theta}{−\sin \theta}=−\cot \theta.\nonumber\]
Cotangent is therefore an odd function, which means that \(\cot(−\theta)=−\cot(\theta)\) for all \(\theta\) 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
\[\csc(−\theta)=\dfrac{1}{\sin(−\theta)}=\dfrac{1}{−\sin \theta}=−\csc \theta. \nonumber\]
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
\[\sec(−\theta)=\dfrac{1}{\cos(−\theta)}=\dfrac{1}{\cos \theta}=\sec \theta. \nonumber\]
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. (Table \(\PageIndex{3}\)). Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry .
| \(\sin \theta=\dfrac{1}{\csc \theta}\) | \(\csc \theta=\dfrac{1}{\sin \theta}\) |
| \(\cos \theta = \dfrac{1}{\sec \theta}\) | \(\sec \theta=\dfrac{1}{\cos \theta}\) |
| \(\tan \theta=\dfrac{1}{\cot \theta}\) | \(\cot \theta=\dfrac{1}{\tan \theta}\) |
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 (Table \(\PageIndex{4}\)).
| \(\tan \theta=\dfrac{\sin \theta}{\cos \theta}\) | \(\cot \theta=\dfrac{\cos \theta}{\sin \theta}\) |
The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.
The Pythagorean identities are based on the properties of a right triangle.
\[{\cos}^2 \theta+{\sin}^2 \theta=1\]
\[1+{\cot}^2 \theta={\csc}^2 \theta\]
\[1+{\tan}^2 \theta={\sec}^2 \theta\]
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.
\[\tan(−\theta)=−\tan \theta\]
\[\cot(−\theta)=−\cot \theta\]
\[\sin(−\theta)=−\sin \theta\]
\[\csc(−\theta)=−\csc \theta\]
\[\cos(−\theta)=\cos \theta\]
\[\sec(−\theta)=\sec \theta\]
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.
\[\tan \theta=\dfrac{\sin \theta}{\cos \theta}\]
\[\cot \theta=\dfrac{\cos \theta}{\sin \theta}\]
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.
- 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.
Verify \(\tan \theta \cos \theta=\sin \theta\).
Solution
We will start on the left side, as it is the more complicated side:
\[ \begin{align*} \tan \theta \cos \theta &=\left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta \\[4pt] &=\sin \theta \end{align*}\]
Analysis
This identity was fairly simple to verify, as it only required writing \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).
Verify the following equivalency using the even-odd identities:
\((1+\sin x)[1+\sin(−x)]={\cos}^2 x\)
Solution
Working on the left side of the equation, we have
\( (1+\sin x)[1+\sin(−x)]=(1+\sin x)(1-\sin x)\)
Since
\[\begin{align*} \sin(-x)&= -\sin x \\ [5pt] &=1-{\sin}^2 x\qquad \text{Difference of squares} \\ [5pt] &={\cos}^2 x \\ {\cos}^2 x&= 1-{\sin}^2 x \\ \end{align*}\]