# 3.1: Solving Trigonometric Equations with 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 (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 heta+{cos}^2 heta=1) (1+{cot}^2 heta={csc}^2 heta) (1+{ an}^2 heta={sec}^2 heta)

The second and third identities can be obtained by manipulating the first. The identity (1+{cot}^2 heta={csc}^2 heta) is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: (1+{cot}^2 heta={csc}^2 heta)

[egin{align*} 1+{cot}^2 heta&= (1+dfrac{{cos}^2}{{sin}^2})qquad ext{Rewrite the left side} &= left(dfrac{{sin}^2}{{sin}^2} ight)+left (dfrac{{cos}^2}{{sin}^2} ight)qquad ext{Write both terms with the common denominator} &= dfrac{{sin}^2+{cos}^2}{{sin}^2} &= dfrac{1}{{sin}^2} &= {csc}^2 end{align*}]

Similarly,(1+{ an}^2 heta={sec}^2 heta)can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

[egin{align*} 1+{ an}^2 heta&= 1+{left(dfrac{sin heta}{cos heta} ight )}^2qquad ext{Rewrite left side} &= {left (dfrac{cos heta}{cos heta} ight )}^2+{left (dfrac{sin heta}{cos heta} ight)}^2qquad ext{Write both terms with the common denominator} &= dfrac{{cos}^2 heta+{sin}^2 heta}{{cos}^2 heta} &= dfrac{1}{{cos}^2 heta} &= {sec}^2 heta 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})).

 ( an(− heta)=− an heta) (sin(− heta)=−sin heta) (cos(− heta)=cos heta) (cot(− heta)=−cot heta) (csc(− heta)=−csc heta) (sec(− heta)=sec heta)

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(− heta)=−sin heta). 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 (sinleft (dfrac{pi}{2} ight )) is opposite the output of (sin left (−dfrac{pi}{2} ight )). Thus,

[egin{align*} sinleft (dfrac{pi}{2} ight)&=1 [4pt] sinleft (-dfrac{pi}{2} ight) &=-sinleft (dfrac{pi}{2} ight) [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(− heta)=cos heta). For example, consider corresponding inputs (dfrac{pi}{4}) and (−dfrac{pi}{4}). The output of (cosleft (dfrac{pi}{4} ight)) is the same as the output of (cosleft (−dfrac{pi}{4} ight)). Thus,

[egin{align*} cosleft (−dfrac{pi}{4} ight ) &=cosleft (dfrac{pi}{4} ight) [4pt] &≈0.707 end{align*}]

See Figure (PageIndex{3}).

For all ( heta) in the domain of the sine and cosine functions, respectively, we can state the following:

• Since (sin(− heta)=−sin heta),sine is an odd function.
• Since (cos(− heta)=cos heta),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,( an(− heta)=− an heta). We can interpret the tangent of a negative angle as

[ an (− heta)=dfrac{sin (− heta)}{cos (− heta)}=dfrac{−sin heta}{cos heta}=− an heta. onumber]

Tangent is therefore an odd function, which means that ( an(− heta)=− an( heta)) for all ( heta) in the domain of the tangent function.

The cotangent identity, (cot(− heta)=−cot heta),also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as

[cot(− heta)=dfrac{cos(− heta)}{sin(− heta)}=dfrac{cos heta}{−sin heta}=−cot heta. onumber]

Cotangent is therefore an odd function, which means that (cot(− heta)=−cot( heta)) for all ( heta) 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(− heta)=dfrac{1}{sin(− heta)}=dfrac{1}{−sin heta}=−csc heta. onumber]

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(− heta)=dfrac{1}{cos(− heta)}=dfrac{1}{cos heta}=sec heta. onumber]

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 heta=dfrac{1}{csc heta}) (csc heta=dfrac{1}{sin heta}) (cos heta = dfrac{1}{sec heta}) (sec heta=dfrac{1}{cos heta}) ( an heta=dfrac{1}{cot heta}) (cot heta=dfrac{1}{ an heta})

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})).

 ( an heta=dfrac{sin heta}{cos heta}) (cot heta=dfrac{cos heta}{sin heta})

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.

[{cos}^2 heta+{sin}^2 heta=1]

[1+{cot}^2 heta={csc}^2 heta]

[1+{ an}^2 heta={sec}^2 heta]

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.

[ an(− heta)=− an heta]

[cot(− heta)=−cot heta]

[sin(− heta)=−sin heta]

[csc(− heta)=−csc heta]

[cos(− heta)=cos heta]

[sec(− heta)=sec heta]

The reciprocal identities define reciprocals of the trigonometric functions.

[sin heta=dfrac{1}{csc heta}]

[cos heta=dfrac{1}{sec heta}]

[ an heta=dfrac{1}{cot heta}]

[csc heta=dfrac{1}{sin heta}]

[sec heta=dfrac{1}{cos heta}]

[cot heta=dfrac{1}{ an heta}]

The quotient identities define the relationship among the trigonometric functions.

[ an heta=dfrac{sin heta}{cos heta}]

[cot heta=dfrac{cos heta}{sin heta}]

Example (PageIndex{1}): Graphing the Equations of an Identity

Graph both sides of the identity (cot heta=dfrac{1}{ an heta}). In other words, on the graphing calculator, graph (y=cot heta) and (y=dfrac{1}{ an heta}).

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.

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 (PageIndex{2}): Verifying a Trigonometric Identity

Verify ( an heta cos heta=sin heta).

Solution

We will start on the left side, as it is the more complicated side:

[ egin{align*} an heta cos heta &=left(dfrac{sin heta}{cos heta} ight)cos heta [4pt] &=sin heta end{align*}]

Analysis

This identity was fairly simple to verify, as it only required writing ( an heta) in terms of (sin heta) and (cos heta).

Exercise (PageIndex{1})

Verify the identity (csc heta cos heta an heta=1).

[ egin{align*} csc heta cos heta an heta=left(dfrac{1}{sin heta} ight)cos hetaleft(dfrac{sin heta}{cos heta} ight) [4pt] & =dfrac{cos heta}{sin heta}(dfrac{sin heta}{cos heta}) [4pt] & =dfrac{sin heta cos heta}{sin heta cos heta} [4pt] &=1 end{align*}]

Example (PageIndex{3A}): Verifying the Equivalency Using the Even-Odd Identities

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

[egin{align*} sin(-x)&= -sin x [5pt] &=1-{sin}^2 xqquad ext{Difference of squares} [5pt] &={cos}^2 x {cos}^2 x&= 1-{sin}^2 x end{align*}]

Example (PageIndex{3B}): Verifying a Trigonometric Identity Involving ({sec}^2 heta)

Verify the identity (dfrac{{sec}^2 heta−1}{{sec}^2 heta}={sin}^2 heta)

Solution

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

[egin{align*}
dfrac{{sec}^2 heta-1}{{sec}^2 heta}&= dfrac{({ an}^2 heta +1)-1}{{sec}^2 heta}
{sec}^2 heta&= { an}^2 heta +1
&= dfrac{{ an}^2 heta}{{sec}^2 heta}
&= { an}^2 hetaleft (dfrac{1}{{sec}^2 heta} ight )
&= { an}^2 heta left ({cos}^2 heta ight )
{cos}^2 heta&= dfrac{1}{{sec}^2 heta}
&= left (dfrac{{sin}^2 heta}{{cos}^2 heta} ight )
{ an}^2 heta&= dfrac{{sin}^2 heta}{{cos}^2 heta}
&= {sin}^2 heta
end{align*}]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

[egin{align*} dfrac{{sec}^2 heta-1}{{sec}^2 heta}&= dfrac{{sec}^2 heta}{{sec}^2 heta}-dfrac{1}{{sec}^2 heta} &= 1-{cos}^2 heta &= {sin}^2 heta end{align*}]

Analysis

In the first method, we used the identity ({sec}^2 heta={ an}^2 heta+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.

Exercise (PageIndex{2})

Show that (dfrac{cot heta}{csc heta}=cos heta).

[egin{align*} dfrac{cot heta}{csc heta}&= dfrac{ frac{cos heta}{sin heta}}{dfrac{1}{sin heta}} &= dfrac{cos heta}{sin heta}cdot dfrac{sin heta}{1} &= cos heta end{align*}]

Example (PageIndex{4}): Creating and Verifying an Identity

Create an identity for the expression (2 an heta sec heta) 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:

[egin{align*} 2 an heta sec heta&= 2left (dfrac{sin heta}{cos heta} ight )left(dfrac{1}{cos heta} ight ) &= dfrac{2sin heta}{{cos}^2 heta} &= dfrac{2sin heta}{1-{sin}^2 heta}qquad ext{Substitute } 1-{sin}^2 heta ext{ for } {cos}^2 heta end{align*}]

Thus,

(2 an heta sec heta=dfrac{2 sin heta}{1−{sin}^2 heta})

Example (PageIndex{5}): Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

(dfrac{{sin}^2(− heta)−{cos}^2(− heta)}{sin(− heta)−cos(− heta)}=cos heta−sin heta)

Solution

[egin{align*} dfrac{{sin}^2(- heta)-{cos}^2(- heta)}{sin(- heta)-cos(- heta)}&= dfrac{{[sin(- heta)]}^2-{[cos(- heta)]}^2}{sin(- heta)-cos(- heta)} &= dfrac{{(-sin heta)}^2-{(cos heta)}^2}{-sin heta -cos heta} ;; ; , sin(-x) = -sinspace x ext { and } cos(-x)=cos space x &= dfrac{{(sin heta)}^2-{(cos heta)}^2}{-sin heta -cos heta}qquad ext{Difference of squares} &= dfrac{(sin heta-cos heta)(sin heta+cos heta)}{-(sin heta+cos heta)} &= cos heta-sin heta end{align*}]

Exercise (PageIndex{3})

Verify the identity (dfrac{{sin}^2 heta−1}{ an heta sin heta− an heta}=dfrac{sin heta+1}{ an heta}).

[egin{align*} dfrac{{sin}^2 heta-1}{ an heta sin heta- an heta}&= dfrac{(sin heta +1)(sin heta -1)}{ an heta(sin heta -1)} &= dfrac{sin heta+1}{ an heta} end{align*}]

Example (PageIndex{6}): Verifying an Identity Involving Cosines and Cotangents

Verify the identity: ((1−{cos}^2 x)(1+{cot}^2 x)=1).

Solution

[egin{align*} (1-{cos}^2 x)(1+{cot}^2 x)&= (1-{cos}^2 x)left(1+dfrac{{cos}^2 x}{{sin}^2 x} ight) &= (1-{cos}^2 x)left(dfrac{{sin}^2 x}{{sin}^2 x}+dfrac{{cos}^2 x}{{sin}^2 x} ight )qquad ext{Find the common denominator} &= (1-{cos}^2 x)left(dfrac{{sin}^2 x +{cos}^2 x}{{sin}^2 x} ight) &= ({sin}^2 x)left (dfrac{1}{{sin}^2 x} ight ) &= 1 end{align*}]

## 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 ((sin x+1)(sin x−1)=0) resembles the equation ((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, (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 (PageIndex{7A}): Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: (2{cos}^2 heta+cos heta−1).

Solution

Notice that the pattern displayed has the same form as a standard quadratic expression,(ax^2+bx+c). Letting (cos heta=x),we can rewrite the expression as follows:

(2x^2+x−1)

This expression can be factored as ((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 (x). At this point, we would replace (x) with (cos heta) and solve for ( heta).

Example (PageIndex{7B}): Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression using the difference of squares: (4{cos}^2 heta−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.

[egin{align*} 4{cos}^2 heta-1&= {(2cos heta)}^2-1 &= (2cos heta-1)(2cos heta+1) end{align*}]

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 (cos heta=x), rewrite the expression as (4x^2−1), and factor ((2x−1)(2x+1)). Then replace (x) with (cos heta) and solve for the angle.

Exercise (PageIndex{4})

Rewrite the trigonometric expression using the difference of squares: (25−9{sin}^2 heta).

This is a difference of squares formula: (25−9{sin}^2 heta=(5−3sin heta)(5+3sin heta)).

Example (PageIndex{8}): Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

({csc}^2 heta−{cot}^2 heta)

Solution

[egin{align*} 1+{cot}^2 heta&= {csc}^2 heta ext{Now we can simplify by substituting } 1+{cot}^2 heta ext{ for } {csc}^2 heta {csc}^2 heta-{cot}^2 heta&= 1+{cot}^2 heta-{cot}^2 heta &= 1 end{align*}]

Exercise (PageIndex{5})

Use algebraic techniques to verify the identity: (dfrac{cos heta}{1+sin heta}=dfrac{1−sin heta}{cos heta}).

(Hint: Multiply the numerator and denominator on the left side by (1−sin heta).)

[egin{align*} dfrac{cos heta}{1+sin heta}left(dfrac{1-sin heta}{1-sin heta} ight)&= dfrac{cos heta (1-sin heta)}{1-{sin}^2 heta} &= dfrac{cos heta (1-sin heta)}{{cos}^2 heta} &= dfrac{1-sin heta}{cos heta} end{align*}]

Media

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

• Fundamental Trigonometric Identities
• Verifying Trigonometric Identities

## 3.1: Solving Trigonometric Equations with Identities

1. Without using a calculator find all the solutions to (4sin left( <3t> ight) = 2).

Show All Steps Hide All Steps

Isolating the sine (with a coefficient of one) on one side of the equation gives,

Because we’re dealing with sine in this problem and we know that the (y)-axis represents sine on a unit circle we’re looking for angles that will have a (y) coordinate of (< extstyle<1 over 2>>). This means we’ll have an angle in the first quadrant and an angle in the second quadrant (that we can use the angle in the first quadrant to find). Here is a unit circle for this situation.

Clearly the angle in the first quadrant is (frac<6>) and by some basic symmetry we can see that the terminal line for the second angle must form an angle of (frac<6>) with the negative (x)-axis as shown above and so it will be : (pi - frac <6>= frac<<5pi >><6>).

From the discussion in the notes for this section we know that once we have these two angles we can get all possible angles by simply adding “( + ,2pi n) for (n = 0, pm 1, pm 2, ldots )” onto each of these.

This then means that we must have,

[3t = frac <6>+ 2pi n hspace<0.5in>< m>hspace<0.5in>3t = frac<<5pi >> <6>+ 2pi n hspace <0.5in>n = 0, pm 1, pm 2, ldots ]

Finally, to get all the solutions to the equation all we need to do is divide both sides by 3.

Understanding the question and drawing the appropriate diagram are the two most important things to be done in solving word problems in trigonometry.

If it is possible, we have to split the given information. Because, when we split the given information in to parts, we can understand them easily.

We have to draw diagram almost for all of the word problems in trigonometry. The diagram we draw for the given information must be correct. Drawing diagram for the given information will give us a clear understanding about the question.

Once we understand the given information clearly and correct diagram is drawn, solving word problems in trigonometry would not be a challenging work.

After having drawn the appropriate diagram based on the given information, we have to give name for each position of the diagram using English alphabets (it is clearly shown in the word problem given below). Giving name for the positions would be easier for us to identify the parts  of the diagram.

Now we have to use one of the three trigonometric ratios (sin, cos and tan) to find the unknown side or angle.

Once the diagram is drawn and we have translated the English Statement (information) given in the question as mathematical equation using trigonometric ratios correctly, 90% of the work will be over. The remaining 10% is just getting the answer. That is solving for the unknown.

These are the most commonly steps involved in solving word problems in trigonometry.

Apart from the stuff given in this section ,    if you need any other stuff in math, please use our google custom search here.

If you have any feedback about our math content, please mail us :

You can also visit the following web pages on different stuff in math.

## 55 Parametric Equations

Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in (Figure). At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time.

In this section, we will consider sets of equations given by and where is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values ofthe orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations : we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations.

### Parameterizing a Curve

When an object moves along a curve—or curvilinear path —in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, bothand vary over time and so are functions of time. For this reason, we add another variable, the parameter , upon which bothandare dependent functions. In the example in the section opener, the parameter is time,Theposition of the moon at time,is represented as the functionand theposition of the moon at time,is represented as the functionTogether, and are called parametric equations, and generate an ordered pairParametric equations primarily describe motion and direction.

When we parameterize a curve, we are translating a single equation in two variables, such asandinto an equivalent pair of equations in three variables,andOne of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time.

When we graph parametric equations, we can observe the individual behaviors ofand ofThere are a number of shapes that cannot be represented in the formmeaning that they are not functions. For example, consider the graph of a circle, given asSolving forgivesor two equations:andIf we graphandtogether, the graph will not pass the vertical line test, as shown in (Figure). Thus, the equation for the graph of a circle is not a function.

However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.

Supposeis a number on an interval,The set of ordered pairs,whereandforms a plane curve based on the parameterThe equationsandare the parametric equations.

Parameterize the curvelettingGraph both equations.

Ifthen to findwe replace the variablewith the expression given inIn other words, Make a table of values similar to (Figure), and sketch the graph.

See the graphs in (Figure). It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated asincreases.

(a) Parametric(b) Rectangular

The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of

Construct a table of values and plot the parametric equations:

Find a pair of parametric equations that models the graph ofusing the parameterPlot some points and sketch the graph.

Ifand we substituteforinto theequation, thenOur pair of parametric equations is

To graph the equations, first we construct a table of values like that in (Figure). We can choose values aroundfromtoThe values in thecolumn will be the same as those in thecolumn becauseCalculate values for the column

The graph ofis a parabola facing downward, as shown in (Figure). We have mapped the curve over the interval shown as a solid line with arrows indicating the orientation of the curve according toOrientation refers to the path traced along the curve in terms of increasing values ofAs this parabola is symmetric with respect to the linethe values ofare reflected across the y-axis.

Parameterize the curve given by

An object travels at a steady rate along a straight path toin the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.

The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x-value of the object starts atmeters and goes to 3 meters. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of orWe can write the x-coordinate as a linear function with respect to time asIn the linear function templateand

Similarly, the y-value of the object starts at 3 and goes towhich is a change in the distance y of −4 meters in 4 seconds, which is a rate of orWe can also write the y-coordinate as the linear functionTogether, these are the parametric equations for the position of the object, where and are expressed in meters and represents time:

Using these equations, we can build a table of values for and (see (Figure)). In this example, we limited values ofto non-negative numbers. In general, any value ofcan be used.

From this table, we can create three graphs, as shown in (Figure).

(a) A graph ofvs.representing the horizontal position over time. (b) A graph of vs. representing the vertical position over time. (c) A graph of vs. representing the position of the object in the plane at time

Again, we see that, in (Figure)(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows.

### Eliminating the Parameter

In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such asandEliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameterfrom a set of parametric equations not every method works for every type of equation. Here we will review the methods for the most common types of equations.

#### Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations

For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve forWe substitute the resulting expression for into the second equation. This gives one equation inand

Givenandeliminate the parameter, and write the parametric equations as a Cartesian equation.

We will begin with the equation forbecause the linear equation is easier to solve for

Next, substituteforin

The Cartesian form is

This is an equation for a parabola in which, in rectangular terms,is dependent onFrom the curve’s vertex atthe graph sweeps out to the right. See (Figure). In this section, we consider sets of equations given by the functionsandwhereis the independent variable of time. Notice, bothandare functions of time so in generalis not a function of

Given the equations below, eliminate the parameter and write as a rectangular equation foras a function
of

Eliminate the parameter and write as a Cartesian equation: and

Isolate

Substitute the expression into

The Cartesian form is

The graph of the parametric equation is shown in (Figure)(a). The domain is restricted toThe Cartesian equation,is shown in (Figure)(b) and has only one restriction on the domain,

Eliminate the parameter and write as a Cartesian equation:and

Solve the first equation for

Then, substitute the expression for into the equation.

The Cartesian form is

To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain onto we restrict the domain ontoThe domain for the parametric equationis restricted to we limit the domain onto

Eliminate the parameter and write as a rectangular equation .

#### Eliminating the Parameter from Trigonometric Equations

Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.

First, we use the identities:

Solving forandwe have

Then, use the Pythagorean Theorem:

Eliminate the parameter from the given pair of trigonometric equations whereand sketch the graph.

Solving forand we have

Next, use the Pythagorean identity and make the substitutions.

The graph for the equation is shown in (Figure).

Applying the general equations for conic sections (introduced in Analytic Geometry, we can identifyas an ellipse centered atNotice that whenthe coordinates areand whenthe coordinates areThis shows the orientation of the curve with increasing values of

Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation:and

### Finding Cartesian Equations from Curves Defined Parametrically

When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such asIn this case, can be any expression. For example, consider the following pair of equations.

Rewriting this set of parametric equations is a matter of substitutingforThus, the Cartesian equation is

Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.

Method 1. First, let’s solve theequation forThen we can substitute the result into the equation.

Now substitute the expression forinto theequation.

Method 2. Solve theequation forand substitute this expression in theequation.

Make the substitution and then solve for

Write the given parametric equations as a Cartesian equation: and

### Finding Parametric Equations for Curves Defined by Rectangular Equations

Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to representand then substitute it into theequation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values foras the domain of the rectangular equation, then the graphs will be different.

Find a set of equivalent parametric equations for

An obvious choice would be to letThen But let’s try something more interesting. What if we letThen we have

The set of parametric equations is

Access these online resources for additional instruction and practice with parametric equations.

### Key Concepts

• Parameterizing a curve involves translating a rectangular equation in two variables,andinto two equations in three variables, x, y, and t. Often, more information is obtained from a set of parametric equations. See (Figure), (Figure), and (Figure).
• Sometimes equations are simpler to graph when written in rectangular form. By eliminatingan equation inandis the result.
• To eliminatesolve one of the equations forand substitute the expression into the second equation. See (Figure), (Figure), (Figure), and (Figure).
• Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve forin one of the equations, and substitute the expression into the second equation. See (Figure).
• There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.
• Find an expression forsuch that the domain of the set of parametric equations remains the same as the original rectangular equation. See (Figure).

### Section Exercises

#### Verbal

What is a system of parametric equations?

A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example,and

Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter.

Explain how to eliminate a parameter given a set of parametric equations.

Choose one equation to solve forsubstitute into the other equation and simplify.

What is a benefit of writing a system of parametric equations as a Cartesian equation?

What is a benefit of using parametric equations?

Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.

Why are there many sets of parametric equations to represent on Cartesian function?

#### Algebraic

For the following exercises, eliminate the parameterto rewrite the parametric equation as a Cartesian equation.

or

For the following exercises, rewrite the parametric equation as a Cartesian equation by building an table.

For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting or by setting

For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using andIdentify the curve.

Ellipse

Circle

Parameterize the line fromtoso that the line is atatand atat

Parameterize the line fromtoso that the line is atatand atat

Parameterize the line fromtoso that the line is atatand atat

Parameterize the line fromtoso that the line is atatand atat

#### Technology

For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect.

yes, at

For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations.

–1
0
1

1
2
3

1 -3 1
2 0 7
3 5 17

-1
0
1
2

#### Extensions

Find two different sets of parametric equations for

Find two different sets of parametric equations for

Find two different sets of parametric equations for

### Glossary

parameter a variable, often representing time, upon whichandare both dependent

## Solving Trigonometric Equations Worksheets

Boost learning parameters with these printable solving trigonometric equation worksheets featuring myriad exercises to solve trig equations in linear and quadratic forms by factoring or by using quadratic formulas. Learn to determine the principal solution of the given trigonometric equations as well. As a precursor to these pdf exercises, the high school students can recapitulate the trigonometric identities charts. Plunge into practice with our free worksheets!

Access these trigonometric worksheets to solve simple trigonometric equations. Each of these level 1 worksheets features trigonometric functions with special angles either in degree or in radians.

Reinforce the concept of solving trigonometric equations by evaluating these equations that include two or more trig functions.

This batch of printable worksheets presents equations with a single trigonometric function. Solve the linear trigonometric equation and obtain principal solutions that lie in the given range.

Modify the trigonometric equation by applying the standard trigonometric identities and express it in terms of a single trig function. Solve the equation and find out the principal solution of the given equation.

This array of trigonometric worksheet pdfs for high school depicts trigonometric equations in the quadratic form. Factorize the expression by combining your algebraic skills and trigonometric identities and then solve the equation.

## Complex Trigonometric Functions

The trigonometric functions are most simply defined using the right triangle. But the right triangle definitions only define the trigonometric functions for angles between 0 and radians.

The six trigonometric functions can also be defined in terms of the unit circle (the circle of radius one centered at the origin). The unit circle definition permits the definition of the trigonometric functions for all positive and negative arguments.

1.2 Definitions via series

The complex trigonometric functions can be represented by the power series:
Sin( z ) =
Cos( z ) =

Other complex trigonometric functions are:
tan( z ) =
cot( z ) =
sec( z ) =
csc( z ) =

1.3 Definitions via complex exponentials

The complex trigonometric functions can be defined algebraically in terms of complex exponentials as:

1.4 Definitions via differential equations

Both the sine and cosine functions satisfy the differential equation

The sine function is the unique solution satisfying the initial conditions y(0) = 0 and y'(0) = 1

The cosine function is the unique solution satisfying the initial conditions y(0) = 1 and y'(0) = 0

The tangent function is the unique solution of the nonlinear differential equation

satisfying the initial conditions y(0) = 0

Then y ' = sec( z ) 2 y 2 = tan( z ) 2

It is an Pythagorean identity.

2) A difference between the real and complex trigonometric functions

There is a big difference between the real and complex trigonometric functions:

1) The real trigonometric functions are not related to the exponential function. But complex trigonometric functions do have Relationship to exponential function

2) The real sine and cosine functions are bounded:
|Sin( x )| 1, |Cos( x )| 1

The complex sine and cosine functions are not bounded if they are defined over the set of all complex numbers.

2.1 Relationship to exponential function

The main distinction between real and complex trigonometric functions is relationship to exponential function.

The complex trigonometric functions can be defined algebraically in terms of complex exponentials as:

It can be shown from the series definitions that the sine and cosine functions are the imaginary and real parts, respectively, of the complex exponential function when its argument is purely imaginary:

This relationship was first noted by Euler and the identity is called Euler's formula.

The relationship between the complex exponential and the trigonometric functions can be expressed as:

2.2 The complex sine and cosine functions are not bounded

There is another distinction between real and complex trigonometric functions. In a case of complex variables

|Sin( z )| 1 and |Cos( z )| 1 are not true .

For example

But the Pythagorean identity is true.

3) Identities

3.1 Periodic identities

All complex trigonometric functions are periodic functions with the same periods as trigonometric function for real variables.
The sine, cosine, secant, and cosecant functions have period :

The tangent and cotangent functions have period :

3.2 Even and odd functions

Cos( z ) is an even function, Sin( z ) is an odd function as trigonometric functions for real variables.

3.3 Pythagorean identities

These identities are based on the Pythagorean theorem.

The second equation is obtained from the first by dividing both sides by .

The third equation is obtained from the first by dividing both sides by .

3.4 The sum and difference formulas

Other key relationships are the sum and difference formulas, which give the sine and cosine of the sum and difference of two angles in terms of sines and cosines of the angles themselves.

Sin( z 1 + z 2 ) = Sin( z 1 )Cos( z 2 ) + Cos( z 1 )Sin( z 2 )

Sin( z 1 - z 2 ) = Sin( z 1 )Cos( z 2 ) - Cos( z 1 )Sin( z 2 )

Cos( z 1 + z 2 ) = Cos( z 1 )Cos( z 2 ) - Sin( z 1 )Sin( z 2 )

Cos( z 1 - z 2 ) = Cos( z 1 )Cos( z 2 ) + Sin( z 1 )Sin( z 2 )

When the two angles are equal, the sum formulas reduce to simpler equations known as the double-angle formulas.

3.5 The double-angle formulas

3.6 More identities

The complex modulus satisfies modulus identity:

If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term.

## Using Trigonometric Identities - Concept

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

When simplifying problems that have reciprocal trig functions, start by substituting in the identities for each. If possible, write tangent in terms of sine and cosine. Use algebra to eliminate any complex fractions, factor, or cancel common terms. When using trigonometric identities, make one side of the equation look like the other or work on both sides of the equation to arrive at an identity (like 1=1).

Trigonometric identities are really important as you've probably already seen if you've used them to graph. One of the things that you'll find in your homework occasionally is problems that ask you to simplify a trig expression.
Well let's recall some basic trig identities that will be need right now. These are some of the reciprocal identities. Secant theta equals 1 over cosine theta, cosecant theta is 1 over sine theta and tangent theta is sine theta over cosine theta. So let's use that to simplify secant theta plus 1 over sine theta plus tangent theta. I'll write that over here. Secant theta plus 1, sine theta plus tangent theta.
A lot of times when you're dealing with um some of the reciprocal trig functions, the best strategy is to switch to sines and cosines. So let me do that first. I'll replace secant theta with 1 over cosine theta. And I'll replace tangent theta with sine theta over cosine theta.
Now, at first glance, this is actually a little more complicated looking than we started but now that we've got sines and cosines we can simplify this using Algebra. Notice that I have a complex fraction here and the denominators in the little fractions are cosine theta so I can get rid of them by multiplying the top and bottom by cosine theta. Now remember when I do this, this cosine theta is going to be multiplied across both of these terms and this cosine theta will be multiplied across both of these terms. So in the numerator, I get cosine times 1 over cosine is 1. Cosine times 1 is cosine. Cosine times sine is sine cosine and sine over cosine times cosine the cosine's cancel and you get sine. And this is really good because notice, I can factor a sine theta out of the bottom and I'm left with cosine theta plus 1 which is exactly what I have in the numerator. And so these things cancel. And I get 1 over sine theta.
Now when you're simplifying a trigonometric expression or any expression, it's best to have as few operations and as few functions in your final answer as possible. I still have the operation of division and I can get rid of that if I just replace this with cosecant theta. So this whole big expression reduces down to the simple cosecant theta. So that's what we mean by simplifying a trigonometric expression.
Don't forget. If you're dealing with reciprocal trig functions like secant tangent, sometimes it's best to switch to sine and cosine. And don't forget this trick, when you have complex fractions, you can do what we call fraction busting. Multiply by the common denominator of the little fractions.

## Study Support: Mathematics

Print Materials to assist with any pre-requisite mathematics required in USQ courses. These materials are based on USQ's Tertiary Preparation Program. Mathematics modules are presented in increasing level of difficulty and complexity from Level A through to Level D.

Comparing numbers (part of module only)

• Comparing quantities of subtraction
• Comparing quantities by division (including percentages)
• Ratios (including ratios in squares, rectangles, circles & triangles)
• Rates

Trigonometry (Level B)

• Sine (including the sine ratio, the sine function, the inverse of the sine function, degrees, minutes and seconds, amplitude and period)
• Cosine (including the cosine ratio, the cosine function, the inverse of the cosine function, amplitude and period)
• Tangent (including the tangent ratio, the tangent function, the inverse of the tangent function, amplitude and period)

Trigonometric functions (Level C)

• Graphs of sine, cosine, and tangent functions
• Modelling using trigonometric functions (including amplitude, vertical shift, the period of trigonometric functions, the phase of trigonometric functions)
• Inverse functions
• Solving trigonometric equations

Trigonometry (Level D)

• Polar Co-Ordinates
• Trigonometric Identities and Multiple Angle Formulae
• Solving Trigonometric Equation
• Periodicity
• Amplitude
• Triangle Solution
• Compound Angles
• Solving Equations Involving Trigonometric Functions

### Video based materials

Introduction to trigonometry (printable handout)

This 20 minute video introduces:

• labeling triangles for trigonometry
• Pythagoras' Theorem
• trigonometric ratios
• worked examples using trigonometric ratios

Solution to triangles - using the Sine and Cosine Rules

This 18 minute video shows how to use the sine and cosine rules (that is, when you do not have right angled triangles.

Using Trigonometric identities

This 11 minute video has a number of worked examples of how to use the trigonometric identities.

You can also get help Mastering your calculator – free, easy to follow online booklets on a range of scientific calculators.

## 3.1: Solving Trigonometric Equations with Identities

On problems 1.) through 8.) find answers WITHOUT using the chain rule.

Some of the following problems require use of the chain rule.

A trigonometric function is appeared with another trigonometric function in division form in some cases but it is not always possible to divide a trigonometric function by another trigonometric function. However, there are two possible cases in which the quotient of two trigonometric functions is also a trigonometric function.

The two possible cases are used as formulas in trigonometry. They are called the quotient trigonometric identities and simply called as quotient identities.

When the angle of a right triangle is represented by theta. The sine, cosine, tangent and cotangent functions are written as $sin< heta>$, $cos< heta>$, $an< heta>$ and $cot< heta>$ respectively.

### Sine by Cosine Identity

The quotient of sine by cosine at an angle is equal to the tangent at that angle.

### Cosine by Sine Identity

The quotient of cosine by sine at an angle is equal to the cotangent of that angle.