# 3.5: Solving Trigonometric Equations

Learning Objectives

• Solve linear trigonometric equations in sine and cosine.
• Solve equations involving a single trigonometric function.
• Solve trigonometric equations using a calculator.
• Solve trigonometric equations that are quadratic in form.
• Solve trigonometric equations using fundamental identities.
• Solve trigonometric equations with multiple angles.
• Solve right triangle problems.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In earlier sections of this chapter, we looked at trigonometric identities. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

## Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is (2pi). In other words, every (2pi) units, the y-values repeat. If we need to find all possible solutions, then we must add (2pi k),where (k) is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is (2pi):

[sin heta=sin( heta pm 2kpi)]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example (PageIndex{1A}): Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation (cos heta=dfrac{1}{2}).

Solution

From the unit circle, we know that

[ egin{align*} cos heta &=dfrac{1}{2} [4pt] heta &=dfrac{pi}{3},space dfrac{5pi}{3} end{align*}]

These are the solutions in the interval ([ 0,2pi ]). All possible solutions are given by

where (k) is an integer.

Example (PageIndex{1B}): Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation (sin t=dfrac{1}{2}).

Solution

Solving for all possible values of (t) means that solutions include angles beyond the period of (2pi). From the section on Sum and Difference Identities, we can see that the solutions are (t=dfrac{pi}{6}) and (t=dfrac{5pi}{6}). But the problem is asking for all possible values that solve the equation. Therefore, the answer is

where (k) is an integer.

Howto: Given a trigonometric equation, solve using algebra

1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
2. Substitute the trigonometric expression with a single variable, such as (x) or (u).
3. Solve the equation the same way an algebraic equation would be solved.
4. Substitute the trigonometric expression back in for the variable in the resulting expressions.
5. Solve for the angle.

Example (PageIndex{2}): Solve the Linear Trigonometric Equation

Solve the equation exactly: (2 cos heta−3=−5), (0≤ heta<2pi).

Solution

Use algebraic techniques to solve the equation.

[egin{align*} 2 cos heta-3&= -5 2 cos heta&= -2 cos heta&= -1 heta&= pi end{align*}]

Exercise (PageIndex{2})

Solve exactly the following linear equation on the interval ([0,2pi)): (2 sin x+1=0).

(x=dfrac{7pi}{6},space dfrac{11pi}{6})

## Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is (pi),not (2pi). Further, the domain of tangent is all real numbers with the exception of odd integer multiples of (dfrac{pi}{2}),unless, of course, a problem places its own restrictions on the domain.

Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: (2 {sin}^2 heta−1=0), (0≤ heta<2pi).

Solution

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate (sin heta). Then we will find the angles.

[egin{align*}
2 {sin}^2 heta-1&= 0
2 {sin}^2 heta&= 1
{sin}^2 heta&= dfrac{1}{2}
sqrt{ {sin}^2 heta }&= pm sqrt{ dfrac{1}{2} }
sin heta&= pm dfrac{1}{sqrt{2}}
&= pm dfrac{sqrt{2}}{2}
heta&= dfrac{pi}{4}, space dfrac{3pi}{4},space dfrac{5pi}{4}, space dfrac{7pi}{4}
end{align*}]

Example (PageIndex{3B}): Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: (csc heta=−2), (0≤ heta<4pi).

Solution

We want all values of ( heta) for which (csc heta=−2) over the interval (0≤ heta<4pi).

[egin{align*} csc heta&= -2 dfrac{1}{sin heta}&= -2 sin heta&= -dfrac{1}{2} heta&= dfrac{7pi}{6},space dfrac{11pi}{6},space dfrac{19pi}{6}, space dfrac{23pi}{6} end{align*}]

Analysis

As (sin heta=−dfrac{1}{2}), notice that all four solutions are in the third and fourth quadrants.

Example (PageIndex{3C}): Solving an Equation Involving Tangent

Solve the equation exactly: ( anleft( heta−dfrac{pi}{2} ight)=1), (0≤ heta<2pi).

Solution

Recall that the tangent function has a period of (pi). On the interval ([ 0,pi )),and at the angle of (dfrac{pi}{4}),the tangent has a value of (1). However, the angle we want is (left( heta−dfrac{pi}{2} ight)). Thus, if ( anleft(dfrac{pi}{4} ight)=1),then

[egin{align*} heta-dfrac{pi}{2}&= dfrac{pi}{4} heta&= dfrac{3pi}{4} pm kpi end{align*}]

Over the interval ([ 0,2pi )),we have two solutions:

( heta=dfrac{3pi}{4}) and ( heta=dfrac{3pi}{4}+pi=dfrac{7pi}{4})

Exercise (PageIndex{3})

Find all solutions for ( an x=sqrt{3}).

(dfrac{pi}{3}pm pi k)

Example (PageIndex{4}): Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation (2( an x+3)=5+ an x), (0≤x<2pi).

Solution

We can solve this equation using only algebra. Isolate the expression ( an x) on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of (−1): ( heta=dfrac{3pi}{4}) and ( heta=dfrac{7pi}{4}).

## Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example (PageIndex{5A}): Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation (sin heta=0.8),where ( heta) is in radians.

Solution

Make sure mode is set to radians. To find ( heta), use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the ({sin}^{−1}) function. What is shown on the screen is ({sin}^{−1}).The calculator is ready for the input within the parentheses. For this problem, we enter ({sin}^{−1}(0.8)), and press ENTER. Thus, to four decimals places,

({sin}^{−1}(0.8)≈0.9273)

The solution is

( heta≈0.9273pm 2pi k)

The angle measurement in degrees is

[egin{align*} heta&approx 53.1^{circ} heta&approx 180^{circ}-53.1^{circ} &approx 126.9^{circ} end{align*}]

Analysis

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using (pi− heta).

Example (PageIndex{5B}): Using a Calculator to Solve a Trigonometric Equation Involving Secant

Solution

We can begin with some algebra.

[egin{align*} sec heta&= -4 dfrac{1}{cos heta}&= -4 cos heta&= -dfrac{1}{4} end{align*}]

Check that the MODE is in radians. Now use the inverse cosine function

[egin{align*}{cos}^{-1}left(-dfrac{1}{4} ight)&approx 1.8235 heta&approx 1.8235+2pi k end{align*}]

Since (dfrac{pi}{2}≈1.57) and (pi≈3.14),(1.8235) is between these two numbers, thus ( heta≈1.8235) is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure (PageIndex{2}). So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is ( heta '≈pi−1.8235≈1.3181). The other solution in quadrant III is ( heta '≈pi+1.3181≈4.4597).

The solutions are ( heta≈1.8235pm 2pi k) and ( heta≈4.4597pm 2pi k).

Exercise (PageIndex{5})

Solve (cos heta=−0.2).

( heta≈1.7722pm 2pi k) and ( heta≈4.5110pm 2pi k)

## Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as (x) or (u). If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example (PageIndex{6A}): Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: ({cos}^2 heta+3 cos heta−1=0), (0≤ heta<2pi).

Solution

We begin by using substitution and replacing (cos heta) with (x). It is not necessary to use substitution, but it may make the problem easier to solve visually. Let (cos heta=x). We have

(x^2+3x−1=0)

The equation cannot be factored, so we will use the quadratic formula: (x=dfrac{−bpm sqrt{b^2−4ac}}{2a}).

[egin{align*} x&= dfrac{ -3pm sqrt{ {(-3)}^2-4 (1) (-1) } }{2} &= dfrac{-3pm sqrt{13}}{2}end{align*}]

Replace (x) with (cos heta ) and solve.

[egin{align*} cos heta&= dfrac{-3pm sqrt{13}}{2} heta&= {cos}^{-1}left(dfrac{-3+sqrt{13}}{2} ight) end{align*}]

Note that only the + sign is used. This is because we get an error when we solve ( heta={cos}^{−1}left(dfrac{−3−sqrt{13}}{2} ight)) on a calculator, since the domain of the inverse cosine function is ([ −1,1 ]). However, there is a second solution:

[egin{align*} heta&= {cos}^{-1}left(dfrac{-3+sqrt{13}}{2} ight) &approx 1.26 end{align*}]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

[egin{align*} heta&= 2pi-{cos}^{-1}left(dfrac{-3+sqrt{13}}{2} ight) &approx 5.02 end{align*}]

Example (PageIndex{6B}): Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: (2 {sin}^2 heta−5 sin heta+3=0), (0≤ heta≤2pi).

Solution

Using grouping, this quadratic can be factored. Either make the real substitution, (sin heta=u),or imagine it, as we factor:

[egin{align*} 2 {sin}^2 heta-5 sin heta+3&= 0 (2 sin heta-3)(sin heta-1)&= 0 qquad ext {Now set each factor equal to zero.} 2 sin heta-3&= 0 2 sin heta&= 3 sin heta&= dfrac{3}{2} sin heta-1&= 0 sin heta&= 1 end{align*}]

Next solve for ( heta): (sin heta≠dfrac{3}{2}), as the range of the sine function is ([ −1,1 ]). However, (sin heta=1), giving the solution ( heta=dfrac{pi}{2}).

Analysis

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise (PageIndex{6})

Solve ({sin}^2 heta=2 cos heta+2), (0≤ heta≤2pi). [Hint: Make a substitution to express the equation only in terms of cosine.]

(cos heta=−1), ( heta=pi)

Example (PageIndex{7A}): Solving a Trigonometric Equation Using Algebra

Solve exactly: (2 {sin}^2 heta+sin heta=0;space 0≤ heta<2pi)

Solution

This problem should appear familiar as it is similar to a quadratic. Let (sin heta=x). The equation becomes (2x^2+x=0). We begin by factoring:

[egin{align*}
2x^2+x&= 0
x(2x+1)&= 0qquad ext {Set each factor equal to zero.}
x&= 0
2x+1&= 0
x&= -dfrac{1}{2} end{align*}]
Then, substitute back into the equation the original expression (sin heta ) for (x). Thus,
[egin{align*} sin heta&= 0
heta&= 0,pi
sin heta&= -dfrac{1}{2}
heta&= dfrac{7pi}{6},dfrac{11pi}{6}
end{align*}]

The solutions within the domain (0≤ heta<2pi) are ( heta=0,pi,dfrac{7pi}{6},dfrac{11pi}{6}).

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

[egin{align*} {sin}^2 heta+sin heta&= 0 sin heta(2sin heta+1)&= 0 sin heta&= 0 heta&= 0,pi 2 sin heta+1&= 0 2sin heta&= -1 sin heta&= -dfrac{1}{2} heta&= dfrac{7pi}{6},dfrac{11pi}{6} end{align*}]

Analysis

We can see the solutions on the graph in Figure (PageIndex{3}). On the interval (0≤ heta<2pi),the graph crosses the (x)-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value. We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example (PageIndex{7B}): Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: (2 {sin}^2 heta−3 sin heta+1=0), (0≤ heta<2pi).

Solution

We can factor using grouping. Solution values of ( heta) can be found on the unit circle.

[egin{align*} (2 sin heta-1)(sin heta-1)&= 0 2 sin heta-1&= 0 sin heta&= dfrac{1}{2} heta&= dfrac{pi}{6}, dfrac{5pi}{6} sin heta&= 1 heta&= dfrac{pi}{2} end{align*}]

Exercise (PageIndex{7})

Solve the quadratic equation (2{cos}^2 heta+cos heta=0).

(dfrac{pi}{2}, space dfrac{2pi}{3}, space dfrac{4pi}{3}, space dfrac{3pi}{2})

## Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example (PageIndex{8A}): Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval (0≤x<2pi).

(cos x cos(2x)+sin x sin(2x)=dfrac{sqrt{3}}{2})

Solution

Notice that the left side of the equation is the difference formula for cosine.

[egin{align*} cos x cos(2x)+sin x sin(2x)&= dfrac{sqrt{3}}{2} cos(x-2x)&= dfrac{sqrt{3}}{2}qquad ext{Difference formula for cosine} cos(-x)&= dfrac{sqrt{3}}{2}qquad ext{Use the negative angle identity.} cos x&= dfrac{sqrt{3}}{2} end{align*}]

From the unit circle in the section on Sum and Difference Identities, we see that (cos x=dfrac{sqrt{3}}{2}) when (x=dfrac{pi}{6},space dfrac{11pi}{6}).

Example (PageIndex{8B}): Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: (cos(2 heta)=cos heta).

Solution

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

[egin{align*} cos(2 heta)&= cos heta 2{cos}^2 heta-1&= cos heta 2 {cos}^2 heta-cos heta-1&= 0 (2 cos heta+1)(cos heta-1)&= 0 2 cos heta+1&= 0 cos heta&= -dfrac{1}{2} cos heta-1&= 0 cos heta&= 1 end{align*}]

So, if (cos heta=−dfrac{1}{2}),then ( heta=dfrac{2pi}{3}pm 2pi k) and ( heta=dfrac{4pi}{3}pm 2pi k); if (cos heta=1),then ( heta=0pm 2pi k).

Example (PageIndex{8C}): Solving an Equation Using an Identity

Solve the equation exactly using an identity: (3 cos heta+3=2 {sin}^2 heta), (0≤ heta<2pi).

Solution

If we rewrite the right side, we can write the equation in terms of cosine:

[egin{align*}
3 cos heta+3&= 2 {sin}^2 heta
3 cos heta+3&= 2(1-{cos}^2 heta)
3 cos heta+3&= 2-2{cos}^2 heta
2 {cos}^2 heta+3 cos heta+1&= 0
(2 cos heta+1)(cos heta+1)&= 0
2 cos heta+1&= 0
cos heta&= -dfrac{1}{2}
heta&= dfrac{2pi}{3},space dfrac{4pi}{3}
cos heta+1&= 0
cos heta&= -1
heta&= pi
end{align*}]

Our solutions are ( heta=dfrac{2pi}{3},space dfrac{4pi}{3},space pi).

## Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as (sin(2x)) or (cos(3x)). When confronted with these equations, recall that (y=sin(2x)) is a horizontal compression by a factor of 2 of the function (y=sin x). On an interval of (2pi),we can graph two periods of (y=sin(2x)),as opposed to one cycle of (y=sin x). This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to (sin(2x)=0) compared to (sin x=0). This information will help us solve the equation.

Example (PageIndex{9}): Solving a Multiple Angle Trigonometric Equation

Solve exactly: (cos(2x)=dfrac{1}{2}) on ([ 0,2pi )).

Solution

We can see that this equation is the standard equation with a multiple of an angle. If (cos(alpha)=dfrac{1}{2}),we know (alpha) is in quadrants I and IV. While ( heta={cos}^{−1} dfrac{1}{2}) will only yield solutions in quadrants I and II, we recognize that the solutions to the equation (cos heta=dfrac{1}{2}) will be in quadrants I and IV.

Therefore, the possible angles are ( heta=dfrac{pi}{3}) and ( heta=dfrac{5pi}{3}). So, (2x=dfrac{pi}{3}) or (2x=dfrac{5pi}{3}), which means that (x=dfrac{pi}{6}) or (x=dfrac{5pi}{6}). Does this make sense? Yes, because (cosleft(2left(dfrac{pi}{6} ight) ight)=cosleft(dfrac{pi}{3} ight)=dfrac{1}{2}).

In quadrant I, (2x=dfrac{pi}{3}), so (x=dfrac{pi}{6}) as noted. Let us revolve around the circle again:

[egin{align*}
2x&= dfrac{pi}{3}+2pi
&= dfrac{pi}{3}+dfrac{6pi}{3}
&= dfrac{7pi}{3}
x&= dfrac{7pi}{6}
ext {One more rotation yields}
2x&= dfrac{pi}{3}+4pi
&= dfrac{pi}{3}+dfrac{12pi}{3}
&= dfrac{13pi}{3}
end{align*}]

(x=dfrac{13pi}{6}>2pi), so this value for (x) is larger than (2pi), so it is not a solution on ([ 0,2pi )).

In quadrant IV, (2x=dfrac{5pi}{3}), so (x=dfrac{5pi}{6}) as noted. Let us revolve around the circle again:

[egin{align*} 2x&= dfrac{5pi}{3}+2pi &= dfrac{5pi}{3}+dfrac{6pi}{3} &= dfrac{11pi}{3} end{align*}]

so (x=dfrac{11pi}{6}).

One more rotation yields

[egin{align*} 2x&= dfrac{5pi}{3}+4pi &= dfrac{5pi}{3}+dfrac{12pi}{3} &= dfrac{17pi}{3} end{align*}]

(x=dfrac{17pi}{6}>2pi),so this value for (x) is larger than (2pi),so it is not a solution on ([ 0,2pi ))﻿.

Our solutions are (x=dfrac{pi}{6}, space dfrac{5pi}{6}, space dfrac{7pi}{6}), and (dfrac{11pi}{6}). Note that whenever we solve a problem in the form of (sin(nx)=c), we must go around the unit circle (n) times.

## Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem,

[a^2+b^2=c^2 label{Pythagorean}]

and model an equation to fit a situation.

Example (PageIndex{10A}): Using the Pythagorean Theorem to Model an Equation

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is (69.5) meters above the ground, and the second anchor on the ground is (23) meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure (PageIndex{4}). Solution

Use the Pythagorean Theorem (Equation ef{Pythagorean}) and the properties of right triangles to model an equation that fits the problem. Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

[egin{align*} a^2+b^2&= c^2 {(23)}^2+{(69.5)}^2&approx 5359 sqrt{5359}&approx 73.2space m end{align*}]

The angle of elevation is ( heta),formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

[egin{align*} an heta&= 69.523 { an}^{-1}(69.523)&approx 1.2522 &approx 71.69^{circ} end{align*}]

The angle of elevation is approximately (71.7°), and the length of the cable is (73.2) meters.

Example (PageIndex{10B}): Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed (1) foot from the wall for every (4) feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

Solution

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be (4a) feet. See Figure (PageIndex{5}). The side adjacent to ( heta) is (a) and the hypotenuse is (4a). Thus,

[egin{align*} cos heta&= dfrac{a}{4a} &= dfrac{1}{4} {cos}^{-1}left (dfrac{1}{4} ight )&approx 75.5^{circ} end{align*}]

The elevation of the ladder forms an angle of (75.5°) with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

[egin{align*} a^2+b^2&= {(4a)}^2 b^2&= {(4a)}^2-a^2 b^2&= 16a^2-a^2 b^2&= 15a^2 b&= asqrt{15} end{align*}]

Thus, the ladder touches the wall at (asqrt{15}) feet from the ground.

Media

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

• Solving Trigonometric Equations I
• Solving Trigonometric Equations II
• Solving Trigonometric Equations III
• Solving Trigonometric Equations IV
• Solving Trigonometric Equations V
• Solving Trigonometric Equations VI

## Key Concepts

• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example (PageIndex{1}), Example (PageIndex{2}), and Example (PageIndex{3}).
• Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example (PageIndex{4}), Example (PageIndex{5}), and Example (PageIndex{6}), and Example (PageIndex{7}).
• We can also solve trigonometric equations using a graphing calculator. See Example (PageIndex{8}) and Example (PageIndex{9}).
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example (PageIndex{10}), Example (PageIndex{11}), Example (PageIndex{12}), and Example (PageIndex{13}).
• We can also use the identities to solve trigonometric equation. See Example (PageIndex{14}), Example (PageIndex{15}), and Example (PageIndex{16}).
• We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example (PageIndex{17}).
• Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example (PageIndex{18}).

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## Solving Trigonometric Equations - Problem 3 Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s talk about another trig equation, this time a secant equation. Now a secant equation involves the reciprocal function secant. Anytime you see a reciprocal function, you want to turn this into a trig equation involving sine cosine or tangent.

So the first thing I’m going to do is call this 1 over cosine of 3x equals 2, because secant equals 1 over cosine. And then I can take the reciprocal of both sides and I get cosine of 3 equals a half. And so you see that I’m in two steps I know that I no longer have to deal with the secant function and this is always true you never have to deal with the reciprocal trig functions when you are solving trig equations. You can go back to one of the familiar ones.

So we have cosine of 3x equals one half, let me make a little substitution here. This is what I usually do when I have something other than just the variable inside my trig function, I’ll call this theta. And so my equation becomes cosine of theta equals a half. And I can find the solutions to this equation on the unit circle. So here is the unit circle, I need to find angles that make the first coordinate of this point one half, because that’s how cosine is defined.

So if you recall pi over 3 does that. Of course this problem asks us to solve for x in the interval from 0 degrees to 360 degrees. This tells us two things, first of all that we are not going to want infinitely many solutions and second that we are working in degrees and not radian. So let me immediately switch to 60 degrees, pi over 3 is 60 degrees, that gives me 1 solution.

How to get another? Remember that sine and cosine each have two solutions usually per period. There’s going to be another solution and with cosine the other solutions due to the fact that cosines are even function, opposite inputs have the same output. You can see that on the unit circle by thinking what are the points on the units circle has an x coordinate of one half? There’s one down here and it's going to be the mirror image of this point. So you are going to have one half common negative y whatever the y coordinate is and see you are going to have the second solution being -60 degrees by symmetry. So for -60 degrees.

Now to get the remainder of the solution we have to use periodicity so we’ve got theta equals plus or minus 60 degrees and when you are dealing with degrees, periodicity means you can add an integer multiple of 360 degrees. Normally we add integer multiple of 2 pi, because 2 pi is one radian but then degrees are one revolution rather. In degrees one revolution is 360 degrees.

So we are about ready to re-substitute, remember that I substituted theta for 3x. So we are going back to 3x and that’s what we get. All we have to do is divide by 3. So dividing by 3 gives me plus or minus 20 degrees plus n times 120.

Now the problem asked me to find only solutions between 0 and 360, so I don’t want to give this as my final answer because I'm not exactly reading the instructions here, so let me start with 20 degrees. Plus 20 degrees plus 0 times 120 that’s the smallest solution I’m going to get. And then let me start with the 20 degrees and keep adding integer multiples until I leave the interval from 0 to 360. So I’m going to add 120 and get 140.

That one is good. I add 120 again and get 260, also good. Add 120 again and get 380 not the interval, so I wouldn’t include that. Now I know that -20 is not a solution but when I add multiples of 120 degrees I will get solutions, so let me add 120 and get 100 degrees that’s the solution, that one will be. Add another 120, 220, add another, 340 and you can see that if you add another 120 you’ll be outside of the interval. So these are your solutions. 20 ,140, 260, 100, 220 and 340.

And notice the way I worked this out I started with the positive 20 and I added multiples of 120. Here I started with -20 and I added multiples of -120 and I'll just chose this as my final answer so the values actually fall in the intervals 0 and 360, that’s your final answer.

## 3.5: Solving Trigonometric Equations

Note: If you would like a review of trigonometry, click on trigonometry.

Example 1: Solve for x in the following equation.

There are an infinite number of solutions to this problem. To solve for x, you must first isolate the tangent term.

If we restrict the domain of the tangent function to , we can use the inverse tangent function to solve for reference angle x ', and then x .

The reference angle is The tangent function is positive in the first quadrant and in the third quadrant and negative in the second and fourth quadrant.

The period of the function is This means that the values will repeat every radians in both directions. Therefore, the exact solutions are and where n is an integer.

The approximate solutions are and where n is an integer.

These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.

Since the left side equals the right side when you substitute 0.52359877for x, then 0.52359877 is a solution.

Since the left side equals the right side when you substitute -0.52359877for x, then -0.52359877 is a solution.

Graph the equation Note that the graph crosses the x-axis many times indicating many solutions.

Note that it crosses at 0.52359877. Since the period is , it crosses again at 0.52359877+3.1415927=3.66519 and at tex 2 html c omment m ark > 0.52359877+2(3.1415927)=6.80678, etc.

Note that it crosses at -0.52359877. Since the period is , it crosses again at -0.52359877+3.1415927=2.617899 and at tex 2 html c omment m ark > -0.52359877+2(3.1415927)=5.759587, etc.

If you would like to work another example, click on Example.

If you would like to test yourself by working some problems similar to this example, click on Problem.

If you would like to go to the next section, click on Next.

## Solving Trigonometric Equations

How would you solve the equation sin θ = 0? You know that θ = 0 is one solution, but this is not the only solution. Any one of the following values of is also a solution.

You can write this infinite solution set as

Solving a Trigonometric Equation

To solve the equation, you should consider that the sine is negative in Quadrants III and IV and that

Thus, you are seeking values of θ in the third and fourth quadrants that have a reference angle of π/3. In the interval [0, 2π], the two angles fitting these criteria are

By adding integer multiples of 2π to each of these solutions, you obtain the following general solution.

Solving a Trigonometric Equation

Solve cos 2 θ = 2 - 3 sin θ, where 0 ≤ θ ≤ 2π.

Using the double-angle identity cos 2 θ = 1 - 2 sin 2 θ, you can rewrite the equation as follows.

 cos 2 θ = 2 - 3 sin θ Given equation 1 - 2 sin 2 θ = 2 - 3 sin θ Trigonometric identity 0 = 2 sin 2 θ - 3sin θ + 1 Quadratic form 0 = (2 sin θ )(sin θ - 1 ) Factor.

If 2 sin θ = 0, then sin θ = 1/2 and θ = π/6 or θ = 5π/6. If sin θ - 1, then sin θ = 1 and θ = π/2. Thus, for 0 ≤ θ ≤ 2π, there are three solutions.

## 3.5: Solving Trigonometric Equations

Note: If you would like a review of trigonometry, click on trigonometry.

Solve for x in the following equation.

There are an infinite number of solutions to this problem.

We can make the solution easier if we convert all the trigonometric terms to like trigonometric terms.

One common trigonometric identity is If we replace the term with , all the trigonometric terms will be tangent terms.

Replace with in the original equation and simplify.

Isolate the tangent term. To do this, rewrite the left side of the equation in an equivalent factored form.

The product of two factors equals zero if at least one of the factors equals zero. This means that if or

We just transformed a difficult problem into two easier problems. To find the solutions to the original equation, , we find the solutions to the equations and

How do we isolate the x in each of these equations? We could take the arctangent of both sides of each equation. However, the tangent function is not a one-to-one function.

Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. The graph of the tangent function is one-to-one on the interval If we restrict the domain of the tangent function to that interval , we can take the arctangent of both sides of each equation.

Since the period of equals , these solutions will repeat every units. The exact solutions are

The approximate values of these solutions are

You can check each solution algebraically by substituting each solution in the original equation. If, after the substitution, the left side of the original equation equals the right side of the original equation, the solution is valid.

You can also check the solutions graphically by graphing the function formed by subtracting the right side of the original equation from the left side of the original equation. The solutions of the original equation are the x-intercepts of this graph.

Since the left side of the original equation equals the right side of the original equation when you substitute 1.249046 for x, then 1.249046 is a solution.

Since the left side of the original equation equals the right side of the original equation when you substitute -0.785398 for x, then -0.785398 is a solution.

We have just verified algebraically that the exact solutions are and and these solutions repeat every units. The approximate values of these solutions are and and these solutions repeat every units.

Graph the equation Note that the graph crosses the x-axis many times indicating many solutions. Let's check a few of these x-intercepts against the solutions we derived.

Verify the graph crosses the x-axis at -0.785398. Since the period is , you can verify that the graph also crosses the x-axis again at -0.785398+3.14159265=2.356195 and at , etc.

Verify the graph crosses the x-axis at 1.249046. Since the period is , you can verify that the graph also crosses the x-axis again at 1.249046+3.14159265=4.39906387 and at , etc.

Note: If the problem were to find the solutions in the interval , then you choose those solutions from the set of infinite solutions that belong to the set 2.356195, 5.497787, 4.39906387, and

If you would like to work another example, click on Example.

If you would like to test yourself by working some problems similar to this example, click on Problem.

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## Lesson Solving trigonometric equations

DEFINITION. A trig equation is an equation containing one or many trig functions of the variable arc x that rotates on the trig unit circle. Solving for x means finding the values of the variable trig arc x whose trig functions make the trig equation true.
Examples of trig equation:
sin ( x + 30 degree) = 0.75 tan (x + pi/3) = 1.5 sin 2x + cos x = 1 tan x + cot x = 1.732
sin x + sin 2x = sin 3x cos x + cos 2x + cos 3x = 0 sin x - sin 3x = cos 2x
Answers, or values of the solution arcs, are expressed in degrees or radians.
Examples: x = 45 degree x = 47.24 degree x = 25.59 degree
x = Pi/5 x = 3Pi /4 x = 7Pi/12

THE TRIG UNIT CIRCLE.
It is a circle with radius R = 1 unit with the origin O. The unit circle defines the trig functions of the variable arc x that rotates counter-clockwise on the trig circle.
When the arc AM, with value x (in radians or degrees), varies on the trig unit circle:
The horizontal axis OAx defines the function f(x) = cos x.
The vertical axis OBy defines the function f(x) = sin x.
The vertical axis At defines the function f(x) = tan x.
The horizontal axis Bu defines the function f(x) = cot x.

THE PERIODIC PROPERTY OF TRIG FUNCTIONS
All trig functions f(x) are periodic meaning they come back to the same value after the arc x rotates counterclockwise one period on the unit circle. Examples:
The trig function f(x) = sin x has 2Pi as period.
The trig function f(x) = tan x has Pi as period.
The trig function f(x) = sin 2x has Pi as period.
The trig function f(x) = cos x/2 has 4Pi as period.

CONCEPT FOR SOLVING TRIG EQUATIONS
To solve a trig equation transform it into one or many basic trig equations. Solving trig equations finally results in solving basic trig equations.

BASIC TRIG EQUATIONS
They are also called "Trig equations in simplest form". There are 4 types of common basic trig equations: sin x = a cos x = a tan x = a cot x = a.
Solving a basic trig equation proceeds by considering the various positions of the arc x on the trig unit circle and by using the trig conversion tables (or calculators).
Example 1. Solve sin x = 0.866. The 2 answers are given by the trig unit circle and calculators:
Answer 1: x1 = Pi/3 Extended answer: x1 = Pi/3 + 2k.Pi
Answer 2: x2 = 2Pi/3 Extended answer: x2 = 2Pi/3 + 2k.Pi
Example 2. Solve: cos x = -1/2. Two answers are given by the unit circle and conversion table:
Answer 1: x1 = 2Pi/3 Extended answer: x1 = 2Pi/3 + 2k.Pi
Answer 2: x2 = -2Pi/3 Extended answer: x2 = -2Pi/3 + 2k.Pi
Example 3. Solve cos x = 0.732. Two answers given by calculator and the unit circle:
Answer 1: x1 = 42.95 degree Extended answer: x1 = 42.95 degree + k.360 degree
Answer 2: x2 = -42.95 degree Extended answer: x2 = -42.95 degree + k.360 degree
Example 4. Solve: cot 2x = 1.732. Trig table and unit circle give:
2x = Pi/6 Extended: 2x = Pi/6 + K.Pi

Example 5. Solve: sin(x - 20 deg.) = 0.5. Trig table and trig unit circle gives:
1) sin (x - 20) = sin 30 deg.-------- 2) sin (x - 20) = sin (180 - 30)
x - 20 = 30 deg. ------------------ x - 20 = 150 deg.
x = 50 deg. ----------------------- x = 170 deg.
Extended x = 50 deg. + k.360 deg. -- Extended x = 170 deg.+ k.360 deg.

Example 6. Solve: sin 2x = cos 3x. The unit circle gives 2 answers:
1) sin 2x = sin (Pi/2 - 3x) ------ 2) sin 2x = sin (Pi - Pi/2 + 3x)
2x = Pi/2 - 3x -------------------- 2x = Pi/2 + 3x
5x = Pi/2 ------------------------- -x = Pi/2
x = Pi/10 ------------------------- x = -Pi/2
Extended x = Pi/10 + k.2Pi --------- Extended: x = -Pi/2 + k.2Pi.

TRANSFORMATION USED IN SOLVING TRIG EQUATIONS.
To transform a trig equation into basic trig equations, use common algebraic transformations (factoring, common factor, polynomial identities. ), definition and properties of trig functions, and trig identities (the most needed). There are 14 common trig identities, called "transformation identities", that are used for the transformation of trig equations. See book titled:"Solving trigonometric equations and inequalities" (Amazon e-book 2010).
Example. The trig equation sin x + sin 2x + sin 3x = 0 can be transformed, using trig identities, into a product of many basic trig equations: 4cos x.sin 3x/2.cos x/2 = 0.
Example. The trig equation cos x + cos 2x + cos 3x = 0 can be transformed, using trig identities, into a product of basic trig equation: cos 2x(2cos x + 1) = 0.
Example. Transform into a product (sin a + cos a).
sin a + cos a = sin a + sin (Pi/2 - a) = 2sin (Pi/4).sin (a + Pi/4)
Example. Transform (sin 2a - sin a) into a product:
sin 2a - sin a = 2sin a.cos a - sin a = sin a.(2cos a - 1).

GRAPHING THE SOLUTION ARCS ON THE UNIT CIRCLE
We can graph to illustrate the solution arcs on the trig unit circle. The terminal points of the solution arcs constitute regular polygons on the trig unit circle.
Example: The terminal points of the solution arcs x = Pi/3 + k.Pi/2 constitute a square on the unit circle.
Example. The solution arcs x = Pi/4 + k.Pi/3 are represented by the vertexes of a regular hexagon on the unit circle.

METHODS FOR SOLVING TRIG EQUATIONS.
There are 2 methods depending on transformation possibilities.

METHOD 1. Transform the given trig equation into a product of basic trig equations. Next, solves these basic trig equations to get all the solution arcs.
Example 7. Solve sin 2x + 2cos x = 0.
Solution: First, transform: sin 2x + 2cos x = 2sin x.cos x + 2cos x = 2cos x(sin x + 1) = 0.
Next, solve the 2 basic trig equations: cos x = 0 and sin x = -1.
Example 8. Solve: cos x + cos 2x + cos 3x = 0.
Solution. First, use trig identities to transform: (cos x + cos 3x) + cos 2x = 2cos 2x.cos x + cos 2x = cos 2x(2cos x + 1) = 0.
Next, solve the 2 basic trig equations: cos 2x = 0 and cos x = -1/2.
Example 9. Solve sin x - sin 3x = cos 2x.
Solution. Using trig identity transform the equation.
(sin x - sin 3x) - cos 2x = 2cos 2x.sin x - cos 2x = cos 2x(2sin x - 1) = 0.
Next, solve the 2 basic trig equations: cos 2x = 0 and sin x = 1/2.

METHOD 2. If the given trig equation contains 2 or more trig functions, transform it into an equation containing only one trig function as variable. There are a few tips on how to select the trig function variable. The common variables to select are: sin x = t cos x = t cos 2x = t tan x = t tan x/2 = t.
Example 10. Solve: 3sin^2 x + 2cos x - 2 = 0. Call cos x = t.
Solution. Replace (sin^2 x) by (1 - cos^2 x) = 1 - t^2.
The equation becomes: 3(1 - t^2) + 2t - 2 = -3t^2 + 2t + 1 = 0.
This is a quadratic equation with 2 real roots: 1 and -1/3. Next, solve the 2 basic trig equations: t = cos x = 1 and t = cos x = -1/3.
Example 11. Solve: tan x + 2tan^2 x - cot x - 2 = 0.
Solution. Call tan x = t, the equation becomes:
t + 2t^2 - 1/t - 2 = 0 = t^2 + 2t^3 - 1 - 2t = t^2(1 + 2t) - (1 + 2t) = (1 + 2t)(t^2 - 1) = 0.
Next, solve the 2 basic trig equations: (t = tan x = -1/2) and (t^2 = tan^2 x = 1).
Example 12. Solve cos 2x - 3sin x - 2 = 0.
Solution. Call sin x = t and replace (cos 2x) by (1 - 2sin^2 x).
(1 - 2t^2) - 3t - 2 = -2t^2 - 3t - 1 = 0. Quadratic equation with 2 real roots: -1 and -1/2.
Next, solve the 2 basic trig equations: t = sin x = -1 and t = sin x = -1/2.

SOLVING SPECIAL TYPES OF TRIG EQUATIONS.
There are a few types of trig equations that require specific transformations. Examples:
a.sin x + b.cos x = c
a(sin x + cos x) + b.sin x.cos x = c
a.sin^2 x + b.sin x.cos x + c.cos^2 x = 0

THE COMMON PERIOD OF A TRIG EQUATION
Unless specified in home-works/tests, the trig equation f(x) = 0 must be solved, at least, within a common period. This means we must find all the solution arcs x within this common period. The common period is the least multiple of all the periods of the trig functions presented in the equation. Examples:
The trig equation f(x) = cos x + 2tan x - 2 = 0 has 2Pi as common period.
The equation f(x) = tan x + 2cot x = 0 has Pi as common period.
The equation f(x) = cos2x + sin x = 0 has 2Pi as common period.
The equation f(x) = sin 2x + cos x - sin x/2 = 0 has 4Pi as common period.

CHECKING ANSWERS BY GRAPHING CALCULATORS AFTER SOLVING.
Solving trig equations is a tricky work that often leads to errors and mistakes. After solving, you may check the answers by using graphing calculators. Using appropriate calculator setup, graph the function f(x). The roots of f(x) = 0 will be given in decimals. For examples, Pi is given as 3.14 360 degree is given as 6.28. For more details, see the last chapter of the trig book mentioned above.

## Solving Trigonometric Equations

Videos and lessons with examples and solutions for High School students on solving trigonometric equations.

In these lessons, we will learn

• how to solve trigonometric equations
• how to solve trigonometric equations by factoring

### Solving Trigonometric Equations

When solving trigonometric equations, we find all the angles that make the equation true. If there is no interval given, use periodicity to show the infinite number of solutions. Two ways to visualize the solutions are (1) the graph in the coordinate plane and (2) the unit circle. The unit circle is the more useful of the two in obtaining an answer.

Solving Trigonometric Equations
2 cos x = 1,
sin(2x) = cos x,
2 + cos 2x = 3 cos x,
sin x = tan x

### Factoring Trigonometric Equations

Solve trigonometric equations that are factorable or in quadratic form

This video solve a trigonometric equation in quadratic form by factoring.

Solving a Trigonometric Equation by Factoring
2 sin 2 x = 1 + cos x Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. ## Math2.org Math Tables:

Under its simplest definition, a trigonometric (lit. "triangle-measuring") function, is one of the many functions that relate one non-right angle of a right triangle to the ratio of the lengths of any two sides of the triangle (or vice versa).

Any trigonometric function (f), therefore, always satisfies either of the following equations:

• If the former equation holds, we can choose any right triangle, then take the measurement of one of the non-right angles, and when we evaluate the trigonometric function at that angle, the result will be the ratio of the lengths of two of the triangle's sides.
• However, if the latter equation holds, we can chose any right triangle, then compute the ratio of the lengths of two specific sides, and when we evaluate the trigonometric function at any that ratio, the result will be measure of one of the triangles non-right angles. (These are called inverse trig functions since they do the inverse, or vice-versa, of the previous trig functions.)

Since there are three sides and two non-right angles in a right triangle, the trigonometric functions will need a way of specifying which sides are related to which angle. (It is not-so-useful to know that the ratio of the lengths of two sides equals 2 if we do not know which of the three sides we are talking about. Likewise, if we determine that one of the angles is 40°, it would be nice to know of which angle this statement is true.

Under a certain convention, we label the sides as opposite, adjacent, and hypotenuse relative to our angle of interest q . full explaination

As mentioned previously, the first type of trigonometric function, which relates an angle to a side ratio, always satisfies the following equation:

 f( q ) = opp/opp f( q ) =opp/adj f( q ) =opp/hyp f( q ) =adj/opp f( q ) = adj/adj f( q ) =adj/hyp f( q ) =hyp/opp f( q ) =hyp/adj f( q ) = hyp/hyp

The three diagonal functions shown in red always equal one. They are degenerate and, therefore, are of no use to us. We therefore remove these degenerate functions and assign labels to the remaining six, usually written in the following order:

 sine( q ) = opp/hyp cosecant( q ) = hyp/opp cosine( q ) = adj/hyp secant( q ) = hyp/adj tangent( q ) = opp/adj cotangent( q ) = adj/opp

Furthermore, the functions are usually abbreviated: sine (sin), cosine (cos), tangent (tan) cosecant (csc), secant (sec), and cotangent (cot).

Do not be overwhelmed. By far, the two most important trig functions to remember are sine and cosine. All the other trig functions of the first kind can be derived from these two funcions. For example, the functions on the right are merely the multiplicative inverse of the corresponding function on the left (that makes them much less useful). Futhermore, the sin(x) / cos(x) = (opp/hyp) / (adj/hyp) = opp / adj = tan(x). Therefore, the tangent function is the same as the quotient of the sine and cosine functions (the tangent function is still fairly handy).

 sin( q ) = opp/hyp csc( q ) = 1/sin( q ) cos( q ) = adj/hyp sec( q ) = 1/cos( q ) tan( q ) = sin( q )/cos( q ) cot( q ) = 1/tan( q )

Let's examine these functions further. You will notice that there are the sine, secant, and tangent functions, and there are corresponding "co"-functions. They get their odd names from various similar ideas in geometry. You may suggest that the cofunctions should be relabeled to be the multiplicative inverses of the corresponding sine, secant, and tangent functions. However, there is a method to this madness. A cofunction of a given trig function (f) is, by definition, the function obtained after the complement its parameter is taken. Since the complement of any angle q is 90° - q , the the fact that the following relations can be shown to hold

The trig functions evaluate differently depending on the units on q , such as degrees, radians, or grads. For example, sin(90°) = 1, while sin(90)=0.89399. explaination

Just as we can define trigonometric functions of the form

inverse functions