Lesson Explainer: Solving Reciprocal Trigonometric Equations | Nagwa Lesson Explainer: Solving Reciprocal Trigonometric Equations | Nagwa

Lesson Explainer: Solving Reciprocal Trigonometric Equations Mathematics

In this explainer, we will learn how to solve trigonometric equations involving secant, cosecant, and cotangent over different intervals in degrees and radians.

Reciprocal trigonometric equations have several real-world applications in various fields, such as physics, engineering, architecture, robotics, music theory, and navigation, to name a few. In physics, they can be used in projectile motion, modeling the mechanics of electromagnetic waves, analyzing alternating and direct currents, and finding the trajectory of a mass around a massive body under the force of gravity.

Let’s begin by recalling the trigonometric functions, whose reciprocals we will examine in this explainer. Consider a right triangle.

The trigonometric functions can be expressed in terms of the ratio of the sides of the triangle as sinOHcosAHtanOA𝜃=,𝜃=,𝜃=.

These functions satisfy the following trigonometric identity: tansincos𝜃=𝜃𝜃.

We note that these trigonometric ratios are defined for acute angles 0<𝜃<90, and the trigonometric functions for all values of 𝜃 are defined on the unit circle using right triangle trigonometry.

Suppose that a point moves along the unit circle in the counterclockwise direction. At a particular position (𝑥,𝑦) on the unit circle with angle 𝜃, the sine function is defined as 𝑦=𝜃sin and the cosine function as 𝑥=𝜃cos, as shown in the diagram above. In other words, the trigonometric functions are defined using the coordinates of the point of intersection of the unit circle with the terminal side of 𝜃 in standard position.

The domain is the set of possible inputs and the range is the set of possible outputs, given its domain. For the trigonometric functions, these are given by the following.

DomainRange
sin𝜃[1,1]
cos𝜃[1,1]
tan𝜃𝜋2+𝑛𝜋,𝑛

Since the tangent function is defined as the ratio of the sine and cosine functions, it is undefined when cos𝜃, the denominator, is zero. In other words, the tangent function has to exclude values of 𝜃 where cos𝜃=0, in order to be well defined. This is why the domain of the tangent function is 𝜋2+𝑛𝜋,𝑛, which just means we subtract the values of 𝜃 where cos𝜃=0 from the set of real numbers in order to exclude this from the input.

The trigonometric functions are periodic, which means if we add an integer multiple of 2𝜋, in radians, or 360 to the angle 𝜃, the value of the function stays the same: sinsincoscostantan(360+𝜃)=𝜃,(360+𝜃)=𝜃,(360+𝜃)=𝜃.

We can see these directly from the unit circle definition of the trigonometric functions. In fact, the tangent function is periodic by 𝜋, in radians, or 180 since we have tantan(180+𝜃)=𝜃.

This fact will be important for finding general solutions for the trigonometric functions. The domains of the trigonometric functions have to be restricted to a particular subset, known as the principle branch, in order to have inverse functions.

The reciprocal trigonometric functions are defined in terms of the standard trigonometric functions as follows.

Definition: Reciprocal Trigonometric Functions

The cosecant, secant, and cotangent functions are defined as cscsinseccoscottancossin𝜃=1𝜃,𝜃=1𝜃,𝜃=1𝜃=𝜃𝜃.

Similarly, for the reciprocal trigonometric functions, we have

DomainRange
csc𝜃[𝑛𝜋,𝑛]],1][1,[
sec𝜃𝜋2+𝑛𝜋,𝑛],1][1,[
cot𝜃[𝑛𝜋,𝑛]

Similar to the case of the tangent function, the cotangent and cosecant functions are defined with sin𝜃 in the denominator, so the values of 𝜃 where sin𝜃=0 should be excluded from the input. The secant function has the same domain as the tangent function, since it contains a cos𝜃 in the denominator.

The reciprocal trigonometric functions are also periodic: csccscsecseccotcot(360+𝜃)=𝜃,(360+𝜃)=𝜃,(360+𝜃)=𝜃.

Similarly to the tangent function, the cotangent function is periodic by 𝜋, in radians, or 180 since we have cotcot(180+𝜃)=𝜃.

The periodicity of the trigonometric functions will be important for finding general solutions for the trigonometric functions. The domains of the trigonometric functions have to be restricted to a particular subset, known as the principle branch, in order to have inverse functions.

The inverse trigonometric functions denoted by sin, cos, and tan are the inverse functions of the trigonometric functions sin, cos, and tan. This means they work in reverse or “go backward” from the usual trigonometric functions. They are defined by 𝑦=𝑥𝑥=𝑦,𝑦=𝑥𝑥=𝑦,𝑦=𝑥𝑥=𝑦.sinsincoscostantan

These can also be written as arcsin𝑥, arccos𝑥, and arctan𝑥. The domain and range for the inverse trigonometric functions are given by the following.

DomainRange
sin𝜃[1,1]𝜋2,𝜋2
cos𝜃[1,1][0,𝜋]
tan𝜃𝜋2,𝜋2

The ranges for these inverse functions in general are when the trigonometric and reciprocal trigonometric functions are restricted to the principal branch. This is to ensure that the functions are one-to-one functions so that the inverse functions evaluate to a single value, known as as the principle value.

For example, if we have a particular trigonometric equation, such as sin𝜃=𝑦, we can find the solutions in the range 𝜃𝜋2,𝜋2 by applying the inverse trigonometric equation: 𝜃=(𝑦).sin

However, if we want to determine all the possible solutions, we need the general solutions given in terms of an integer 𝑛, which we can obtain from the CAST diagram and the periodicity of the trigonometric functions.

Let’s recall the CAST diagram.

The Cast Diagram

  • In the first quadrant, all trigonometric functions are positive.
  • In the second quadrant, the sine function is positive.
  • In the third quadrant, the tangent function is positive.
  • In the fourth quadrant, the cosine function is positive.

Let’s recall how we can find the solutions to trigonometric equations.

Solutions to Trigonometric Equations

The CAST diagram helps us to remember the signs of the trigonometric functions for each quadrant.

In particular, the CAST diagram tells us that solutions to the trigonometric equations are given by the following.

  • If sin𝜃=𝑥 and 1𝑥1, 𝜃=𝑥𝜃=180𝑥,sinsin for 𝜃[90,270], or, in radians, 𝜃=𝑥𝜃=𝜋𝑥,sinsin for 𝜃𝜋2,3𝜋2.
  • If cos𝜃=𝑥 and 1𝑥1, then we can express the angle 𝜃 in terms of the inverse cosine function in degrees as 𝜃=𝑥𝜃=360𝑥,coscos for 𝜃[0,360] or in radians as 𝜃=𝑥𝜃=2𝜋𝑥,coscos for 𝜃[0,2𝜋].
  • If tan𝜃=𝑥, then we can express the angle 𝜃 in terms of the inverse tangent function in degrees as 𝜃=𝑥𝜃=180+𝑥,tantan for 𝜃]90,90[]90,270[ or in radians as 𝜃=𝑥𝜃=𝜋+𝑥,tantan for 𝜃𝜋2,𝜋2𝜋2,3𝜋2.

The ranges given for 𝜃 follow from the ranges of the inverse trigonometric functions.

We can also see this from the unit circle as shown.

The inverse reciprocal trigonometric functions denoted by csc, sec, and cot are the inverse functions of the reciprocal trigonometric functions sec, csc, and cot. They are defined by 𝑦=𝑥𝑥=𝑦,𝑦=𝑥𝑥=𝑦,𝑦=𝑥𝑥=𝑦.csccscsecseccotcot

The domain and range for the inverse reciprocal trigonometric functions are given by the following.

DomainRange
csc𝜃],1][1,[𝜋2,00,𝜋2
sec𝜃],1][1,[0,𝜋2𝜋2,𝜋
cot𝜃]0,𝜋[

Similiarly, if we have a reciprocal trigonometric equation, such as cscfor𝜃=𝑥,𝑥0, we can find the solutions in the range 𝜃𝜋2,00,𝜋2 by applying the inverse trigonometric equation: 𝜃=(𝑥).csc

We can also rewrite the equation in terms of the standard trigonometric equations and use the general solutions for them, csccscsin𝜃=𝑥,1𝜃=1𝑥,𝜃=1𝑥, which follow from the definition of the cosecant function. We can find the solutions in the range 𝜃𝜋2,00,𝜋2 by applying the inverse sine function; we can write a solution as 𝜃=1𝑥.sin

The reason we do not have solutions for the original trigonometric equation in full range for the inverse sine function (i.e., 𝜃𝜋2,𝜋2), is because the cosecant function is defined as the reciprocal of the sine function, so we want to exclude the solution points with sin𝜃=0, which appears in the denominator.

Thus, using the solutions to the trigonometric equations, we can also determine the solutions to the reciprocal trigonometric equations either by using their definitions directly or by using the inverse trigonometric and inverse reciprocal trigonometric functions, which are related by sincsccossectancotifcotif𝑥=1𝑥,𝑥=1𝑥,𝑥=1𝑥𝑥>0,1𝑥𝜋𝑥<0, where 𝑥0. This means we can use the CAST diagram for the reciprocal trigonometric functions.

Now, let’s consider an example where we determine the solutions to a trigonometric equation inolving reciprocal trigonometric function using the solutions given by the CAST diagram.

Example 1: Solving Trigonometric Equations Involving Special Angles and Quotient Identities

Find the set of values satisfying sincot𝜃𝜃=12, where 0𝜃90.

Answer

In this example, we will find the set of values satisfying a trigonometric equation.

In order to solve the trigometric equation, we will use the definition of the cotangent function: cotcossin𝜃=𝜃𝜃.

Thus, using this with the given trigonometric equation, we have sincotsincossincos𝜃𝜃=𝜃×𝜃𝜃=𝜃.

Now, we have to solve the trigonometric equation cos𝜃=12.

The solutions can be found from the inverse cosine function and the CAST diagram to be 𝜃=12=120cos and 𝜃=36012=360120=240.cos

There are are no solutions in the required range. Thus, the set of values is the empty set: .

As mentioned, while these give solutions to the trigonometric equations, we have to take extra care with the reciprocal trigonometric equations, since we have to avoid solutions with sin𝜃=0 or cos𝜃=0 when these functions are in the denominator of an expression involving reciprocal trigonometric functions. We can use the same solutions in the steps to solve a trigonometric equation; we just have to make sure that we take this into account for the final solution.

Example 2: Solving Trigonometric Equations Involving Special Angles

Find the value of 𝜃 that satisfies csc𝜃2=0, where 𝜃0,𝜋2.

Answer

In this example, we will solve a trigonometric equation involving special angles.

We can rearrange the given trigonometric equation as csc𝜃=2.

We can solve this equation using the definition of the cosecant function in terms of the sine function, cscsin𝜃=1𝜃, to instead find solutions to sin𝜃=22.

Since we require 𝜃0,𝜋2, which contains acute solutions and is a subset of the range of the inverse cosecant function, we can apply the inverse cosecant function directly to obtain 𝜃=22=45.sin

Thus, the value of 𝜃 is 45.

In the next example, we will solve a trigonometric equation by changing the interval over which solutions exist.

Example 3: Solving Trigonometric Equations Involving Special Angles

Find the solution set of 𝜃 that satisfies 3(90𝜃)2=0csc, where 𝜃[0,180].

Answer

In this example, we will solve a trigonometric equation involving special angles.

If we let 𝛼=90𝜃, then we can determine the solution to be csc𝛼=23,

We can solve this equation using the definition of the cosecant function in terms of the sine function, cscsin𝛼=1𝛼, to instead find solutions to sin𝛼=32.

We can apply the inverse sine function for 𝛼[90,90] directly as 𝛼=32=60.sin

Now, we can determine the value of 𝜃 from 𝜃=90𝛼 to be 𝜃=9060=30.

Thus, the solution set is {30}.

The general solutions to the trigonometric equations can be found from the solutions we obtain from the CAST diagram or inverse trigonometric functions, 𝜃, by adding a integer multiple of 360 or 2𝜋, in radians. We do this for all the solutions we obtain, since the trigonometric functions are periodic. Thus, the general solution, ̂𝜃, for 𝑛, is ̂𝜃=𝜃+360𝑛 in degrees or ̂𝜃=𝜃+2𝜋𝑛 in radians.

When solving trigonometric equations, we are usually given a particular range for the angle 𝜃 to determine the solutions, which means we may only need to consider a few values of 𝑛, where appropriate. A solution set is the set of values that contains solutions to the trigonometric equation in the required range.

Now, let’s consider another example where we find the solutions using the inverse sine function, the CAST diagram, and the periodicity of the functions.

Example 4: Solving Trigonometric Equations Involving Special Angles and Reciprocal Identities

Find the set of values satisfying 2𝜃+𝜃𝜃=0sincossec, where 0𝜃<360.

Answer

In this example, we will find the set of values satisfying a trigonometric equation.

In order to solve the trigometric equation, we will use the definition of the secant function: seccos𝜃=1𝜃.

Thus, using this with the given trigonometric equation, we have 2𝜃+𝜃𝜃=2𝜃+𝜃×1𝜃=2𝜃+1.sincossecsincoscossin

Now, we have to solve the trigonometric equation 2𝜃+1=0𝜃=12.sinsin

The general solutions can be found from the inverse sine function and the CAST diagram to be 𝜃=12+360𝑛=30+360𝑛sin and 𝜃=18012=180+30+360𝑛=210+360𝑛.sin

The first expression gives 𝜃=330, for 𝑛=1, and the second expression 𝑥=210, for 𝑛=0. For other integers 𝑛, we would obtain angles outside the required range.

Thus, the set of values is {210,330}.

Now, let’s consider another example where we use the CAST diagram to determine the general solutions to a reciprocal trigonometric equation, but for a specific range.

Example 5: Solving Trigonometric Equations Involving Special Angles

Find the set of values satisfying 3𝜃=1cot given 0<𝜃<360.

Answer

In this example, we will solve a trigonometric equation involving special angles.

We can rearrange the given trigonometric equation as cot𝜃=13.

We can solve this equation using the definition of the cotangent function in terms of the tangent function, cottan𝜃=1𝜃, to instead find solutions to tan𝜃=3.

The general solutions can be found from the inverse tangent function and the CAST diagram to be 𝜃=3+360𝑛=60+360𝑛tan and 𝜃=3+180+360𝑛=60+180+360𝑛=240+360𝑛.tan

The first expression gives 𝜃=60 and the second expression 𝜃=240, for 𝑛=0. For other integers 𝑛, we would obtain angles outside the required range.

Therefore, the set of values is {60,240}.

Now, let’s consider an example where we determine the smallest positive angle using the CAST diagram for a reciprocal trigonometric equation.

Example 6: Solving Trigonometric Equations Involving Special Angles and Periodic Identities

Find 𝜃 in degrees given sec(180+𝜃)=233, where 𝜃 is the smallest positive angle.

Answer

In this example, we will solve a trigonometric equation in which 𝜃 is the smallest angle.

We can solve this equation using the definition of the secant function in terms of the cosine function, seccos𝜃=1𝜃, to instead find solutions to cos(180+𝜃)=32.

The solutions can be found by applying the inverse cosine function and the cast diagram to be 180+𝜃=32+360𝑛=150+360𝑛𝜃=150180+360𝑛=30+360𝑛cos and 180+𝜃=36032+360𝑛=360150+360𝑛=210+360𝑛𝜃=210180+360𝑛=30+360𝑛.cos

Since 𝜃 is the smallest positive angle, the second expression gives 𝜃=30, for 𝑛=0. For the first expression and other integers 𝑛, we would obtain angles that are greater or negative.

Thus, the smallest positive angle for the given trigonometric equation is 30.

Finally, let’s consider an example where we have to first simplify a given trigonometric equation and then determine the general solutions for a recprical trigonometric equation, but for a specific range.

Example 7: Solving Trigonometric Equations Involving Special Angles

Find the set of values satisfying 2(𝜃)(𝜃)+(𝜃)(𝜃)=0sincscseccot given that 0𝜃<360.

Answer

In this example, we will solve a trigonometric equation involving special angles.

In order to solve the trigometric equation, we will use the definition of the cosecant, secant, and cotangent functions: cscsinseccoscotcossin𝜃=1𝜃,𝜃=1𝜃,𝜃=𝜃𝜃.

Thus, using these with the given trigonometric equation, we have 2𝜃𝜃+𝜃𝜃=2𝜃×1𝜃+1𝜃×𝜃𝜃=2+𝜃.sincscseccotsinsincoscossincsc

Thus, we have to solve the trigonometric equation 2+𝜃=0𝜃=2.csccsc

We can solve this equation using the definition of the cosecant function in terms of the sine function to instead find solutions to sin𝜃=12.

The general solutions can be found from the inverse cosine function and the CAST diagram to be 𝜃=12+360𝑛=30+360𝑛sin and 𝜃=18012=180+30+360𝑛=210+360𝑛.sin

The first expression gives 𝜃=330, for 𝑛=1, and the second expression 𝑥=210, for 𝑛=0. For other integers 𝑛, we would obtain angles outside the required range.

Thus, the set of values is {210,330}.

Let us finish by recapping a few important key points from this explainer.

Key Points

  • In order to solve reciprocal trigonometric equations, we can use the definitions of the cosecant, secant, and cotangent functions: cscsinseccoscottancossin𝜃=1𝜃,𝜃=1𝜃,𝜃=1𝜃=𝜃𝜃.
  • For reciprocal trigonometric equations, we can find a solution by either using the inverse reciprocal trigonometric equation or by relating them to the standard trigonometric functions and applying the inverse trigonometric function.
  • After finding the principal value solution, in degrees or radians, we can find the general solution for the trigonometric functions, for 𝑛, using the CAST diagram and the periodicity of the trigonometric functions.
  • The inverse trigonometric and inverse reciprocal trigonometric functions are also related, which means we can use the same CAST diagram for their solutions.

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