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CK-12 Trigonometry

Page 19

by CK-12 Foundation


  Solution:

  Let .

  The solution can be verified graphically:

  Lesson 7

  Solving Trigonometric Equations Using Inverse Notation

  Learning objectives

  A student will be able to:

  Solve trigonometric equations using inverse notation

  Introduction

  Many trigonometric equations use inverse trigonometric functions to obtain a solution. An inverse trigonometric function can be written by using as an exponent for the function or by using the word ‘arc’ before the function. When solving equations, to avoid confusing the exponent of as meaning a reciprocal function, it is recommended that arccos, arcsin, arctan, etc. be used.

  It is often necessary to express a functional relationship with in terms of . For example:

  a. is read as “ is the angle whose cosine is .” In this case, .

  b. is read as “ is the angle whose tangent is .” In this case, .

  c. is read as “ is the angle whose cosecant is .” In this case, , or

  Following the above examples, means that . Using this relationship means that there is an unlimited number of possible values for for a given value of in . For , we know that and . In fact, for , and , just to name a few. To have a properly defined function, there must be only one value of the dependent variable for a given value of the independent variable. In order to have only one value of for each value of in the domain of the inverse trigonometric functions, it is not possible to include all values of in the range. For this reason, the range of each of the inverse trigonometric functions is defined as:

  Therefore, is the only value of the function that is acceptable since it is the only one that lies within the defined range. The value is outside the defined range for . A second quadrant angle cannot be chosen for , since its sine is also positive and this would lead to ambiguity. The sine is negative for fourth quadrant angles, and to have a continuous range of values, all angles in this quadrant would be expressed in the form of negative angles. The same concept applies to determining values for . However, the range for cannot be chosen on this way since the cosine of a fourth quadrant angle is also positive. To maintain a continuous range of values for , the second quadrant angles are chosen for negative values of .

  The following graphs of the inverse trigonometric functions will show the domains and ranges. The graph of the inverse sine function is obtained by first graphing along the axis and then highlighting the section of the curve within the restricted range . The following graphs are simply another view of the inverse trigonometric functions:

  The following examples will help to develop a clearer understanding of the values and the meanings of the inverse trigonometric functions.

  Example 1:

  Evaluate each of the following expressions without using technology. The unit circle (special angles) can be used.

  a.

  b.

  c.

  Solution:

  a. An angle in the fourth quadrant that lies within the restricted range is or

  b. An angle in the first quadrant that lies within the restricted range is or

  c. An angle in the second quadrant that lies within the restricted range is or

  Example 2:

  Using technology, find the value in radian measure, of each of the following:

  a.

  b.

  c.

  Solution:

  a.

  b.

  c.

  Example 3:

  Find the value of and of

  Solution:

  arcsin is a first quadrant angle equal to . The next step is to find .

  is .

  which is a first quadrant angle. The next step is to find

  is .

  Lesson Summary

  In this lesson you learned that inverse trigonometric equations could be solved by using function notation. In order to determine the correct values when evaluating, the restricted ranges of the inverse trigonometric functions must be considered. The restricted ranges were also presented graphically to enhance your understanding of the inverse trigonometric functions.

  Points to Consider

  Is it possible to use other trigonometric identities to sole inverse trigonometric equations?

  Review Questions

  Given , solve for .

  Find

  Solve the trigonometric equation over the interval

  Solve the trigonometric equation such that

  Review Answers

  Since the values of are restricted, so are the values of .

  A first quadrant angle Or using technology

  The graph of the cosine function is one-to-one on the interval

  If we restrict the domain of the cosine function to that interval , we can take the arccosine of both sides of each equation.

  This is the solution within the restricted range.

  However, this is the reference angle and we know that cosine is also positive in the fourth quadrant. This gives another answer of that is in the interval To include all real solutions, the solutions would repeat every Therefore the solutions would be written as or where is any integer.

  The graph of the cosine function is one-to-one on the interval . If we restrict the domain of the cosine function to that interval, we can take the arccosine of both sides of each equation.

  However, this is the reference angle and we know that cosine is also positive in the fourth quadrant. This gives another answer of that is in the interval .

  To include all real solutions, the solutions would repeat every . Therefore the solutions would be written as or , where is any integer.

  Solving Trigonometric Equations Using Inverse Functions

  Learning objectives

  A student will be able to:

  Solve trigonometric equations using inverse functions

  Introduction

  In this lesson, you will use the inverse trigonometric functions of arcsine, arccosine and arctangent to solve trigonometric equations. These types of questions have been presented in other lessons of the chapter, but it as good idea to practice more of these problems.

  Example 1:

  Solve the following trigonometric equation

  Solution:

  or

  Example 2:

  Solve for .

  Solution:

  Example 3:

  Use inverse trigonometric functions to solve the following equation in terms of .

  Solution:

  Lesson Summary

  In this lesson you have solved trigonometric equations by using the inverse trigonometric functions. As you have seen, it is not always necessary to obtain a numerical value as an answer. The second and third examples display this fact very well. These examples are also very good problems for keeping your formula manipulation skills keen.

  Points to Consider

  Is it possible to solve trigonometric equations by using trigonometric identities?

  Review Questions

  The electric current in a certain circuit is given by Solve for .

  Review Answers

  Solving Inverse Equations Using Trigonometric Identities

  Learning objectives

  A student will be able to:

  Solve trigonometric equations using identities

  Introduction

  When solving an inverse trigonometric equation, it is often necessary to apply one or more of the trigonometric identities that you have studied in previous lessons. Applying these identities involves making a substitution for one or more terms in the given equation. Once the substitutions have been made, the equation can be readily solved.

  Example 1:

  Use the triangle to find .

  Solution:

  From the triangle

  Example 2:

  Solve for over the interval

  Solution:

  The solution is over the interval

  Lesson Summary

  In this lesson you le
arned that substituting a trigonometric identity into an equation made it possible to solve. Without these identities, a solution would be impossible. In order to be successful when solving these equations, you will have to remember the various identities.

  Points to Consider

  Is solving these equations by using trigonometric identities applicable to real world problems?

  Review Questions

  The intensity of a certain type of polarized light is given by the equation . Solve for .

  The following diagram represents the ends of a water-trough. The ends are actually isosceles trapezoids. Determine the maximum value of the trough and the value of that maximizes the volume.

  Review Answers

  Hint: The volume is times the area of the end. The end consists of two congruent right triangles and one rectangle. The area of each right triangle is and that of the rectangle is .

  The maximum value is approximately and occurs when .

  Chapter 5: Triangles and Vectors

  The Law of Cosines

  Learning Objectives

  A student will be able to:

  Understand how the Law of Cosines is derived.

  Apply the Law of Cosines when you know two sides and the included of an oblique (non-right) triangle (SAS).

  Apply the Law of Cosines when you know all three sides of an oblique triangle.

  Identify accurate drawings of oblique triangles.

  Use the Law of Cosines in real-world and applied problems.

  Introduction

  Real-World Application:

  An architect is designing a kitchen for a client. When designing a kitchen, the architect must pay special attention to the placement of the stove, sink, and refrigerator. In order for a kitchen to be utilized effectively, these three amenities must form a triangle with each other. This is known as the “work triangle.” By design, the three parts of the work triangle must be no less than apart and no more than apart. Based on the dimensions of the current kitchen, the architect has determined that the sink will be away from the stove and away from the refrigerator. If the sink forms a angle with the stove and the refrigerator, will the distance between the stove and the refrigerator remain within the confines of the work triangle? If he moves the stove so that it is from the sink and makes the fridge from the stove, how does this affect the angle the sink forms with the stove and the refrigerator?

  Up until this point, we have only looked at how to solve problems involving right triangles. We learned to use the Pythagorean Theorem and the trigonometry functions such as sine, cosine, and tangent, to find missing pieces in right triangles. However, not every situation we encounter in life involves a right triangle. Right triangles are really a special case of all triangles. Faced with problems that deal with generalized triangles, including triangles with all acute angles or ones with an obtuse angle and two acute angles we need other, more general, tools. In the application above, we have an obtuse triangle, which means that we cannot use the Theorem of Pythagoras to solve this problem.

  We will refer back to the above application in a little while.

  The Law of Cosines is one tool we can use in certain situations involving all triangles: right, obtuse, and acute.

  The Law of Cosines is a general statement relating the lengths of the sides of any general triangle to the cosine of one of its angles. There are two situations in which we can and want to use the Law of Cosines:

  When we know two sides and the included angle in an oblique triangle and want to find the third side (SAS)

  When we know all three sides in an oblique triangle and want to find one of the angles (SSS)

  In this lesson, we will learn more about the Law of Cosines, how it is derived, and how to apply it to different problems and situations. We will also look at applications involving the Law of Cosines and how it can be useful in finding angles and lengths when other methods (such as measuring) can be unreliable.

  Derive the Law of Cosines

  contains an altitude that extends from and intersects . We will refer to the length of as . The sides of measure and If is long, then measures

  Using the Theorem of Pythagoras, we know that

  We can use a similar process to derive all three forms of the Law of Cosines:

  Note that if either or is then and the Law of Cosines is identical to the Pythagorean Theorem.

  Side of an Oblique Triangle (given the other two sides)

  One case where we can use the Law of Cosines is when we know two sides and the included angle in a triangle (SAS) and want to find the third side.

  Since isn’t a right triangle, we cannot use the Theorem of Pythagoras or trigonometry functions to find the third side. However, we can use the Law of Cosines. First we will look at how to use the Law of Cosines in this situation. Then, we will look back at the baseball application from earlier.

  Example 1:

  Using from above, , and . Find .

  Note that the negative answer is thrown out as having no geometric meaning in this case.

  We will now refer back to the Real-World Application at the beginning of the section.

  Part 1: In order to find the distance from the sink to the refrigerator, we need to know side . To find side , we will need to use the Law of Cosines since we are dealing with an obtuse triangle and thus have no right angles to work with. We know the length two sides: the sink to the stove and the sink to the refrigerator. We also know the included angle (the angle the sink forms with the fridge and the stove) is . This means we have the SAS case and can apply the Law of Cosines.

  Answer: No, the sink and the refrigerator are too far apart by

  Part 2: In order to find how the angle is affected, we will again need to utilize the Law of Cosines since we do not know the measures of any of the angles.

  Answer: The new angle would be , which means it would be less than the original angle.

  Any Angle of a Triangle (given three sides)

  Another situation where we can apply the Law of Cosines is when we know all three sides in a triangle (SSS) and we need to find one of the angles. The Law of Cosines allows us to find any of the three angles in the triangle. First, we will look at how to apply the Law of Cosines in this case, and then we will look at a real-world application.

  Example 2:

  In oblique , , and . Find .

  Since we know all three sides of the triangle, we can use the Law of Cosines to find . It is important to note that we could use the Law of Cosines to find or also.

  Answer: The measure of .

  Real-World Application: Sam is building a retaining wall for a garden that he plans on putting in the back corner of his yard. Due to the placement of some trees, the dimensions of his wall need to be as follows: side , side , and side . At what angle do side and side need to be? Side and side ? Side and side ?

  Part 1: Since we know the measures of all three sides of the retaining wall, we can use the Law of Cosines to find the measures of the angles formed by adjacent walls. We will refer to the angle formed by side and side as , the angle formed by side and side as , and the angle formed by side and side as . First, we will find . How far from will angle be?

  Answer: .

  Part 2: Next we will find the measure of using the Law of Cosines.

  Answer: The measure of .

  Part 3: Now that we know two of the angles, we can find the third angle using the Triangle Sum Theorem. Remember that all three angles in a triangle must add up to . We will now find the measure of

  Answer: The measure of .

  Identify Accurate Drawings of General Triangles

  The Law of Cosines can also be used to verify that drawings of oblique triangles are accurate. In a right triangle, we might use the Theorem of Pythagoras to verify that all three sides are the correct length, or we might use trigonometric rations to verify an angle measurement. However, when dealing with an obtuse or acute triangle, we must rely on the Law of Cosines.

  Example 3: In at the right, ,
, And . Is the drawing accurate if it labels as ? If not, what should measure?

  Part 1: We will use the Law of Cosines to check whether or not is .

  Answer: Since , we know that is not

  Part 2: We will know use the Law of Cosines to figure out the correct measurement of .

  Answer: should measure .

  For some situations, it will be necessary to utilize not only the Law of Cosines, but also the Theorem of Pythagoras and trigonometric ratios to verify that a triangle or quadrilateral has been drawn accurately.

  Real-World Application: A builder received plans for the construction of a second-story addition on a house. At the right is the diagram of how the architect wants the roof framed. The builder decides to add a perpendicular support beam from the peak of the roof to the base. He estimates that new beam should be high, but he wants to double-check before he begins construction. Is the builder’s estimate of for the new beam correct? If not, how far off is he?

 

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