CK-12 Trigonometry
Page 17
The value of in the equation is and this represents the amplitude of the graph. Notice that the graph of has undergone a vertical stretch of and now has an amplitude of this same value. The amplitude of a cosine curve is one half of the difference between the maximum value and the minimum value of the graph. In this particular graph the amplitude is . The amplitude is always the .
The value of will affect the phase shift of the graph. In other words, the first value will no longer be zero. If is negative, the graph will undergo a horizontal translation to the right. If is positive, the graph will undergo a horizontal translation to the left. The entire curve will slide horizontally along the axis.
Notice that in the graph of , the graph of was translated horizontally and to the right along the axis. The translation is to the right because will only remain negative if the value of is indeed positive. Therefore the phase shift of is .
As the graph of can be stretched or shrunk vertically, it can also be stretched or shrunk horizontally. This means that the period of can be increased or decreased. This transformation depends upon the value of .
Notice that when the period was increased to , the graph of became two cycles of the curve. However, when the graph of was drawn, only one cycle of the curve was produced. This is due to the fact that the one cycle of the curve was horizontally stretched by a factor of . The period of is the period of one cycle divided by the value of which in this case is . This yields a period of .
The final transformation that exists is a vertical reflection. The graph is reflected in the axis. This transformation is denoted by a negative sign before .
In the above figure, the graph of is vertically reflected across the axis. This reflection is indicated by the negative sign in the equation . This negative sign can be thought of as being before the coefficient of the letter , which in this case is understood as being one. The reflection has no affect on the amplitude, the sinusoidal axis, the phase shift or the period.
All of these graphs were drawn using values in degrees. The same results would occur for values in radian measure. It is time to apply these concepts to solve a problem.
Example 1:
Sketch the graph of .
Solution:
From the graph it can be seen that the period is
The equation of the sinusoidal axis is
The phase shift is
The amplitude is
These parts of the sinusoidal curve are directly related to the transformations of .
Amplitude
Sinusoidal axis
Phase Shift
Example 2:
For the following graph, list the transformations of and write an equation to model the graph.
Solution:
An equation, in standard form, to model this graph is . Now, we will revisit the problem that was posed at the beginning of the lesson.
A remote control helicopter was being tested for its consistent flying ability. Under correct monitoring; the helicopter could fly up and down in a sinusoidal pattern. To demonstrate this movement, a graph was drawn to show the helicopter’s height at various times. The graph showed that the helicopter reached its maximum height of in and at it was at its minimum height of . Write an equation to model this problem.
Solution:
The first step in writing an equation to model this situation would be to sketch a graph of the given information.
Table of Values of Equation
The five critical points that are used to plot the graph of the equation are highlighted in blue in the above table of values. The table of values was generated by using technology(software program Autograph).
Some students will find it necessary to draw the entire graph while others may need only to plot the given points as shown below:
Whichever graph is drawn, the students will now have to obtain an equation that will model this situation. The equation can be presented in degree measure or in radian measure. The results will be the same.
or
The following is the graph drawn in radian measure.
Lesson Summary
In this lesson, you have reviewed the sinusoidal curve of and the transformations associated with it. In addition to reviewing these concepts, you have also graphed and all of the transformations individually to determine the affect that and has on the graph of . To demonstrate your understanding of these affects, you applied your knowledge to a real world problem.
Points to Consider
Is it possible to solve the for in terms of in order to obtain values for given values for ?
Review Questions
For the following graph, list the transformations of and write an equation to model the graph.
Review Answers
The equation that would model this graph is
Solving for Particular Values in Trigonometric Equations
Learning objectives
A student will be able to:
Use the equation to solve for in terms of .
Determine the value of when given a specific value for .
Introduction
The equation is an equation that is used to determine values for when specific values of are given. However, in real world applications, it is often necessary to determine values for when specific values for are given. To explore this further, we will return to the problem presented in lesson 5.1. As an extension to the problem, we will calculate the first time at which the helicopter reaches an altitude of .
Once again, here is the problem with the added extension:
A remote control helicopter was being tested for its consistent flying ability. Under correct monitoring; the helicopter could fly up and down in a sinusoidal pattern. To demonstrate this movement, a graph was drawn to show the helicopter’s height at various times. The graph showed that the helicopter reached its maximum height of in and at it was at its minimum height of . At what time will the helicopter first reach an altitude of ?
Using the equation y = c + a cos b(x - d) to solve for x in terms of y.
The equation used to model this problem was in the form and was determined to be . The format of the equation is such that any value for can be readily determined given any specific value for . To calculate , you need only substitute the given value for and proceed with the calculations as shown in the equation. However, such is not the case to determine a value for . To calculate a value of using involves a great deal of mathematical manipulation. Before beginning the task of calculating the value of , the equation will be solved for in terms of . With the equation expressed in terms of , the chance of making errors in the calculations should be minimized.
Similar manipulations can be done to solve for in the general forms of the other trigonometric functions such as
Now that we have manipulated the equation to facilitate solving for , we shall proceed to deal with the extension of the previous problem.
Example 1:
Solution:
The helicopter reaches a height of for the first time at approximately . Due to the fact that not all of the given decimal places were used while performing the calculations, the answer is an approximate, equal answer.
The result can be checked by using the graphing calculator.
Lesson Summary
This lesson was intended to demonstrate the advantage of solving the equation for in terms of . When this task was completed, you were given the opportunity to work with this formula to determine a specific value for given a specific value for .
Points to Consider
Can the same process be applied to problems that have measurements given in radian measure?
Review Questions
Geothermal energy is an important natural resource of Iceland. A geology student, doing field work, noticed that steam from a vent flowed in a sinusoidal nature. Twelve seconds into his recording he noted that the steam plume reached its maximum height of above the vent and four seconds later it subsided to its lowest plume of above the vent. Use this informa
tion to determine an equation to model this problem and then use the equation to determine when the steam plume would first reach a height of ?
Review Answers
Although the problem does not request that you sketch a graph, it is often the first step in obtaining a solution. The graph can be obtained quickly using a graphing calculator. From the graph; the transformations can be used to determine the equation to model the situation and the values can be used to calculate a value of .
Using the trace function on the calculator gives an estimate of the value.
The equation that models this problem is
Calculations done on the TI-83
Applications, Technological Tools
Learning objectives
A student will be able to:
Solve real world problems using the equation and the equation solved for in terms of
Introduction
In the previous lesson you learned how to apply to determine a value for given a specific value for . In addition to using the formula, you also explored the use of the graphing calculator to represent the problem graphically as well as to confirm your answer. In this lesson, you will solve another real world problem using the techniques previously presented.
Examples
Example 1:
While on a cruise in the Caribbean, I noticed a dolphin swimming along side of the ship. He was consistently reaching a height of out of the water while diving to a depth of only . I decided to begin timing his jumping actions. At four seconds, he was at the top of his leap and every three seconds thereafter. Find a model to represent the jumping pattern of the dolphin and use the model to determine the time that the dolphin was at a height of .
Equation:
Using the trace function of the calculator, the dolphin was at a height of at .
Calculations done on the TI-83
Lesson 5
Solving Trigonometric Equations Analytically
Learning objectives
A student will be able to:
Use the fundamental trigonometric identities to solve trigonometric equations and to express trigonometric expressions in simplest form.
Introduction
By now we have seen trigonometric functions represented in many ways: Ratios between the side lengths of right triangles, as functions of coordinates as one travels along the unit circle and as abstract functions with inverses and graphs. The applications thus far have been mainly computational. Now it is time to make use of the properties of the trigonometric functions to gain knowledge of the connections between the functions themselves. The patterns of these connections can be applied to simplify trigonometric expressions and to solve trigonometric equations.
Example 1:
Simplify the following expressions using the basic trigonometric identities:
a.
b.
c.
Solution:
a.
b.
c.
In the above examples, the given expressions were simplified by applying the patterns of the basic trigonometric identities.
Example 2:
Without the use of technology, find all solutions to the following equations such that
a.
Solution:
a.
The tangent function is also positive in the third quadrant.
Therefore
Likewise, the tangent function is negative in Quadrants and .
Therefore and
b.
The cosine function is also positive in the fourth quadrant.
Therefore
Likewise the sine function is also positive in the second quadrant.
Therefore
In the above examples, exact values were obtained for the solutions of the equations. These solutions were within the domain that was specified.
Lesson Summary
In this lesson you have learned that the trigonometric functions have relationships that can be applied to both simplifying expressions and to solving trigonometric equations. The results were obtained by applying previously learned trigonometric identities as well as the necessary skills for solving equations.
Points to Consider
Are there other methods for solving equations that can be adapted to solving trigonometric equations?
Will any of the trigonometric equations involve solving quadratic equations?
Review Questions
Solve the equation for .
Solve the equation over the interval
Solve the trigonometric equation for all values of such that
Solve the trigonometric equation such that
Review Answers
Because the problem deals with , the domain values must be doubled, making the domain
The reference angle is
The angles for will be in Quadrants
The values for are needed so the above values must be divided by .
The results can readily be checked by graphing the function. The four results are reasonable since they are the only results indicated on the graph that satisfy
However, cosine is also positive in the fourth quadrant, so the other possible solution for is
In addition, cosine is also negative in the third quadrant, so the other possible solution for is
Now we can confirm the results by graphing the function . For graphing purposes, the function was entered as
Our results have been confirmed from the graph such that and .
By graphing the function the above results can be confirmed.
Confirming the values for may be easier if the values were converted to decimal form.
Since the interval is , we must consider the values for
Once again, the values for can be confirmed by graphing the function
Solve Trig Equations (Factoring)
Learning objectives
A student will be able to:
Solve trigonometric equations by factoring
Introduction
A trigonometric equation is an equation involving a trigonometric function. If the equation is true for all values of the variable, it is an identity. The same methods that are applied to solve other equations are used to solve trigonometric equations. The algebra skills like factoring and substitution that are used to solve various equations are very useful when solving trigonometric equations. As with algebraic expressions, one must be careful to avoid dividing by zero during these maneuvers. Most often these equations are solved for principal values of the variable. These are the values for the variable that are in the domain of the trigonometric function.
Example 1:
Solve for principal values of
Solution:
Example 2:
Solve for all values of .
Solution:
where is any integer
Example 3:
Solve for principal values of .
Solution:
There is no solution since sin is in the interval
Some trigonometric equations have no solutions. This means that there is no replacement for the variable that will result in a true expression.
Lesson Summary
In this lesson you learned how to apply the strategies used in algebra to solve equations to solving trigonometric equations. This lesson dealt with applying the skills required to factor both linear and quadratic expressions.
Points to Consider
Is there a way to solve a trigonometric equation that will not factor?
Is substitution of a function with an identity a feasible approach to solving a trigonometric equation?
Review Questions
Solve for principal values of .
Solve for principal values of .
Find all the solutions for the trigonometric equation over the interval
Solve the trigonometric equation over the interval
Review Answers
or
and
and
Cosine is positive in Quadrants
and
Does not exist
The sine function is also positive in the second quadrant. Therefore the value of is also
Vocabulary
Principal Value - Values for the variable that are in the domain of the trigonometric function.
Solve Equations (Using Identities)
Learning objectives
A student will be able to:
Use trigonometric identities to write trigonometric expressions in terms of one trigonometric function by using the identities for the purpose of solving the equation.