Chapter 8: Trig Equations and Inverse Trig Functions

Transcription

1 Haberman MTH Section I: The Trigonometric Functions Chapter 8: Trig Equations and Inverse Trig Functions EXAMPLE : Solve the equations below: a sin( t) b sin( t) 0 sin a Based on our experience with the sine function, we know that know that t is a solution to sin( t), so we We also know that the sine function is periodic with period, so it s values repeat every units Thus, t is also a solution to sin( t) In fact, t k is a solution for every integer k Since t k is a solution for every integer, this represents infinitely many solutions, but it still doesn t represent all of the solutions; see Figure y The green dots represent points with horizontal coordinates of the form Figure : The graph of y sin( t) and the line t k, k The other instances where the blue and purple graphs intersect are also solutions to the equation sin( t) but they are NOT represented by t k It should be clear after studying Figure that we are missing lots of solutions Notice that one of the solutions we are missing is just as close to as our original solution, t, is to 0 (Recall the identity sin( t) sin t that we first noticed in Chapter )

2 Thus, t is a solution, and the rest of the solutions have the form t k, k ; see Figure So the complete solution is t k for all k or t k for all k Figure : The red dots represent points with horizontal coordinates of the form t k, k, while the green dots represent points with horizontal coordinates of the form t k, k The orange and green dots together represent all of the solutions to sin( t) b Unlike in part a, we don t know what input for sin( t) is related to the output 0 Clearly the solutions to sin( t) 0 are not standard values for which we have memorized the sine values In order to solve sin( t) 0, we need a way to undo sine To clarify this concept, let s consider an analogous situation: How do we solve x 0? In order to solve the cube-root: x 0 we need a way to undo cubing In this case, we can use x 0 x x 0 0 Notice that the cube-root function is the inverse of the cubing function; of course this makes sense since inverse functions undo each other (Inverse functions are studied in MTH ) So we need to construct an inverse for the sine function in order to solve equations like sin( t) 0

3 The reason we need to construct the inverse of the sine function is that the sine function isn t one-to-one since it does not pass the horizontal line test; in Figure, above, notice how the graph of y sin( t) intersects the horizontal line y many times Since y sin( t) isn t one-to-one, it doesn t have an inverse function Since we really want an inverse sine function so that we can solve equations like sin( t) 0, we will restrict the domain of the sine function so that the result is a one-to-one function We want to choose an interval of the domain that contains a oneto-one portion of the graph, and we want to choose and interval that utilizes the entire range of the sine function Following tradition, we well choose the interval, In Figure, this interval of the sine function is highlighted; notice that this on this interval, the function is one-to-one and has the same range as the complete sine function Figure : The interval, of the graph of sin( ) y t ; on this interval, the sine function is one-to-one and has the same range as the complete sine function Recall that when we construct the inverse of a function we need to reverse the rolls of the inputs and the outputs, so that the inputs for the origin function become the outputs for the inverse function, and the outputs for the original become the inputs for the inverse DEFINITION: The inverse sine function, denoted following: y sin ( t), is defined by the If y and sin( y) t, then y sin ( t) By construction, the range of y sin ( t) is,, and the domain is the same as the range of the sine function:, Note that the inverse sine function is often called the arcsine function and denoted y arcsin( t)

4 Key Point: As we ve discussed in Chapter, we can denote powers of trigonometric functions by putting the exponent between the function name and the input variable; for example, sin( t) sin ( t) The definition above implies that inverse function notation looks like the sine function raised to the power (ie, the reciprocal of the sine function), but the reciprocal of a function isn t the same as it s inverse! In order to avoid ambiguous notation, the notation sin ( t) always refers to the inverse function If you want to denote the reciprocal of the sine function, you need to use the notation sin( t) : t sin( ) csc( t) sin( t) but csc( t) sin ( x)! Now we can solve sin( t) 0 and finish part b of the Example We were trying to solve this equation when we realized that we needed to construct the inverse sine function We now possess the tools we ll need to solve the equation, so let s solve it: sin( t) 0 sin( t) sin 0 t sin 0 sin t 088 Apply the sine inverse function to both sides (Note that we can use a calculator to obtain an approximation for should find a button on your calculator labeled sin ) sin 0 ; you Although we've found a solution to the equation, we aren t don t yet! Since it's one-toone, the sine inverse function only gives us one value, but we know that the period nature of the sine function suggests that there are infinitely many solutions to an equation like this (See Figure ; notice how many times the sine function reaches the output 0 ) Figure : The graph of y sin( t) intersecting the line y 0 many, many times Each point of intersection represents a solution to sin( t ) 0

5 We can find all of the solutions by using the solution the inverse sine function gave us as well as the fact that the sine function has period Since the sine function has period units, we know that the outputs repeat every units So if t 088 is a solution, t 088 k for all k must also be solutions This gives us LOTS of solutions, but we are still missing half of them (Recall we had the same problem in part a of the first example in this chapter) In order to get the rest of the solutions, can use the identity sin( t) sin t, and subtract our original solution ( t 088 ) from : t ( 088) k, k Thus, the complete solution is t 088 k for all k or t 088 k for all k EXAMPLE : Solve co s( t) 0 Now we need a way to undo cosine so that we can solve this equation as we solved sin( t) 0 in Example We need an inverse cosine function but, like the sine function, cosine is NOT one-to-one so it doesn t have an inverse function (See Figure ) Figure : The graph of y cos( t) and the line y 0 Clearly, the cosine function is not one-to-one Since we really want an inverse cosine function, we will restrict the domain of the cosine function so that the result is a one-to-one function We want to choose an interval of the domain that contains a one-to-one portion of the graph, and we want to choose and interval that utilizes the entire range of the cosine function Following tradition, we well choose the interval 0, In Figure (below), this interval of the cosine function is highlighted; notice that this on this interval, the function is one-to-one and has the same range as the complete cosine function

6 Figure : The interval 0, of the graph of y cos( t) ; on this interval, the cosine function is one-to-one and has the same range as the complete cosine function Again, recall that when we construct the inverse of a function we need to reverse the rolls of the inputs and the outputs, so that the inputs for the origin function become the outputs for the inverse function, and the outputs for the original become the inputs for the inverse DEFINITION: The inverse cosine function, denoted the following: y cos ( t), is defined by If 0 y and cos( y) t, then y cos ( t) By construction, the range of y cos ( t) is 0,, and the domain is the same as the range of the cosine function:, Note that the inverse cosine function is often called the arccosine function and denoted y arccos( t) Now we can finish Example by solving cos( t) 0 : cos cos( t) 0 t cos t cos( ) 0 t cos 0 (Note that we can use a calculator to obtain an approximation for find a button on your calculator labeled cos ) cos (0) ; you should Although we have found a solution to the given equation, we aren t don t yet! Since it is one-to-one, the cosine inverse function only gives us one value, but we know that the period nature of the cosine function suggests that there are infinitely many solutions to an equation like this

7 7 Since the period of the cosine function is units, we can find another solution by adding any integer-multiple of the solution we found above Thus, t k, k, represents infinitely many solutions to the given equation but it doesn t represent all of the solutions; see Figure 7 below Figure 7: Graph of y cos( t) and the line y 0 The red dots represent points with horizontal coordinates of the form t k, k The other instances where the blue and purple graphs intersect are solutions to the equation co s( t) 0 but they are NOT represented by t k, k It should be clear after studying Figure that we are missing solutions and that one of the solutions we are missing is on the left side of the y-axis just as close to y-axis as our original solution, t (Recall the identity cos( t) cost that we noticed in Chapter ) It should be clear that this solution is t, so we can represent the rest of the solutions with t k Thus, the complete solution to co s( t) 0 is t k or t k for all k Now let s define the inverse tangent function Recall that the tangent function is one-to-one on the interval, ; since the period of tangent is units, this interval represents a complete period of tangent In order to construct the inverse tangent function, we restrict the tangent function to the interval, DEFINITION: The inverse tangent function, denoted y tan ( t), is defined by the following: If y and tan( y) t, then y tan ( t) tan ( t) is, By construction, the range of y, and the domain is the same as the range of the tangent function: Note that the inverse tangent function is often called the arctangent function and denoted y arctan( t)

8 EXAMPLE : a Evaluate sin 8 b Evaluate c Evaluate a To evaluate sin( p) cos (0) tan () sin, we need to find a value, p, such that Our experience tells us that p and p Thus, sin b To evaluate cos (0), we need to find a value, p, such that cos( p) 0 Our experience tells us that p Thus, cos (0) c To evaluate tan (), we need to find a value, p, such that tan( p) Our experience tells us that p Thus, tan () 0 p and p and EXAMPLE : a Evaluate sin sin b Evaluate cos cos c Evaluate tan tan a To evaluate sin sin, we need to first evaluate find sin find a value, p, such that p and that p Thus, sin sin( p), so we need to Our experience tells us Now we can evaluate sin sin si sin sin n :

9 b To evaluate cos cos find a value, p, such that 0 p and p Thus, cos, we need to first evaluate find cos cos( p) Now we can evaluate cos cos cos cs o cos, so we need to Our experience tells us that : 9 c To evaluate tan tan, we need to first evaluate find to find a value, p, such that us that p p tan, so we need and tan( p) Our experience tells Thus, tan Now we can evaluate tan tan tan tan tan : Notice that the answers to all three parts of this example are exactly what we should have expected the answers to be since inverse functions undo each other (We should have studied inverse functions in our previous course-work) But we have to be careful since the inverse sine, cosine, and tangent functions are NOT the inverses of the complete sine, cosine, and tangent functions The next example should help explain why we need to be careful with the inverse trigonometric functions EXAMPLE : a Evaluate sin sin b Evaluate cos cos( ) c Evaluate tan tan

10 0 a Based on what we noticed in the last example and what we know about how inverse functions undo each other, we might assume that sin sin is equal to, but this is NOT true (Notice that it can t possibly be true since the answer to this question is an output for the inverse sine function and isn t in the range of this function!) So sin sin sin sin sin sin since since sin and is equal to, not, since isn t in the range of y sin ( t) b Since inverse functions undo each other, we might assume that cos cos equal to, but this is NOT true (Notice that it can t possibly be true since the answer to this question is an output for the inverse cosine function and isn t in the range of this function!) cos cos co cos since cos 0 since s 0 and 0 0 is cos cos So is equal to 0, not, since isn t in the range of y cos ( t) c Since inverse functions undo each other, we might assume that tan tan equal to, but this is NOT true (Notice that it can t possibly be true since the answer to this question is an output for the inverse tangent function and isn t in the range of this function!) tan n tan tan tan ta since since and is So tan tan y tan ( t) is equal to, not, since isn t in the range of

11 EXAMPLE : Solve the equations below: a cos( t) b sin( x) c tan( x) a cos( t) cos( t) cos( t) co cs o cos For the next step, note that s t k or t k for all k See the solution to Example, above, to review how to find all of the solutions to an equation involving cosine b sin( x) sin( x) sin( x) sin or sin x sin k x sin k for all k See the solution to parts a and b of the Example, above, to review how to find all of the solutions to an equation involving sine c tan( x) tan( x) tan( x) n For the next step, note t an tan ta hat t x k for all k Notice that we add k (rather than k ) to our solutions since, unlike sine and cosine, the period of tangent is units

12 EXAMPLE 7: Solve the equations below: a sin( x) b cos( t) c 0 tan( t) 0 a Notice that the trigonometric function involved in the given equation is sin( x ), and recall that sin( x ) has period units, ie, the values for sin( x ) repeat every units This means that once we find a solution to the given equation we ll be able to add to it any integer multiple of and obtain another solution Thus, we should expect the phrase k for all k to be involved in our solutions You ll see in the work below that we add k to our solutions after applying the inverse sine function since the period of the sine function is units In the last step, we finish solving for x and obtain the desired period-shift of k units sin( x) sin( x) sin( x) in in or x k x k for all k or or 8 8 sin s For the next step, note that s x k x k for all k x k x k for all k b Notice that the trigonometric function involved in the given equation is co s( t), and recall that cos( t) has period units, ie, the values for co s( t) repeat every units This means that once we find a solution to the given equation we ll be able to add to it any integer multiple of and obtain another solution Thus, we should expect the phrase k for all k to be involved in our solutions You ll see in the work below that we add k to our solutions after applying the inverse cosine function since the period of the cosine function is units In the last step, we finish solving for t and obtain the desired period-shift of k units

13 cos( t) cos( t) cos( t) or t k t k for all k or k or k 9 9 cos cos For the next step, note that cos t k t k for all k t t for all k c Notice that the trigonometric function involved in the given equation is tan( t), and recall that tan( t) has period units, ie, the values for tan( t) repeat every units This means that once we find a solution to the given equation we ll be able to add to it any integer multiple of and obtain another solution Thus, we should expect the phrase k for all k to be involved in our solutions You ll see in the work below that we add k to our solutions after applying the inverse tangent function since the period of the tangent function is units In the last step, we finish solving for t and obtain the desired period-shift of k units 0 tan( t) 0 tan( t) tan tan( t) tan t tan k for all k tan k t for all k EXAMPLE 8: Solve the equation cos( x) 0 for x cos( x) 0 cos cos( x) cos( x) cos

14 or x cos k x cos k for all k or x cos k x cos k for all k x cos k x cos k for all k or Notice that we were asked to find the solutions in the interval x, so we need to find which of the infinitely many solutions we ve found are in the interval It might help if we approximate the values we found above: x cos k 0 k x cos k 0 k for all k or We know that so we need to find values that satisfy the equation as above and are between and k : x 0 ( ) or x 0 ( ) 7 Only 7 is in the desired interval k 0 : x 0 0 or x Both of these values are in the desired interval k : x 0 or x 0 7 Only 7 is in the desired interval k : x 0 or x Neither of these values is in the desired interval We could try more k-values, but we can tell from the work we ve done thus far that no other k-values will give us solutions that are in the interval x Thus, the solution to cos( x) 0 for x is x 7, x 0, x 0, or x 7