INVERSE TRIGONOMETRIC FUNCTIONS

Transcription

1 INVERSE TRIGONOMETRIC FUNCTIONS Inverse Functions Intuitivel, a function f() has an inverse, if the function can be undone That is, if for an we want to find the value of for which f() = and we get one and onl one value of, the function is said to have an inverse Eample Let s start b looking at the functions f() = and g() = Suppose I wanted to find the value that gave f() = and g() = Notice that in the case of f() there is one and onl one value that gives f() =, namel = On the other hand, for g() there are two possible values of that g() =, namel = ±8 So, we see that the function f() does have an inverse (there is, after all, nothing special about the value = ) while the function g() does not have an inverse There is a wa ou can tell if a function has an inverse b looking at its graph First recall the Vertical Line Test which sas that a graph is a function if an line drawn parallel to the -ais intersects the graph at most once From this ou can see that in Eample both f() and g() are functions whereas the graph of = ± is not a function Suppose now that we can graph a function f() and want to graph its inverse Remember to find the inverse of a function we find the -values that come from a given -value Notice this is switching the normal roles of and : normall we find values from a given value Graphicall, we can model this b switching the and coordinates of the points on the graph For this graph to be an inverse, this graph has to be a function, ie, it has to pass the Vertical Line Test Eample Like in Eample let f() = and g() = We alread know that f() does have an inverse function ce we get one and onl one value for each fied ) and the g() does not since at least one value could come from two places Let s check these results with our graphical test Notice, first, that switching and coordinates is the same as reflecting the graph across the line = (ou should convince ourself of this) The reflected graph of f() = looks like a stretchedout S (see Figure ) and passes the Vertical Line Test and so is an inverse On the other hand, the reflected graph of g() is a parabola rotated 90 degrees to the right(see Figure ) This is not an inverse since it fails the vertical line test If a function f() has an inverse we denote the inverse b f () Notation Realize that f () is not the same as [f()] = f() Now that ou have this intuition we can make this Definition Let f : A B be a function where A and B are subsets of R If there eists another function f () : B A for which f(f ()) = and f (f()) = then f () is called the inverse function of f()

2 INVERSE TRIG 0 Figure On the left is the graph of g() = and on the right is that graph reflected around the line = Figure On the left is the graph of f() = and on the right is that graph reflected around the line = Since the Vertical Line Test tells us whether or not a graph is a function, it would be nice to have a similar test that tells us more quickl whether or not the graph of a function has an inverse Let s eamine what we have been doing thus far We have a graph, reflect it across the line = and appl the Vertical Line Test If, on the other hand, we were to draw a horizontal line on the original, nonreflected

3 INVERSE TRIG graph, we would get the same information This is because when we reflect the graph and the horizontal line across = we get a vertical line going through the reflected graph The number of times this horizontal line intersects the original graph, is the same as the number of times the resulting vertical line intersects the reflected graph In short, the Horizontal Line Test tells us a quick wa to tell if a graph has an inverse: if an horizontal line intersects a graph no more than once, the function that generated that graph has an inverse Eample Before we go on, we note that while g() = does not have an inverse function (see Eample ), if we define G() = for positive we see that G() does have an inverse, since when G() passes the Horizontal Line Test In general, then, if we can find a part of the graph that does pass the Horizontal Line Test, and restrict our attention to that domain, we can invert functions Now, as the title suggests, we will investigate sin(), cos(), tan() and their inverses We ll treat the case of sin() intuitivel and epect ou to do the same with cos() and tan() Since sin() does not pass the Horizontal Line Test (see Figure ), it does not have an inverse All is not lost, however, because of what we did in Eample Remember, in the eample we limited our attention to a part of f() s domain that would have an inverse: ie, a part that would pass the Horizontal Line Test There are several parts to which we could limit our attention For eample, values between and π, or values between π and π, etc B convention, we limit [ our attention to the first interval, π ] So, we define Sin() = sin() for in [, π ] This function Sin() has an inverse and we denote it b Sin () Similar arguments and conventions allow us to make this Definition The inverse trigonometric functions Sin (), Cos () and Tan () are defined as [ = Sin () for in [, ] if and onl if sin() = for in, π ] = Cos () for in [, ] if and onl if cos() = for in [0,[ π] = Tan () for in (, ) if and onl if Tan() = for in, π ] Let s do some eamples: Eample Evaluate each epression: () Sin (0), ( () Sin ) ( )) () Sin ( )) 5π () Sin Solution () Let = Sin (0) for in [, π ] Then sin() = 0 so = 0

4 INVERSE TRIG Figure On the left is the graph of h() = sin() and on the right is that graph reflected around the line = ( () Similarl, Sin ) = π ( ) () Note that sin = Then Sin ( sin ( )) 5π () Almost identicall, we get Sin ( )) ( ) = Sin = = Remark Notice for the last two eamples above we entered different values into the function, but got the same thing out This is an artifact of the wa we limited our domain Notice, also, that what these inverse functions do is take a ratio and give an angle Eample 5 Show that cos ( Sin () ) = [ Solution Let = Sin () for in, π ] Then cos ( Sin () ) = cos() Since is either in the first or fourth quadrant, cos() 0 Using the Pthagorean Identit cos () + sin () = and the fact that for this particular sin() = we obtain cos ( Sin () ) = cos() = + sin () = + We chose the positive part of the square root function because of the wa we chose the at the beginning of the eample Remark Remember that cos () = [cos()] but Cos () [Cos()]

5 INVERSE TRIG 5 Remark Other notation for Sin () is commonl used Eg, Sin () is often written as one of sin (), Arcsin() or arcsin() Evaluate without a calculator ( ) () Sin () Tan () () Cos () ( )) π () Sin ( ( )) (5) cos Sin ( ( 5 )) () tan Cos Problems Epress without using trig or inverse trig functions (see Eample 5) () sin(cos ( ()) ( )) () sec Cos