10.6 The Inverse Trigonometric Functions

Transcription

1 0.6 The Inverse Trigonometric Functions The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with finding inverses of the (circular) trigonometric functions. Our immediate roblem is that, owing to their eriodic nature, none of the si circular functions is one-to-one. To remed this, we restrict the domains of the circular functions in the same wa we restricted the domain of the quadratic function in Eamle.. in Section. to obtain a one-to-one function. We first consider f() = cos(). Choosing the interval [0,] allows us to kee the range as [, ] as well as the roerties of being smooth and continuous. Restricting the domain of f() = cos() to[0,]. Recall from Section. that the inverse of a function f is ticall denoted f. For this reason, some tetbooks use the notation f () = cos () for the inverse of f() = cos(). The obvious itfall here is our convention of writing (cos()) as cos (), (cos()) as cos () and so on. It is far too eas to confuse cos () with cos() =sec() so we will not use this notation in our tet. Instead, we use the notation f () = arccos(), read arc-cosine of. To understand the arc in arccosine, recall that an inverse function, b definition, reverses the rocess of the original function. The function f(t) = cos(t) takes a real number inut t, associates it with the angle = t radians, and returns the value cos(). Digging deeer, we have that cos() = cos(t) isthe -coordinate of the terminal oint on the Unit Circle of an oriented arc of length t whose initial oint is (, 0). Hence, we ma view the inuts to f(t) = cos(t) as oriented arcs and the oututs as -coordinates on the Unit Circle. The function f, then, would take -coordinates on the Unit Circle and return oriented arcs, hence the arc in arccosine. Below are the grahs of f() = cos() and f () = arccos(), where we obtain the latter from the former b reflecting it across the line =, in accordance with Theorem.. f() =cos(), 0 ale ale reflect across =! switch and coordinates f () =arccos(). But be aware that man books do! As alwas, be sure to check the contet! See age 70 if ou need a review of how we associate real numbers with angles in radian measure.

2 80 Foundations of Trigonometr We restrict g() =sin() in a similar manner, although the interval of choice is,. Restricting the domain of f() =sin() to,. It should be no surrise that we call g () = arcsin(), which is read arc-sine of. g() =sin(), ale ale. reflect across =! switch and coordinates g () =arcsin(). We list some imortant facts about the arccosine and arcsine functions in the following theorem. Theorem 0.6. Proerties of the Arccosine and Arcsine Functions Proerties of F () = arccos() Domain: [, ] Range: [0,] arccos() =t if and onl if 0 ale t ale and cos(t) = cos(arccos()) = rovided ale ale arccos(cos()) = rovided 0 ale ale Proerties of G() = arcsin() Domain: [, ] Range:, arcsin() =t if and onl if ale t ale and sin(t) = sin(arcsin()) = rovided ale ale arcsin(sin()) = rovided ale ale additionall, arcsine is odd

3 0.6 The Inverse Trigonometric Functions 8 Everthing in Theorem 0.6 is a direct consequence of the facts that f() = cos() for 0 ale ale and F () = arccos() are inverses of each other as are g() = sin() for ale ale and G() = arcsin(). It s about time for an eamle. Eamle Find the eact values of the following. (a) arccos (c) arccos (e) arccos cos 6 (b) arcsin (d) arcsin (f) arccos cos 6 (g) cos arccos (h) sin arccos. Rewrite the following as algebraic eressions of and state the domain on which the equivalence is valid. (a) tan (arccos ()) (b) cos ( arcsin()) Solution.. (a) To find arccos, we need to find the real number t (or, equivalentl, an angle measuring t radians) which lies between 0 and with cos(t) =. We know t = meets these criteria, so arccos =. (b) The value of arcsin is a real number t between and with sin(t) =. The number we seek is t =. Hence, arcsin =. (c) The number t = arccos lies in the interval [0,] with cos(t) =. Our answer is arccos =. (d) To find arcsin,weseekthenumbert in the interval, with sin(t) =.The answer is t = 6 so that arcsin = 6. (e) Since 0 ale 6 ale, we could siml invoke Theorem 0.6 to get arccos cos 6 = 6. However, in order to make sure we understand wh this is the case, we choose to work the eamle through using the definition of arccosine. Working from the inside out, arccos cos 6 = arccos. Now, arccos is the real number t with 0 ale t ale and cos(t) =.Wefindt = 6, so that arccos cos 6 = 6.

4 8 Foundations of Trigonometr (f) Since 6 does not fall between 0 and, Theorem 0.6 does not al. We are forced to work through from the inside out starting with arccos cos 6 = arccos. From the revious roblem, we know arccos = 6. Hence, arccos cos 6 = 6. (g) One wa to simlif cos arccos is to use Theorem 0.6 directl. Since is between and, we have that cos arccos = and we are done. However, as before, to reall understand wh this cancellation occurs, we let t = arccos.then, b definition, cos(t) =. Hence, cos arccos = cos(t) =, and we are finished in (nearl) the same amount of time. (h) As in the revious eamle, we let t = arccos so that cos(t) = for some t where 0 ale t ale. Since cos(t) < 0, we can narrow this down a bit and conclude that <t<, so that t corresonds to an angle in Quadrant II. In terms of t, then,weneedtofind sin arccos =sin(t). Using the Pthagorean Identit cos (t)+sin (t) =, we get +sin (t) = or sin(t) =±. Since t corresonds to a Quadrants II angle, we choose sin(t) =. Hence, sin arccos =.. (a) We begin this roblem in the same manner we began the revious two roblems. To hel us see the forest for the trees, we let t = arccos(), so our goal is to find a wa to eress tan (arccos ()) = tan(t) in terms of. Sincet = arccos(), we know cos(t) = where 0 ale t ale, but since we are after an eression for tan(t), we know we need to throw out t = from consideration. Hence, either 0 ale t< or <tale so that, geometricall, t corresonds to an angle in Quadrant I or Quadrant II. One aroach to finding tan(t) is to use the quotient identit tan(t) = sin(t) cos(t). Substituting cos(t) = into the Pthagorean Identit cos (t)+sin (t) = gives +sin (t) =, from which we get sin(t) =±.Sincetcorresonds to angles in Quadrants I and II, sin(t) 0, so we choose sin(t) =.Thus, tan(t) = sin(t) cos(t) = To determine the values of for which this equivalence is valid, we consider our substitution t = arccos(). Since the domain of arccos() is[, ], we know we must restrict ale ale. Additionall, since we had to discard t =, we need to discard = cos = 0. Hence, tan (arccos ()) = is valid for in [, 0) [ (0, ]. (b) We roceed as in the revious roblem b writing t = arcsin() so that t lies in the interval, with sin(t) =. We aim to eress cos ( arcsin()) = cos(t) interms of. Since cos(t) is defined everwhere, we get no additional restrictions on t as we did in the revious roblem. We have three choices for rewriting cos(t): cos (t) sin (t), cos (t) and sin (t). Since we know =sin(t), it is easiest to use the last form: cos ( arcsin()) = cos(t) = sin (t) = Alternativel, we could use the identit: + tan (t) =sec (t). Since =cos(t), sec(t) = =. The reader cos(t) is invited to work through this aroach to see what, if an, di culties arise.

5 0.6 The Inverse Trigonometric Functions 8 To find the restrictions on, we once again aeal to our substitution t = arcsin(). Since arcsin() is defined onl for ale ale, the equivalence cos ( arcsin()) = is valid onl on [, ]. A few remarks about Eamle 0.6. are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that the are one-to-one. One instance of this henomenon is the fact that arccos cos 6 = 6 as oosed to 6. This is the eact same henomenon discussed in Section. when we saw ( ) = as oosed to. Additionall, even though the eression we arrived at in art b above, namel, is defined for all real numbers, the equivalence cos ( arcsin()) = is valid for onl ale ale. This is akin to the fact that while the eression is defined for all real numbers, the equivalence ( ) = is valid onl for 0. For this reason, it as to be careful when we determine the intervals where such equivalences are valid. The net air of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, resectivel. First, we restrict f() = tan() toits fundamental ccle on, to obtain f () = arctan(). Among other things, note that the vertical asmtotes = and = of the grah of f() = tan() become the horizontal asmtotes = and = of the grah of f () = arctan(). f() =tan(), <<. reflect across =! switch and coordinates f () =arctan(). Net, we restrict g() = cot() to its fundamental ccle on (0,) to obtain g () = arccot(). Once again, the vertical asmtotes = 0 and = of the grah of g() = cot() become the horizontal asmtotes = 0 and = of the grah of g () = arccot(). We show these grahs on the net age and list some of the basic roerties of the arctangent and arccotangent functions.

6 8 Foundations of Trigonometr g() =cot(), 0 <<. reflect across =! switch and coordinates g () =arccot(). Theorem 0.7. Proerties of the Arctangent and Arccotangent Functions Proerties of F () = arctan() Domain: (, ) Range:, as!, arctan()! + ; as!, arctan()! arctan() =t if and onl if <t< and tan(t) = arctan() = arccot for >0 tan (arctan()) = for all real numbers arctan(tan()) = rovided << additionall, arctangent is odd Proerties of G() = arccot() Domain: (, ) Range: (0,) as!, arccot()! ; as!, arccot()! 0 + arccot() =t if and onl if 0 <t<and cot(t) = arccot() = arctan for >0 cot (arccot()) = for all real numbers arccot(cot()) = rovided 0 < <

7 0.6 The Inverse Trigonometric Functions 8 Eamle Find the eact values of the following. (a) arctan( ) (b) arccot( ) (c) cot(arccot( )) (d) sin arctan. Rewrite the following as algebraic eressions of and state the domain on which the equivalence is valid. (a) tan( arctan()) (b) cos(arccot()) Solution.. (a) We know arctan( ) is the real number t between and with tan(t) =. We find t =, so arctan( ) =. (b) The real number t = arccot( ) lies in the interval (0,) with cot(t) =. We get arccot( ) = 6. (c) We can al Theorem 0.7 directl and obtain cot(arccot( )) =. However, working it through rovides us with et another oortunit to understand wh this is the case. Letting t = arccot( ), we have that t belongs to the interval (0,) and cot(t) =. Hence, cot(arccot( )) = cot(t) =. (d) We start simlifing sin arctan b letting t = arctan. Then tan(t) = for some <t<. Since tan(t) < 0, we know, in fact, <t<0. One wa to roceed is to use The Pthagorean Identit, +cot (t) =csc (t), since this relates the recirocals of tan(t) and sin(t) and is valid for all t under consideration. From tan(t) =,we get cot(t) =. Substituting, we get + =csc (t) so that csc(t) =±. Since <t<0, we choose csc(t) =,sosin(t) =. Hence, sin arctan =.. (a) If we let t = arctan(), then <t< and tan(t) =. We look for a wa to eress tan( arctan()) = tan(t) in terms of. Before we get started using identities, we note that tan(t) isundefinedwhent = + k for integers k. Dividing both sides of this equation b tells us we need to eclude values of t where t = + k,wherek is an integer. The onl members of this famil which lie in, are t = ±,which means the values of t under consideration are, [, [,.Returning to arctan(t), we note the double angle identit tan(t) = tan(t), is valid for all the tan (t) values of t under consideration, hence we get tan( arctan()) = tan(t) = tan(t) tan (t) = It s alwas a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Eamle Were the identities we used there valid for all t under consideration? A edantic oint, to be sure, but what else do ou eect from this book?

8 86 Foundations of Trigonometr To find where this equivalence is valid we check back with our substitution t = arctan(). Since the domain of arctan() is all real numbers, the onl eclusions come from the values of t we discarded earlier, t = ±. Since = tan(t), this means we eclude = tan ± = ±. Hence, the equivalence tan( arctan()) = holds for all in (, ) [ (, ) [ (, ). (b) To get started, we let t = arccot() so that cot(t) = where 0 <t<. In terms of t, cos(arccot()) = cos(t), and our goal is to eress the latter in terms of. Since cos(t) is alwas defined, there are no additional restrictions on t, so we can begin using identities to relate cot(t) to cos(t). The identit cot(t) = cos(t) sin(t) is valid for t in (0,), so our strateg is to obtain sin(t) in terms of, then write cos(t) = cot(t)sin(t). The identit + cot (t) =csc (t) holds for all t in (0,) and relates cot(t) and csc(t) = sin(t). Substituting cot(t) =, we get + () =csc (t), or csc(t) =± +. Since t is between 0 and, csc(t) > 0, so csc(t) = + which gives sin(t) =. Hence, + cos(arccot()) = cos(t) = cot(t) sin(t) = + Since arccot() is defined for all real numbers and we encountered no additional restrictions on t, we have cos (arccot()) = for all real numbers. + The last two functions to invert are secant and cosecant. A ortion of each of their grahs, which were first discussed in Subsection 0.., are given below with the fundamental ccles highlighted. The grah of =sec(). The grah of =csc(). It is clear from the grah of secant that we cannot find one single continuous iece of its grah which covers its entire range of (, ][[, ) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant functions, we must settle for a iecewise aroach wherein we choose one iece to cover the to of the range, namel [, ), and another iece to cover the bottom, namel (, ]. There are two generall acceted was make these choices which restrict the domains of these functions so that the are one-to-one. One aroach simlifies the Trigonometr associated with the inverse functions, but comlicates the Calculus; the other makes the Calculus easier, but the Trigonometr less so. We resent both oints of view.

9 0.6 The Inverse Trigonometric Functions Inverses of Secant and Cosecant: Trigonometr Friendl Aroach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, resectivel. For f() =sec(), we restrict the domain to 0, [, f() =sec() on 0, [, reflect across =! switch and coordinates f () =arcsec() and we restrict g() =csc() to, 0 [ 0,. g() =csc() on, 0 [ 0, reflect across =! switch and coordinates g () =arccsc() Note that for both arcsecant and arccosecant, the domain is (, ] [ [, ). Taking a age from Section., we can rewrite this as { : }. This is often done in Calculus tetbooks, so we include it here for comleteness. Using these definitions, we get the following roerties of the arcsecant and arccosecant functions.

10 88 Foundations of Trigonometr Theorem 0.8. Proerties of the Arcsecant and Arccosecant Functions a Proerties of F () = arcsec() Domain: { : } =(, ] [ [, ) Range: 0, [, as!, arcsec()! + ; as!, arcsec()! arcsec() =t if and onl if 0 ale t< or <tale and sec(t) = arcsec() = arccos rovided sec (arcsec()) = rovided arcsec(sec()) = rovided 0 ale < or <ale Proerties of G() = arccsc() Domain: { : } =(, ] [ [, ) Range:, 0 [ 0, as!, arccsc()! 0 ; as!, arccsc()! 0 + arccsc() =t if and onl if ale t<0 or 0 <tale arccsc() = arcsin rovided csc (arccsc()) = rovided arccsc(csc()) = rovided ale <0 or 0 <ale additionall, arccosecant is odd a... assuming the Trigonometr Friendl ranges are used. and csc(t) = Eamle Find the eact values of the following. (a) arcsec() (b) arccsc( ) (c) arcsec sec (d) cot (arccsc ( )). Rewrite the following as algebraic eressions of and state the domain on which the equivalence is valid. (a) tan(arcsec()) (b) cos(arccsc())

11 0.6 The Inverse Trigonometric Functions 89 Solution.. (a) Using Theorem 0.8, we have arcsec() = arccos =. (b) Once again, Theorem 0.8 comes to our aid giving arccsc( ) = arcsin = 6. (c) Since doesn t fall between 0 and or and, we cannot use the inverse roert stated in Theorem 0.8. We can, nevertheless, begin b working inside out which ields arcsec sec = arcsec( ) = arccos =. (d) One wa to begin to simlif cot (arccsc ( )) is to let t = arccsc( ). Then, csc(t) = and, since this is negative, we have that t lies in the interval, 0. We are after cot (arccsc ( )) = cot(t), so we use the Pthagorean Identit + cot (t) =csc (t). Substituting, we have + cot (t) =( ), or cot(t) =± 8=±. Since ale t<0, cot(t) < 0, so we get cot (arccsc ( )) =.. (a) We begin simlifing tan(arcsec()) b letting t = arcsec(). Then, sec(t) = for t in 0, [,, and we seek a formula for tan(t). Since tan(t) is defined for all t values under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identit +tan (t) =sec (t). This is valid for all values of t under consideration, and when we substitute sec(t) =, we get + tan (t) =. Hence, tan(t) =±. If t belongs to 0, then tan(t) 0; if, on the the other hand, t belongs to, then tan(t) ale 0. As a result, we get a iecewise defined function for tan(t) tan(t) = (, if 0 ale t<, if <tale Now we need to determine what these conditions on t mean for. Since =sec(t), when 0 ale t<,, and when <tale, ale. Since we encountered no further restrictions on t, the equivalence below holds for all in (, ] [ [, ). tan(arcsec()) = (, if, if ale (b) To simlif cos(arccsc()), we start b letting t = arccsc(). Then csc(t) = for t in, 0 [ 0,, and we now set about finding an eression for cos(arccsc()) = cos(t). Since cos(t) is defined for all t, we do not encounter an additional restrictions on t. From csc(t) =, we get sin(t) =, so to find cos(t), we can make use if the identit cos (t)+sin (t) =. Substituting sin(t) = gives cos (t)+ =. Solving, we get r 6 6 cos(t) =± 6 = ± Since t belongs to, 0 [ 0,, we know cos(t) 0, so we choose cos(t) = 6. (The absolute values here are necessar, since could be negative.) To find the values for

12 80 Foundations of Trigonometr which this equivalence is valid, we look back at our original substution, t = arccsc(). Since the domain of arccsc() requires its argument to satisf, the domain of arccsc() requires. Using Theorem., we rewrite this inequalit and solve to get ale or. Since we had no additional restrictions on t, the equivalence cos(arccsc()) = 6 holds for all in, [, Inverses of Secant and Cosecant: Calculus Friendl Aroach In this subsection, we restrict f() =sec() to 0, [, f() =sec() on 0, [, reflect across =! switch and coordinates f () =arcsec() and we restrict g() =csc() to 0, [,. g() =csc() on 0, [, reflect across =! switch and coordinates g () =arccsc() Using these definitions, we get the following result.

13 0.6 The Inverse Trigonometric Functions 8 Theorem 0.9. Proerties of the Arcsecant and Arccosecant Functions a Proerties of F () = arcsec() Domain: { : } =(, ] [ [, ) Range: 0, [, as!, arcsec()! ; as!, arcsec()! arcsec() =t if and onl if 0 ale t< or ale t< and sec(t) = arcsec() = arccos for onl b sec (arcsec()) = rovided arcsec(sec()) = rovided 0 ale < or ale < Proerties of G() = arccsc() Domain: { : } =(, ] [ [, ) Range: 0, [, as!, arccsc()! + ; as!, arccsc()! 0 + arccsc() =t if and onl if 0 <tale or <tale arccsc() = arcsin for onl c csc (arccsc()) = rovided arccsc(csc()) = rovided 0 <ale or <ale a... assuming the Calculus Friendl ranges are used. b Comare this with the similar result in Theorem 0.8. c Comare this with the similar result in Theorem 0.8. and csc(t) = Our net eamle is a dulicate of Eamle The interested reader is invited to comare and contrast the solution to each. Eamle Find the eact values of the following. (a) arcsec() (b) arccsc( ) (c) arcsec sec (d) cot (arccsc ( )). Rewrite the following as algebraic eressions of and state the domain on which the equivalence is valid. (a) tan(arcsec()) (b) cos(arccsc())

14 8 Foundations of Trigonometr Solution.. (a) Since, we ma invoke Theorem 0.9 to get arcsec() = arccos =. (b) Unfortunatel, is not greater to or equal to, so we cannot al Theorem 0.9 to arccsc( ) and convert this into an arcsine roblem. Instead, we aeal to the definition. The real number t = arccsc( ) lies in 0, [, and satisfies csc(t) =. The t we re after is t = 7 6, so arccsc( ) = 7 6. (c) Since lies between and, we ma al Theorem 0.9 directl to simlif arcsec sec =. We encourage the reader to work this through using the definition as we have done in the revious eamles to see how it goes. (d) To simlif cot (arccsc ( )) we let t = arccsc ( ) so that cot (arccsc ( )) = cot(t). We know csc(t) =, and since this is negative, t lies in,. Using the identit + cot (t) =csc (t), we find + cot (t) =( ) so that cot(t) =± 8=±. Since t is in the interval,, we know cot(t) > 0. Our answer is cot (arccsc ( )) =.. (a) We begin simlifing tan(arcsec()) b letting t = arcsec(). Then, sec(t) = for t in 0, [,, and we seek a formula for tan(t). Since tan(t) is defined for all t values under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identit +tan (t) =sec (t). This is valid for all values of t under consideration, and when we substitute sec(t) =, we get + tan (t) =. Hence, tan(t) =±. Since t lies in 0, [,, tan(t) 0, so we choose tan(t) =. Since we found no additional restrictions on t, the equivalence tan(arcsec()) = holds for all in the domain of t = arcsec(), namel (, ] [ [, ). (b) To simlif cos(arccsc()), we start b letting t = arccsc(). Then csc(t) = for t in 0, [,, and we now set about finding an eression for cos(arccsc()) = cos(t). Since cos(t) is defined for all t, we do not encounter an additional restrictions on t. From csc(t) =, we get sin(t) =, so to find cos(t), we can make use if the identit cos (t)+sin (t) =. Substituting sin(t) = gives cos (t)+ =. Solving, we get r 6 6 cos(t) =± 6 = ± If t lies in 0,, then cos(t) 0, and we choose cos(t) = 6.Otherwise,tbelongs to, in which case cos(t) ale 0, so, we choose cos(t) = 6 This leads us to a (momentaril) iecewise defined function for cos(t) 8 >< cos(t) = >: 6, if 0 ale t ale 6, if <tale

15 0.6 The Inverse Trigonometric Functions 8 We now see what these restrictions mean in terms of. Since =csc(t), we get that for 0 ale t ale,, or. In this case, we can simlif = so cos(t) = 6 = 6 Similarl, for <tale, we get ale, or ale. In this case, =, so we also get cos(t) = = = ( ) 6 Hence, in all cases, cos(arccsc()) =, and this equivalence is valid for all in the domain of t = arccsc(), namel, [, 0.6. Calculators and the Inverse Circular Functions. In the sections to come, we will have need to aroimate the values of the inverse circular functions. On most calculators, onl the arcsine, arccosine and arctangent functions are available and the are usuall labeled as sin, cos and tan, resectivel. If we are asked to aroimate these values, it is a simle matter to unch u the aroriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to emlo some ingenuit, as our net eamle illustrates. Eamle Use a calculator to aroimate the following values to four decimal laces. (a) arccot() (b) arcsec() (c) arccot( ) (d) arccsc. Find the domain and range of the following functions. Check our answers using a calculator. (a) f() = arccos (b) f() = arctan (). (c) f() = arccot + Solution.. (a) Since > 0, we can use the roert listed in Theorem 0.7 to rewrite arccot() as arccot() = arctan. In radian mode, we find arccot() = arctan (b) Since, we can use the roert from either Theorem 0.8 or Theorem 0.9 to write arcsec() = arccos.69.

16 8 Foundations of Trigonometr (c) Since the argument is negative, we cannot directl al Theorem 0.7 to hel us find arccot( ). Let t = arccot( ). Then t is a real number such that 0 <t< and cot(t) =. Moreover, since cot(t) < 0, we know <t<. Geometricall, this means t corresonds to a Quadrant II angle = t radians. This allows us to roceed using a reference angle aroach. Consider, the reference angle for, as ictured below. B definition, is an acute angle so 0 < <, and the Reference Angle Theorem, Theorem 0., tells us that cot( ) =. This means = arccot() radians. Since the argument of arccotangent is now a ositive, we can use Theorem 0.7 to get = arccot() = arctan radians. Since = = arctan.6779 radians, we get arccot( ) =arccot( ) radians Another wa to attack the roblem is to use arctan. B definition, the real number t = arctan satisfies tan(t) = with <t<. Since tan(t) < 0, we know more secificall that <t<0, so t corresonds to an angle in Quadrant IV. To find the value of arccot( ), we once again visualize the angle = arccot( ) radians and note that it is a Quadrant II angle with tan() =. This means it is eactl units awa from, and we get = + = + arctan.6779 radians. Hence, as before, arccot( ).6779.

17 0.6 The Inverse Trigonometric Functions 8 =arccot( ) radians (d) If the range of arccosecant is taken to be, 0 [ 0,, we can use Theorem 0.8 to get arccsc = arcsin If, on the other hand, the range of arccosecant is taken to be 0, [,, then we roceed as in the revious roblem b letting t = arccsc. Then t is a real number with csc(t) =. Since csc(t) < 0, we have that <ale,sotcorresonds to a Quadrant III angle,. As above, we let be the reference angle for. Then0< < and csc( ) =, which means = arccsc radians. Since the argument of arccosecant is now ositive, we ma use Theorem 0.9 to get = arccsc = arcsin radians. Since = + = + arcsin.87 radians, arccsc.87. =arccsc radians

18 86 Foundations of Trigonometr. (a) Since the domain of F () = arccos() is ale ale, we can find the domain of f() = arccos b setting the argument of the arccosine, in this case,between and. Solving ale ale gives ale ale, so the domain is [, ]. To determine the range of f, we take a cue from Section.7. Three ke oints on the grah of F () = arccos() are (,), 0, and (, 0). Following the rocedure outlined in Theorem.7, we track these oints to,,(0, 0) and,. Plotting these values tells us that the range of f is,. Our grah confirms our results. (b) To find the domain and range of f() = arctan (), we note that since the domain of F () = arctan() is all real numbers, the onl restrictions, if an, on the domain of f() = arctan () come from the argument of the arctangent, in this case,. Since is defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can, once again, aeal to Theorem.7. Choosing our ke oint to be (0, 0) and tracking the horizontal asmtotes = and =, we find that the grah of = f() = arctan () di ers from the grah of = F () = arctan() b a horizontal comression b a factor of and a vertical stretch b a factor of. It is the latter which a ects the range, roducing a range of,. We confirm our findings on the calculator below. = f() = arccos = f() = arctan () (c) To find the domain of g() = arccot +, we roceed as above. Since the domain of G() = arccot() is(, ), and is defined for all, we get that the domain of g is (, ) as well. As for the range, we note that the range of G() = arccot(), like that of F () = arctan(), is limited b a air of horizontal asmtotes, in this case =0 and =. Following Theorem.7, we grah = g() = arccot + starting with = G() = arccot() and first erforming a horizontal eansion b a factor of and following that with a vertical shift uwards b. This latter transformation is the one which a ects the range, making it now (, ). To check this grahicall, we encounter a bit of a roblem, since on man calculators, there is no shortcut button corresonding to the arccotangent function. Taking a cue from number c, we attemt to rewrite g() = arccot + in terms of the arctangent function. Using Theorem 0.7, wehave that arccot = arctan when > 0, or, in this case, when >0. Hence, for >0, we have g() = arctan +. When < 0, we can use the same argument in number c that gave us arccot( ) = + arctan to give us arccot = + arctan. It also confirms our domain!

19 0.6 The Inverse Trigonometric Functions 87 Hence, for <0, g() = + arctan + = arctan +. What about = 0? We know g(0) = arccot(0) + =, and neither of the formulas for g involving arctangent will roduce this result. 6 Hence, in order to grah = g() on our calculators, we need to write it as a iecewise defined function: 8 >< g() = arccot + = >: We show the inut and the result below. arctan arctan +, if <0, if =0 +, if >0 = g() in terms of arctangent = g() = arccot + The inverse trigonometric functions are ticall found in alications whenever the measure of an angle is required. One such scenario is resented in the following eamle. Eamle The roof on the house below has a 6/ itch. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of feet. Find the angle of inclination from the bottom of the roof to the to of the roof. Eress our answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hotenuse of a right triangle with legs of length 6 feet and feet. Using Theorem 0.0, we 6 Without Calculus, of course... 7 The authors would like to thank Dan Stitz for this roblem and associated grahics.

20 88 Foundations of Trigonometr find the angle of inclination, labeled below, satisfies tan() = 6 =.Sinceisan acute angle, we can use the arctangent function and we find = arctan radians feet feet 0.6. Solving Equations Using the Inverse Trigonometric Functions. In Sections 0. and 0., we learned how to solve equations like sin() = for angles and tan(t) = for real numbers t. In each case, we ultimatel aealed to the Unit Circle and relied on the fact that the answers corresonded to a set of common angles listed on age 7. If, on the other hand, we had been asked to find all angles with sin() = or solve tan(t) = for real numbers t, we would have been hard-ressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a osition to solve these equations. A good arallel to kee in mind is how the square root function can be used to solve certain quadratic equations. The equation = is a lot like sin() = in that it has friendl, common value answers = ±. The equation = 7, on the other hand, is a lot like sin() =.Weknow8 there are answers, but we can t eress them using friendl numbers. 9 To solve = 7, we make use of the square root function and write = ± 7. We can certainl aroimate these answers using a calculator, but as far as eact answers go, we leave them as = ± 7. In the same wa, we will use the arcsine function to solve sin() =, as seen in the following eamle. Eamle Solve the following equations.. Find all angles for which sin() =.. Find all real numbers t for which tan(t) =. Solve sec() = for. Solution.. If sin() =, then the terminal side of, when lotted in standard osition, intersects the Unit Circle at =. Geometricall, we see that this haens at two laces: in Quadrant I and Quadrant II. If we let denote the acute solution to the equation, then all the solutions 8 How do we know this again? 9 This is all, of course, a matter of oinion. For the record, the authors find ± 7 just as nice as ±.

21 0.6 The Inverse Trigonometric Functions 89 to this equation in Quadrant I are coterminal with, and serves as the reference angle for all of the solutions to this equation in Quadrant II. =arcsin radians Since isn t the sine of an of the common angles discussed earlier, we use the arcsine functions to eress our answers. The real number t = arcsin is defined so it satisfies 0 <t< with sin(t) =. Hence, = arcsin radians. Since the solutions in Quadrant I are all coterminal with, we get art of our solution to be = +k = arcsin +k for integers k. Turning our attention to Quadrant II, we get one solution to be. Hence, the Quadrant II solutions are = +k = arcsin +k, for integers k.. We ma visualize the solutions to tan(t) = as angles with tan() =. Since tangent is negative onl in Quadrants II and IV, we focus our e orts there. =arctan( ) radians Since isn t the tangent of an of the common angles, we need to use the arctangent function to eress our answers. The real number t = arctan( ) satisfies tan(t) = and <t<0. If we let = arctan( ) radians, we see that all of the Quadrant IV solutions

22 80 Foundations of Trigonometr to tan() = are coterminal with. Moreover, the solutions from Quadrant II di er b eactl units from the solutions in Quadrant IV, so all the solutions to tan() = are of the form = + k = arctan( ) + k for some integer k. Switching back to the variable t, we record our final answer to tan(t) = as t = arctan( ) + k for integers k.. The last equation we are asked to solve, sec() =, oses two immediate roblems. First, we are not told whether or not reresents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universall acceted range of the arcsecant function. For that reason, we adot the advice given in Section 0. and convert this to the cosine roblem cos() =. Adoting an angle aroach, we consider the equation cos() = and note that our solutions lie in Quadrants II and III. Since isn t the cosine of an of the common angles, we ll need to eress our solutions in terms of the arccosine function. The real number t = arccos is defined so that <t<with cos(t) =. If we let = arccos radians, we see that is a Quadrant II angle. To obtain a Quadrant III angle solution, we ma siml use = arccos. Since all angle solutions are coterminal with or, we get our solutions to cos() = to be = +k = arccos +k or = +k = arccos + k for integers k. Switching back to the variable, we record our final answer to sec() = as = arccos +k or = arccos +k for integers k. =arccos radians =arccos radians = arccos radians The reader is encouraged to check the answers found in Eamle both analticall and with the calculator (see Section 0.6.). With ractice, the inverse trigonometric functions will become as familiar to ou as the square root function. Seaking of ractice...

23 0.6 The Inverse Trigonometric Functions Eercises In Eercises - 0, find the eact value.!. arcsin ( ). arcsin. arcsin (0) 6. arcsin. arcsin 7. arcsin 9. arcsin () 0. arccos ( ). arccos. arccos 7. arccos. arctan!!. arctan 6. arccot 9. arccot (0) 0. arccot. arccos (0). arccos 8. arccos () 9. arctan. arctan (0). arctan!!!. arcsin 8. arcsin. arccos 6. arccos!!! 0. arctan ( )!. arctan ()! 7. arccot ( ) 8. arccot!. arccot (). arccot. arcsec (). arccsc (). arcsec 6. arccsc 7. arcsec! 8. arccsc! 9. arcsec () 0. arccsc () In Eercises - 8, assume that the range of arcsecant is 0, [, and that the range of arccosecant is 0, [, when finding the eact value.!. arcsec ( ). arcsec. arcsec. arcsec ( ). arccsc ( ) 6. arccsc! 7. arccsc 8. arccsc ( )

24 8 Foundations of Trigonometr In Eercises 9-6, assume that the range of arcsecant is 0, [, and that the range of arccosecant is, 0 [ 0, when finding the eact value.! 9. arcsec ( ) 0. arcsec. arcsec. arcsec ( )!. arccsc ( ). arccsc. arccsc 6. arccsc ( ) In Eercises 7-86, find the eact value or state that it is undefined.!! 7. sin arcsin 8. sin arcsin 60. sin (arcsin ( 0.)) 6. sin arcsin 6. cos arccos 6. cos arccos 9. sin arcsin 6. cos arccos!! 6. cos (arccos ( 0.998)) 66. cos (arccos ()) 67. tan (arctan ( )) 68. tan arctan 69. tan arctan 70. tan (arctan (0.96)) 7. tan (arctan ()) 7. cot (arccot ()) 7. cot arccot 7 7. cot (arccot ( 0.00)) 76. cot arccot 78. sec (arcsec ( )) 79. sec arcsec 7 7. cot arccot 77. sec (arcsec ()) 80. sec (arcsec (0.7)) 8. sec (arcsec (7)) 8. csc arccsc 8. csc arccsc 8. csc arccsc!! 8. csc (arccsc (.000)) 86. csc arccsc In Eercises 87-06, find the eact value or state that it is undefined. 87. arcsin sin arcsin sin 89. arcsin!! sin

25 0.6 The Inverse Trigonometric Functions arcsin sin 6 9. arccos cos 96. arccos cos 9. arcsin sin 9. arccos cos 97. arctan tan 99. arctan (tan ()) 00. arctan tan 0. arccot cot 0. arccot cot 0. arccot cot 06. arccot cot 9. arccos cos 9. arccos cos arctan tan 0. arctan tan 0. arccot (cot ()) In Eercises 07-8, assume that the range of arcsecant is 0, [, and that the range of arccosecant is 0, [, when finding the eact value. 07. arcsec sec 0. arcsec sec. arccsc csc 6. arccsc csc arcsec sec. arcsec sec. arccsc csc 7. arcsec sec 09. arcsec sec 6. arccsc csc 6. arccsc csc 9 8. arccsc csc 8 In Eercises 9-0, assume that the range of arcsecant is 0, [, and that the range of arccosecant is, 0 [ 0, when finding the eact value. 9. arcsec sec 0. arcsec sec. arcsec sec 6. arcsec sec. arcsec sec. arccsc csc 6. arccsc csc 8. arccsc csc 6 6. arccsc csc 9. arcsec sec 7. arccsc csc 9 0. arccsc csc 8

26 8 Foundations of Trigonometr In Eercises -, find the eact value or state that it is undefined.. sin arccos. sin arccos. sin (arctan ( )). sin arccot. sin (arccsc ( )) 6. cos arcsin 7. cos arctan 7 8. cos (arccot ()) 9. cos (arcsec ()) 0. tan arcsin!!. tan arccos. tan (arccot ()). cot arcsin. tan arcsec. cot arccos 6. cot arccsc 7. cot (arctan (0.)) 8. sec arccos 9. sec arcsin. csc (arccot (9)). csc arcsin 0. sec (arctan (0)). sec arccot In Eercises - 6, find the eact value or state that it is undefined.. sin arcsin + 7. tan arctan() + arccos 9. sin arccsc 6. cos arcsin 6. cos arccot!!!!!! 0 0. csc arctan 6. cos (arcsec() + arctan()) 8. sin arcsin 60. sin ( arctan ()) 6. cos arcsec 7 arctan() 6. sin

27 0.6 The Inverse Trigonometric Functions 8 In Eercises 6-8, rewrite the quantit as algebraic eressions of and state the domain on which the equivalence is valid. 6. sin (arccos ()) 66. cos (arctan ()) 67. tan (arcsin ()) 68. sec (arctan ()) 69. csc (arccos ()) 70. sin ( arctan ()) 7. sin ( arccos ()) 7. cos ( arctan ()) 7. sin(arccos()) 7. sin arccos 7. cos arcsin 76. cos (arctan ())!! 77. sin( arcsin(7)) 78. sin arcsin 79. cos( arcsin()) 80. sec(arctan()) tan(arctan()) 8. sin (arcsin() + arccos()) 8. cos (arcsin() + arctan()) 8. tan ( arcsin()) 8. sin arctan() 8. If sin() = for <<, find an eression for +sin() in terms of. 86. If tan() = 7 for <<, find an eression for sin() in terms of. 87. If sec() = for 0 <<, find an eression for tan() in terms of. In Eercises 88-07, solve the equation using the techniques discussed in Eamle then aroimate the solutions which lie in the interval [0, ) to four decimal laces. 88. sin() = cos() = sin() = cos() = sin() = cos() = tan() = 7 9. cot() = 96. sec() = 97. csc() = cos() = tan() = sin() = 8 0. tan() = sin() =0.0

28 86 Foundations of Trigonometr 0. sin() = cos() = cos() = cot() = tan() = In Eercises 08-0, find the two acute angles in the right triangle whose sides have the given lengths. Eress our answers using degree measure rounded to two decimal laces. 08., and 09., and 0. 6, 7 and 6. A gu wire 000 feet long is attached to the to of a tower. When ulled taut it touches level ground 60 feet from the base of the tower. What angle does the wire make with the ground? Eress our answer using degree measure rounded to one decimal lace.. At Cli s of Insanit Point, The Great Sasquatch Canon is 77 feet dee. From that oint, a fire is seen at a location known to be 0 miles awa from the base of the sheer canon wall. What angle of deression is made b the line of sight from the canon edge to the fire? Eress our answer using degree measure rounded to one decimal lace.. Shelving is being built at the Utilit Mu n Research Librar which is to be inches dee. An 8-inch rod will be attached to the wall and the underside of the shelf at its edge awa from the wall, forming a right triangle under the shelf to suort it. What angle, to the nearest degree, will the rod make with the wall?. A arasailor is being ulled b a boat on Lake Iizuti. The cable is 00 feet long and the arasailor is 00 feet above the surface of the water. What is the angle of elevation from the boat to the arasailor? Eress our answer using degree measure rounded to one decimal lace.. A tag-and-release rogram to stud the Sasquatch oulation of the eonmous Sasquatch National Park is begun. From a 00 foot tall tower, a ranger sots a Sasquatch lumbering through the wilderness directl towards the tower. Let denote the angle of deression from the to of the tower to a oint on the ground. If the range of the rifle with a tranquilizer dart is 00 feet, find the smallest value of for which the corresonding oint on the ground is in range of the rifle. Round our answer to the nearest hundreth of a degree. In Eercises 6 -, rewrite the given function as a sinusoid of the form S() =A sin(! + ) using Eercises and 6 in Section 0. for reference. Aroimate the value of (which is in radians, of course) to four decimal laces. 6. f() =sin() + cos() 7. f() = cos()+sin() 8. f() = cos() sin() 9. f() = 7 sin(0) cos(0)

29 0.6 The Inverse Trigonometric Functions f() = cos() sin(). f() =sin() cos() In Eercises -, find the domain of the given function. Write our answers in interval notation.. f() = arcsin(). f() = arccos. f() = arcsin. f() = arccos 6. f() = arctan() 7. f() = arccot 9 8. f() = arctan(ln( )) 9. f() = arccot( ) 0. f() = arcsec(). f() = arccsc( + ). f() = arcsec. f() = arccsc e 8 h. Show that arcsec() = arccos for as long as we use 0, i [, as the range of f() = arcsec().. Show that arccsc() = arcsin of f() = arccsc(). for 6. Show that arcsin() + arccos() = 7. Discuss with our classmates wh arcsin h as long as we use for ale ale. 6= 0., 0 [ 0, i as the range 8. Use the following icture and the series of eercises on the net age to show that arctan() + arctan() + arctan() = D(, ) A(0, ) O(0, 0) B(, 0) C(, 0)

30 88 Foundations of Trigonometr (a) Clearl AOB and BCD are right triangles because the line through O and A and the line through C and D are erendicular to the -ais. Use the distance formula to show that BAD is also a right triangle (with \BAD being the right angle) b showing that the sides of the triangle satisf the Pthagorean Theorem. (b) Use AOB to show that = arctan() (c) Use BAD to show that (d) Use BCD to show that = arctan() = arctan() (e) Use the fact that O, B and C all lie on the -ais to conclude that + + =. Thus arctan() + arctan() + arctan() =.

31 0.6 The Inverse Trigonometric Functions Answers. arcsin ( ) =!. arcsin =!. arcsin =. arcsin =. arcsin (0) = 0 6. arcsin = 6 6! 7. arcsin =! 8. arcsin = 9. arcsin () =! 0. arccos ( ) =. arccos =!. arccos = 6. arccos =. arccos (0) =. arccos =! 6. arccos =! 7. arccos = 8. arccos () = arctan = 0. arctan ( ) =!. arctan = 6!. arctan (0) = 0. arctan =. arctan () = 6. arctan =! 8. arccot = 6. arccot 9. arccot (0) = = 6 7. arccot ( ) =! 0. arccot =. arccot () =. arccot = 6. arcsec () =. arccsc () = 6! 7. arcsec = 6. arcsec =! 8. arccsc = 6. arccsc = 9. arcsec () = 0 0. arccsc () =. arcsec ( ) =. arcsec =

32 80 Foundations of Trigonometr. arcsec 6. arccsc! = 7 6 = 9. arcsec ( ) =. arcsec ( ) =. arccsc ( ) = arccsc 0. arcsec! = =. arcsec ( ) =. arccsc ( ) = 6!. arccsc = 6. arccsc ( ) = 7. sin arcsin = 8. sin arcsin 9. sin arcsin = 6. sin arcsin is undefined. 6. cos arccos 6. cos arccos = 8. arccsc ( ) =!. arcsec = 6. arccsc!! = 60. sin (arcsin ( 0.)) = 0.!! = 6. cos arccos = 6. cos (arccos ( 0.998)) = cos (arccos ()) is undefined. 67. tan (arctan ( )) = 68. tan arctan = 69. tan arctan = 70. tan (arctan (0.96)) = tan (arctan ()) = 7. cot (arccot ()) = 7 7. cot arccot = 7. cot arccot = cot (arccot ( 0.00)) = cot arccot = sec (arcsec ()) = 78. sec (arcsec ( )) = =

33 0.6 The Inverse Trigonometric Functions sec arcsec is undefined. 80. sec (arcsec (0.7)) is undefined. 8. sec (arcsec (7)) = 7 8. csc arccsc =!! 8. csc arccsc =!! 8. csc arccsc is undefined. 8. csc (arccsc (.000)) = csc arccsc is undefined. 87. arcsin sin = 88. arcsin sin = arcsin sin = 90. arcsin sin = arcsin sin = 9. arccos cos = 9. arccos cos = 9. arccos cos = 9. arccos cos = 96. arccos cos = arctan tan = 98. arctan tan = 99. arctan (tan ()) = arctan tan is undefined 0. arctan tan 0. arccot cot = = 0. arccot cot = 07. arcsec sec = 09. arcsec sec = arcsec sec = 0. arccot cot = 0. arccot (cot ()) is undefined 06. arccot cot 08. arcsec sec = = 0. arcsec sec is undefined.. arccsc csc = 6 6

34 8 Foundations of Trigonometr. arccsc csc =. arccsc csc = 7. arcsec sec = 9. arcsec sec =. arcsec sec = 6 6. arcsec sec =. arccsc csc = 7. arccsc csc = 9. arcsec sec =. sin arccos =. sin (arctan ( )) =. sin (arccsc ( )) =. arccsc csc 6. arccsc csc 8. arccsc csc 0. arcsec sec = = = =. arcsec sec is undefined.. arccsc csc = arccsc csc = 8. arccsc csc = arccsc csc = 8 8. sin arccos =. sin arccot 6 = 6 6. cos arcsin = 7. cos arctan 7 = 9. cos (arcsec ()) =. tan arccos. tan (arccot ()) = =. tan arcsec 8. cos (arccot ()) = 0 0!! 0. tan arcsin = =. cot arcsin =

35 0.6 The Inverse Trigonometric Functions 8. cot arccos!! = 6. cot arccsc = 7. cot (arctan (0.)) = 8. sec arccos!! 9. sec arcsin = 0. sec (arctan (0)) = 0!! 0. sec arccot = 0. csc (arccot (9)) = 8. csc arcsin =. sin arcsin 7. tan arctan() + arccos 9. sin arccsc 6. cos arcsin 6. cos arccot + = 7 6 = 0 69 = 7 = = 6. sin (arccos ()) = for ale ale 66. cos (arctan ()) = 67. tan (arcsin ()) = + for all for << 68. sec (arctan ()) = + for all 69. csc (arccos ()) = 70. sin ( arctan ()) = + for << for all 7. sin ( arccos ()) = for ale ale. csc arctan = = 6. cos (arcsec() + arctan()) = 8. sin arcsin = 60. sin ( arctan ()) = 6. cos arcsec = r arctan() 6. sin = 0 0

36 8 Foundations of Trigonometr 7. cos ( arctan ()) = for all + 7. sin(arccos()) = for ale ale 7. sin arccos = for ale ale 7. cos arcsin = for ale ale 76. cos (arctan ()) = +9 for all 77. sin( arcsin(7)) = 9 for!! 78. sin arcsin = 79. cos( arcsin()) = for 7 ale ale 7 for ale ale ale ale 80. sec(arctan()) tan(arctan()) = + for all 8. sin (arcsin() + arccos()) = for ale ale 8. cos (arcsin() + arctan()) = tan ( arcsin()) = for in, 8. sin arctan() = 8 >< >: s + + s + + for ale ale for 0 for <0! [,! [!, 8. If sin() = for <<,then +sin() = arcsin If tan() = 7 for <<,then sin() = arctan The equivalence for = ± can be verified indeendentl of the derivation of the formula, but Calculus is required to full understand what is haening at those values. You ll see what we mean when ou work through the details of the identit for tan(t). For now, we eclude = ± from our answer.

37 0.6 The Inverse Trigonometric Functions If sec() = for 0 <<, then tan() = 6 arcsec = arcsin 89. = arccos 9 +k or = +k or = 7 arcsin arccos 9 +k, in[0, ), ,.8 +k, in[0, ),.799, = + arcsin(0.69) + k or = arcsin(0.69) + k, in[0, ),.769, = arccos(0.7) + k or = arccos(0.7) + k, in[0, ),., = arcsin(0.008) + k or = arcsin(0.008) + k, in[0, ), , = arccos +k or = arccos +k, in[0, ), 0.076, = arctan(7) + k, in[0, ),.6, = arctan + k, in[0, ),.08, = arccos +k or = arccos = + arcsin +k or = arcsin = arctan 0 + k, in[0, ),.877, = arcsin +k or = arcsin = arccos +k or = arccos = arctan(0.0) + k, in[0, ), 0.000,.76 +k, in[0, ), 0.8,. 7 +k, in[0, ),.6, k, in[0, ), 0.8,.77 +k, in[0, ),.06, = arcsin(0.0) + k or = arcsin(0.0) + k, in[0, ), 0.78, = + arcsin(0.7) + k or = arcsin(0.7) + k, in[0, ),.968, = arccos(0.98) + k or = arccos(0.98) + k, in[0, ), 0.879, = arccos( 0.67) + k or = arccos( 0.67) + k, in[0, ),.697,. 06. = arctan(7) + k, in[0, ),.6, = arctan( 0.609) + k, in[0, ),.9,.78

38 86 Foundations of Trigonometr and and and f() =sin() + cos() = sin + arcsin sin( +.760) 7. f() = cos()+sin() =sin + arcsin sin( +0.6) 8. f() = cos() sin() =!! 0 0 sin + arccos 0 sin( +.898) 0 9. f() = 7 sin(0) cos(0) = sin 0 + arcsin 0. f() = cos() sin() =sin + + arcsin. f() =sin() cos() = sin + arcsin sin(0.870) sin( +.8)!! sin( 0.66).. ale, " #,. ale,. (, ] [ [, ] [ [, ) 6. (, ) 7. (, ) [ (, ) [ (, ) ale 8., 9., ale 0., [,. (, 6] [ [, ). (, ] [ [, ). [0, )