Analtic Trigonometr Chapter Analtic Trigonometr Inverse Trigonometric Functions The trigonometric functions act as an operator on the variable (angle, resulting in an output value Suppose this process is reversed: given a -value, is it possible to work backward to determine the angle that produced? In simplest terms, we wish to solve equations such as, tan, sec, etc In some cases the value can be determined b inspection For eample, the simple algebraic equation implies that is one possible solution However, this method does not work this eas in general: for eample, b inspection what is the solution to tan? Review of Inverse Graphs and Inverse Functions A function f ( generates a set of points {(, f ( } that are plotted on a Cartesian (- coordinate ais sstem The result is the graph of the function The inverse graph of a function f ( is a graph of the set of points {(, f ( } In simplest terms, the inverse graph of f ( is a new graph in which each ordered pair (, of f ( has its coordinates reversed to (, The points (, and (, are smmetrical across the line ; this allows a simple wa to sketch the inverse graph of a function f ( However, in most cases the inverse graph will not be a function ce it fails the vertical line test To ensure that the inverse graph of a function f ( is itself a function, the given function must be one-to-one A function is one-to-one if f ( a f ( b implies that a b Visuall, a function that is one-to-one passes the horizontal line test To summarize, a function that passes the horizontal line test is said to be one-to-one, and if this condition is met, its inverse graph will be a function as well If this is true, then the inverse function is denoted as f ( The Inverse Sine Function (Arce Consider the function f ( Its graph is given in the net page: 9
Analtic Trigonometr The e function is not one-to-one ce it does not pass the horizontal line test However, if the domain of f ( is restricted to the interval, then it is one-to-one and its inverse graph is therefore a function: Therefore, the inverse e function (also called the arce function, written arc(, is defined to be the inverse graph of the function f ( on the closed interval The domain of the inverse e function is and the range is f ( The inverse e function returns values in the st quadrant (if or the th quadrant (if Eample : Determine values for (a arc, (b arc, (c arc, (d arc( and (e arc( Solutions: The results for (a, (b and (c are based on the normal measures of the e function in the first and fourth quadrants Thus, arc, arc and arc For (d, arc( and for (e, the solution is not defined ce is outside the domain Eample : Ug a calculator set to degree mode, what is arc(? Solution: The solution to two decimal places is the angle 8 degrees 7
Analtic Trigonometr Eample : Ug a calculator in radian mode, what is arc(? Solution: To two decimal places, the result is 7 radian The Inverse Coe Function (Arcine The ine function f ( ( is not one-to-one over its usual domain of the Real numbers However, when restricted to the closed interval, the ine function is one-to-one, and hence its inverse graph is also a function, which will be defined at the inverse ine function (also written arc( The inverse ine graph f ( arc( is defined on the domain with a range of f ( It returns values in the st quadrant (if or the nd quadrant (if Eample : Determine values for (a arc, (b arc, (c arc arc( and (e arc(, (d Solutions: The results for (a, (b and (c are based on the normal measures of the ine function in the first and second quadrants Thus, arc, arc and arc For (d, arc( and for (e, the solution is not defined ce is outside the domain Eample : What radian measure solves arc(? Solution: The answer is radians Eample : If (, find ( Solution: This is identical to the question ( (arc On a right triangle, the opposite measure of one of the non-right angles is and the hpotenuse is With these 7
Analtic Trigonometr facts, the Pthagorean Theorem allows for the adjacent leg to be solved: Therefore, the ine of this angle is The Inverse Tangent Function (Arctangent Like the e and ine function, the tangent function f ( tan( is not one-to-one unless restricted to a smaller domain, which is usuall chosen to be Thus restricted, the inverse tangent function (written arctan( is defined as the inverse graph of the tangent function The domain of the arctangent function is and its range is f ( It returns values in the st quadrant (if or the th quadrant (if Notabl, it has two horizontal asmptotes at as, and as Eample 7: Determine the values to (a arctan(, (b arctan( and (c arctan( Solutions: For (a, arctan(, for (b, arctan( and for (c, the result is undefined, although it trends to positive infinit as Eample 8: Evaluate (arctan( 7
Analtic Trigonometr Solution: The arctan of is an angle whose opposite leg is and adjacent leg is Therefore, the hpotenuse is 7 Thus, the e of this angle is the opposite 7 divided b the hpotenuse: (arctan ( 7 7 Important Results ( t ( t, tan ( t tan ( t, ( t ( t Eample 9: Evaluate the following epressions a ( b ( c ( / d f g tan ( h Solution: a ( (( / / b c d e f g h i ( ( / tan ( i ( / (/ (( / / ( ( ( ( / / tan ( tan (tan( / / tan ( tan ( tan (tan( / / tan tan tan tan( / / Common Errors to Avoid ( e tan ( Students sometimes make the erroneous assumption that the arc cancels the, for eample, arc( (, which is not necessaril true! Pa special attention to the domain appropriate to the particular function being evaluated Consider these following eamples: 7
Analtic Trigonometr Eample : Is it true that arc( (? Solution: Yes, ce Eample : Is it true that arc( (? Solution: No The argument is outside the domain of arce The arce function returns the smallest such angle corresponding to the input Since lies in the rd quadrant, the arce function will return an angle in the th quadrant equivalent to The correct result is arc( ( Eample a: Is it true that arc( (? Solution: Yes The argument is within the domain of the arcine function Eample b: Is it true that (arc (? Solution: No The value is outside the domain of the arcine function This epression is undefined Generalized Epressions The argument can be left as an independent variable and epressions combined to derive relationships between the e, ine and tangent operations with the arce, arcine and arctangent inverse operations Eample : Simplif the epression (arc( Solution: The arc( suggests a result such that ( Sketching a right triangle with at the opposite leg relative to angle, and at the adjacent leg to, the Pthagorean Theorem gives the hpotenuse as having length Therefore, the ine of this angle ( is the adjacent leg divided b the hpotenuse: ( (arc( Eample : Simplif the epression arc(( Solution: Since, rewrite arc(( as arc(( Since is on the domain of the arcine function, the epression reduces to arc( ( 7
Analtic Trigonometr Eample : Solve the equation in the interval Solution: Factor the equation: ( ( Set each factor equal to zero and solve for : arc( arc( Note that the arce function returns values in quadrants and onl Furthermore, other solutions ma need to be inferred from the ones provided b the arce operation: Since is a solution, there also eists a solution in quadrant b smmetr Therefore, is also a solution to the equation The solution provided b the arce operation is correct but outside the desired bounds This is easil remedied b selecting as the equivalent solution Therefore, the solution set to this equation is,, } { Eample : Solve the equation tan tan in the interval Solution: Factor the equation: tan tan (tan (tan Each factor is set equal to zero and solved for the variable: tan tan arctan( 9 radians tan tan arctan( radians The solution set (in radians is { 9, } 7
Analtic Trigonometr Trigonometric Identities An identit is an equation that is alwas true for all values in its domain For eample, the equation a b ( a b( a b is an identit ce it is true for all a and b In a similar manner, man trigonometric epressions can be epressed in different forms, thus forming trigonometric identities Reciprocal Identities From the definitions of the si trigonometric functions, the tangent, cotangent, secant and ecant functions can all be epressed as identities involving the e and/or ine functions: tan cot sec csc Simple algebraic manipulations can generate more identities As an eample, ce tan, it is possible to generate a new identit b multipling the and deriving tan However, restrictions placed on the domain are maintained throughout an further manipulations Thus, the epression (n tan is true as long as, n N, ce these restrictions are in place because of the presence of the tangent function The Pthagorean Identities The e and ine functions are defined on the unit circle and are related b the Pthagorean identit: With simple algebra, new corollar identities can be formed: These identities are true for all Furthermore, the Pthagorean Identit can be divided through b generate more identities: or to Divide b : cot Divide b : tan These identities are true for all for which the epressions are defined csc sec 7
Analtic Trigonometr General Identities and Demonstrations The purpose of emploing identities is to reduce a large trigonometric epression into something smaller and easier to manipulate The net few eamples illustrate the method to demonstrate the truthfulness of each identit Until the proof is demonstrated, the (potential identit is called a theorem The general method is to use whatever identit is appropriate at an given step this often takes practice to recognize the right identit Furthermore, there ma be more than one wa to demonstrate the truthfulness of the identit The onl requirement that must not be violated is that each side of the identit must be handled separatel An algebraic maneuver that assumes the equalit is true cannot be used ce the truthfulness of the equation has et to be shown! In other words, ou cannot use what ou are tring to prove In most cases, this removes the method of cross-multiplication from consideration Otherwise, normal algebraic techniques are used as needed Alwas work from the most complicated form first Eample : Prove that Solution: The left side of the theorem shows two rational epressions being summed Rewrite the left side as a sum b ug a common denominator The common denominator of and is their product, The numerators must be adjusted accordingl: ( The epression on the right side can be simplified b distributing the : ( Since, the numerator reduces to Thus, we have shown b transitivit the following: ( Therefore, the left side of the theorem is equal to the right side The theorem is proven; it is an identit 77
Analtic Trigonometr Eample : Prove that sec csc sec Solution: The left side of the theorem contains more epressions with which to manipulate Convert all functions into equivalent forms in terms of e and ine: sec csc The denominators of the two small epressions in the denominator of the main epression can be combined b summing over a common denominator: Reciprocating the combined epression in the main denominator, the epressions cancel: ( Since, the epression now becomes sec sec sec Thus, the theorem is proven Notice that we never once manipulated the epression sec directl in this proof We were able to show the relationship entirel b manipulating the epressions from the left side of the original theorem Shift and Reflection Identities The e and ine functions are identical in shape and differ onl b a horizontal shift Specificall, the e function is the ine function shifted units to the right Conversel, the ine function is the e function shifted units to the left Therefore, two useful identities can be stated: ( ( Furthermore, the e function shifted units to the left or right results in an inverted e function The same is true for the ine function These two identities are summarized below: ( ( 78
Analtic Trigonometr Shifting the e and ine functions units in opposite directions result in inverted ine and e functions: ( ( Of course, the e and ine functions have a period of identities can be stated: Therefore, two more ( ( The tangent function has a period of so its shift identit is tan( tan Shift identities for the secant, ecant and cotangent functions can be derived b converting into e and ine functions A vertical reflection of an function can be determined b replacing the argument with The result is a graph that is reflected across the -ais The ine function is alread smmetrical to the -ais so it will remain unchanged The e and tangent functions, both smmetrical to the origin, will invert across the -ais Thus, the following reflection identities can be stated: ( ( tan( tan For now, the proofs of these identities are done visuall In the net section, a method will be developed to analticall prove all such shift and reflection identities Eample : Prove that ( Solution: Factor the negative within the argument of the e function: ( ( ( The leading negative within the argument can be moved to in front of the e function: ( ( ( Since (, then ( ( 79
Analtic Trigonometr Sum and Difference Formulas The sum and difference forms of the e and ine functions are the following four forms: (, (, ( and ( The trigonometric function can be viewed as operators on the variable(s and, but the are not linear operators The function/operators cannot be distributed across addition and subtraction Hence, statements like ( are false Unfortunatel these are common mistakes students make when considering the sum and difference forms of the trigonometric functions In this section we will present a geometric proof for all these four forms We prove that ( Suppose that OX be the initial side and the terminal side OY makes an angle with initial side and the terminal side OZ makes angles with the side OY Take a point P on the side OZ and make perpendicular lines PQ and PS on OX and OY respectivel Further we make perpendicular line ST and SR according to the adjacent diagram Now observe that the angle TPS is equal to the angle The angle POR is equal to + And ( PQ PT TQ OP OP PT TQ PT SR OP OP OP OP PT PS SR OS PS OP OS OP ( ( ( Now we have to prove the result for ine: Remember that T Y S O Q R X, (, ( P Z 8
Analtic Trigonometr ( ( ( (, (,, ( Also ( ( ( We now present the algebraic proof The sum and difference forms can be derived from an analsis of the angles drawn on a unit circle, and the familiar distance formula We consider the case ( first, from which the other three forms can easil be derived as corollaries Proof of ( = + Consider the unit circle in the first quadrant (without loss of generalit Let ra r be drawn with angle, and r be drawn with angle, and let Therefore, ra r intersects the circle at point (, and ra r intersects the circle at point (, Rigidl rotate the circle clockwise through an angle of, so that the original ra r sits along the positive -ais while original ra r now has an angle of elevation therefore intersects the unit circle at the point ((,( and The core of the proof is to show that the distance between points (, and (, is the same as the distance from point ((,( to the point (, See the following diagram: Figure Diagram showing ras r and r (left, and the rotation through an angle (right The distance between (, and (, is found b the distance formula: 8
Analtic Trigonometr D((,,(, ( ( The Pthagorean identit was used twice in the second-to-third step The distance between ((,( and (, is D(((,(,(, (( ( (( ( ( ( The Pthagorean identit ( ( was used in the second-to-third step Since the two distances are equal, relate them b equalit: ( Squaring both sides removes the radicals The constants cancel, and the above equation algebraicall reduces to: ( (A Therefore, the difference formula for the ine function is proved Proofs of the Remaining Forms The other three forms can be derived ug the difference form of the ine function (A above For eample, ug the shift identities ( and (, we can make the substitution for into the above form (A to get ( ( ( (B The left side of (B can be re-arranged as ( (, which equals ( On the right side of (B, we use the substitutions ( and ( Therefore, equation (B can be written as ( This proves the difference formula for the e function 8
Analtic Trigonometr The sum forms can be proven ug the smmetr identities ( and ( Therefore, the term is substituted with in equation (A to get ( ( ( ( ( This proves the sum form for the ine function The same substitution is made in equation (B: ( ( ( ( ( This proves the sum form for the e function The Sum and Difference Identities for the Sine and Coe Functions ( ( ( ( The following eamples illustrate some of the was these formulas can be used: Eample : Calculate ( eactl Solution: The angle is the difference of angles and : Therefore, ( ( ( ( ( ( ( ( ( ( Eample : Calculate ( eactl Solution: degrees can be written as the sum of and degrees Therefore, ( ( ( ( ( ( ( ( ( ( Eample : Prove that ( Solution: Use the sum form of the ine function: ( ( ( 8
Analtic Trigonometr Eample : Evaluate tan( eactl Solution: The angle is the sum of angles and Therefore, Alternate proof: Sum and Difference Angle Formulas For Tangent The sum and difference formulas for the tangent function can be derived ug the sum and difference forms for the e and ine function For tan(, we rewrite this in terms of e and ine: tan( ( ( The quotient is rewritten ug the sum forms: ( ( Multipl the numerator and denominator b the epression and simplif: ( ( ( ( tan tan tan tan Therefore, the sum formula for the tangent function is: tan( tan tan tan tan The difference formula is found b replacing with : 8
Analtic Trigonometr tan( tan tan( tan tan( tan tan tan tan The two forms can be summarized together as tan( tan tan tan tan 7 Eample : Evaluate tan( eactl 7 Solution: The angle is the sum of angles and Therefore, tan( 7 tan( tan tan tan tan ( In the final step, the denominator was rationalized b multiplication of the conjugate 8
Analtic Trigonometr Double and Half-Angle Formulas A natural etension of the sum and difference formulas (section are the double-angle formulas For eample, the epression ( is a double-angle epression, ce the argument is interpreted as double the angle Ug the sum formula for, the epression ( can be rewritten as: ( ( This is an etremel useful identit known as the double-angle identit for e The double-angle identit for the ine function is derived similarl: ( ( Ug the Pthagorean identities (section for or, the double-angle identit for ine can be rewritten in two corollar forms: ( ( ( ( For the tangent function, its double-angle identit is tan tan tan tan( tan( tan tan tan All forms can be summarized as the double angle identities: These formulas are ver useful and can be incorporated into man of the common identit proofs that are encountered The Double-Angle Identities ( ( tan tan( tan Eample : Prove that ( ( Solution: Epand the left side b multiplication: ( The first and third terms sum to b the Pthagorean Identit The middle term is ( Therefore, the identit is proven 8
Analtic Trigonometr Eample : Let angle be given in the following diagram Calculate ( and ( Solution: The length of the hpotenuse is found b the Pthagorean formula: hp 7 Therefore, for (, we use its double angle formula: ( ( 7 ( For (, an one of the three double-angle formulas ma be used: ( ( 7 ( 9 The Pthagorean identit could also have been used to determine ( : ( ( ( 89 ( 89 It is interesting to note that the numbers, and form an integer solution to the Pthagorean formula: In fact, the double-angle formulas for e and ine alwas generate a Pthagorean triple of integer solutions, as the net eample illustrates: Eample : Suppose a right triangle has angle acute angle as one of its interior angles Let the opposite leg be a and the adjacent leg be b Determine ( and ( and show that the numerators and the denominator form an integer Pthagorean triple Solution: The hpotenuse is hp a b For (, we get Similarl, ( The epressions ab, b a and the Pthagorean formula: a b ( ( ( a b a b ab a b b a b a ( ( a b a b b a are integers and form an integer solution to a b 87
Analtic Trigonometr (ab a b b True ( b a b b a a a? ( b? b ( b a a a? ( b a The equalit is true This is an interesting wa to generate integer solutions to the Pthagorean formula, and shows also that the number of possible integer solutions to the Pthagorean formula is infinite Eample : If and is in the first quadrant, find sec( Solution: We will need the ine of Since the opposite leg is and the hpotenuse, the adjacent leg is adj Therefore, To determine sec(, rewrite it as sec( ( The double-angle formula for ine gives 7 ( ( ( Therefore, sec( is 7 The Half-Angle Formulas Ug the double-angle formulas for e and ine, we can derive similar formulas for the half-angle of e and ine The usual approach is to start with the double-angle formula for ine, entirel in terms of ine Let v be a temporar dumm variable: (v v Let v, so therefore v The above formula becomes ( Then algebraicall solve for the term ( : ( ( ( This establishes the half-angle formula for ine The sign is selected depending on the quadrant in which the angle is located To determine the half-angle formula for e, use the Pthagorean identit ( ( : ( ( ( ( 88
Analtic Trigonometr Be careful to note that the two formulas both contain and the look quite similar The half-angle formula for tangent is derived b the following steps: tan( ( ( This is one common wa to write the half-angle formula for tangent Another wa to write the formula is to rationalize the denominator b multipling b the conjugate: tan( ( ( ( ( The half-angle formulas are summarized below: The Half-Angle Formulas ( ( tan( Eample : Let angle be given in the following diagram Calculate ( and ( Solution: The hpotenuse is Therefore, ( 7 7 7 In the final step, the internal fraction was rationalized b multipling top and bottom b The value for ( is found via a similar calculation: ( 7 7 7 89
Analtic Trigonometr Product-to-Sum and Sum-to-Product Formulas Occasionall it is desirable to convert the product of unlike e and/or ine terms into the sum or difference of two separate e or ine terms, and occasionall the reverse is true In an case, the product-to-sum and sum-to-product identities allow us to make these conversions These formulas are especiall useful in integral calculus The Product-to-Sum Identities The product-to-sum identities arise from a simple algebraic manipulation of the sum and difference formulas for the e and ine function For eample, we write both the sum and difference formulas for the ine function: ( ( Subtracted, the terms cancel and we get: ( ( Divide out the and we have one such identit: [( ( ] Summed, the terms cancel, and diving b, we get another similar identit: [( ( ] The process is repeated with the sum and difference formulas for the e function, and two more identities are derived in a similar manner The set of product-to-sum identities are given in the net page The Product-to-Sum Identities [( ( ] [( ( ] [( ( ] [( ( ] Consider the following eample: Eample : Write as the sum of two individual trigonometric terms Solution: The identit [( ( ] is used here We get the following: [( ( ] [( ( ] 9
Analtic Trigonometr Recall that the ine absorbs negatives, so that the final result is [( (] The Sum-to-Product Identities These identities are proven b performing the product-to-sum identities in reverse Consider the identit u v [( u v ( u v ] Multipl the to remove the fraction, and let u v and u v Solving for u and v in terms of and ields the following: u and v Thus, the equation becomes: u v [( u v ( u v] The other three identities are solved in a similar manner The four sum-to-product identities are: The Sum-to-Product Identities Eample : Rewrite the sum of functions Solution: The identit produces the following: as the product of two trigonometric ( ( ( ( B appling the appropriate product-to-sum identit, the latter result can easil be rewritten into its sum form These formulas can be used to prove various identities: Eample : Prove the identit: cot ( 9
Analtic Trigonometr Solution: The appropriate sum-to-product identities are used on both sides of the equation: cot ( cot (( cot ((( (( ((( (( (( Carefull follow each step and notice the cancellation taking place as well as the absorption of negative signs b the ine function Eample : Prove the identit: tan p q p p q q Solution: It s probabl easiest to work the right side of this equation first: p p q q p q p q p q p q The p q factors cancel as well as the s The identit is thus proven: p p q q p q p q p q p q p q p q tan p q In the net section we will see an eample how these formulas can be used to solve a trigonometric equation Solution of Trigonometric Equations An equation is that involves one or more of the various trigonometric functions is called a trigonometric equation The variable is in the argument position of the trigonometric functions, and normall polnomial and eponential terms are not included For eample, 9
Analtic Trigonometr (, ( and are all considered trigonometric equations, while is not considered a trigonometric equation ce it involves (in this case polnomial terms Equations of this last tpe are usuall impossible to solve b purel algebraic means We will not consider these latter forms in this section Linear Trigonometric Equations A linear trigonometric equation involves just one trigonometric function, usuall itself a linear term In these cases, solutions are easil found b the usual rules of algebra, and the correct use of the inverse trigonometric function at the appropriate time Bear in mind all solutions will be in radians, and that ou will be responsible for locating all solutions of the equation taking into account smmetries Eample : Solve for : State the solution sets as follows: (a all solutions, (b solutions within Solution: Adding the and diving the results in this equivalent equation: Now is the correct time to emplo the inverse e function as an operator: ( ( From our knowledge of the inverse e function, the primar solution is However, there is another solution Recall the e is positive in quadrants I and II; b smmetr, the secondar solution is Thus, for (a, the set of all solutions is found b adding integer multiples of to each solution: n, n }, n Q { where set Q represents the integers For (b, the restriction simpl allows us to avoid listing all (infinitel man solutions of this equation The solution set is therefore just the set, } { Eample : Solve for : ( State the solution sets as follows: (a all solutions, (b solutions within ( Solution: Isolate the ine term: ( 9
Analtic Trigonometr Now we use the inverse ine as an operator: (( ( ( The value for ( is (primar, quadrant I, and b smmetr, (secondar, quadrant IV We now have two equations: Both solve the same wa, ielding n & n n & n Thus, to answer (a, the set of all solutions is { n, n } To find the solution restricted to, we need to evaluate for values of n When n =, we get the two principal solutions & ; these two solutions lie in quadrants I and II, respectivel When n =, we generate two more solutions 7 &, which lie in quadrants III and IV, respectivel When n =, we generate two more solutions, but the place us back to the original solutions in quadrants I and II Therefore, to answer part (b, the solution set of this equation when restricted to is 7,,, } { Note that there was no need to evaluate for n ce doing so generates solutions redundantl Eample : Solve for : Give all solutions in the interval Solution: In this case we can factor : ( Each factor can be solved for separatel Setting ields the solutions and Setting ields which in turn ields solutions and Therefore, the solution set to this function is,,, } { Comment: the restriction is common in trigonometric equations ce it requests onl those solutions in the first four quadrants, and no more (ie no 9
Analtic Trigonometr redundancies Note that this restriction includes but ecludes ce at this latter value, we are repeating the unit circle once again Thus, in the previous eample, there was no need to state the solution ce it is redundant to the solution Comment: In the preceding eample, do NOT divide b In doing so, ou remove all solutions that the factor provides, and ou potentiall divide b zero, which is never a legal arithmetic maneuver Non-linear Trigonometric Equations If a trigonometric term is raised to a power other than one, it is considered non-linear The usual rules of algebra still appl, and we solve these forms ug whatever methods are appropriate Eample : Solve for in the interval : Solution: The e term is squared, so we take the radical, including both positive and negative solutions: Thus, we need to solve two equations for : and In the first case, the inverse e function as an operator ields ( as the primar solution, and b smmetr, as the secondar solution In the second case, the inverse e operator ields ( as the primar solution and b smmetr, as a solution Thus, there are four solutions and the solution set is,,, } { Eample : Solve for in the interval : Solution: This equation involves both e and ine terms, as well as a squared term In cases like this, it is usuall best to get all trigonometric terms into a common form Thus, we re-arrange this equation a bit, then replace the term ug the identit This will result in a quadratic equation in terms of ine onl: 9
Analtic Trigonometr ( The last equation is now a quadratic equation with as the unknown Factor: Each factor is then solved for zero: ( (, and Therefore, the solution set is {,, } Other Forms There is no limit to the number of possible equations involving trigonometric functions Some ma be eas to solve, some not so eas, and some impossible You should alwas look for possible identities to simplif the equation, collecting terms, and avoiding the dividing out of a trigonometric function The following eample is one of man possible trigonometric equations: Eample : Solve for in the interval : sec csctan Solution: Replacing each of these forms in terms of e and ine, we can greatl simplif this equation and write it in terms of a gle trigonometric function: sec csc tan Cross-multipling and collecting the terms to one side ields We factor: ( Each factor is solved for its solutions: 9
Analtic Trigonometr, However, none of them are acceptable! Wh is this? Note that the sec term is not defined when, so the two solutions must be discarded, and the csc term is not defined when, so it too must be discarded The solution set is empt: Eample 7: Solve for in the interval : Solution: These two e terms are unlike ; the cannot be combined convenientl b addition Therefore, we use an appropriate sum-to-product identit (section to assist: ( ( Therefore, each factor can be set to and solved We eplore the solutions generated b the factor ( first: ( & n n ( n & ( n The solutions are generated b evaluating for n,,, etc B direct evaluation we get these solutions: 8 n, n, n The last solution is the same as and can be ignored, and there is no need to evaluate for higher values of n ce this will simpl repeat our known solutions again The solutions generated b the factor (, we get the following: ( & n n ( n & ( n The onl meaningful solution that results from these two equations is All others are outside the restriction Therefore, the solution set to this equation is: {,, 8,,, } If f (, then the solution set can be illustrated as the -intercepts of this graph in the interval : 97
Analtic Trigonometr 7 Law of Sines and Coes Figure: Graph for eample 7 Law of Coes Suppose that a triangle has sides of lengths a, b, and c corresponding to the angles,, and as shown in the figure Then a b c bc C b a b c a ca c a b ab A c B And Law of Sines given below Eamples A B C a b c A triangle has sides of length 8, and, and the angle between these sides is degrees Find the length of the third side Solution We use the formula (Use our calculator for simplification A a b c bc 8 a 8 (8( The third side has the length equals a B C Use law of Sines to find the indicated side CB =, where angle CAB = degrees and angle CBA = degrees, AB 7 cm C Solution We have angle ACB = 8 (+ = 9 9 7 A B (7 9 98