Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions we consruced in Chaper can be used o solve equaions like sin( ) bu he fracion is a friendly sine value so we don need o use he inverse sin, so we know ha is a soluion o sin( ) We also know ha he sine funcion is periodic wih period, so is values repea every unis, so angles like sine funcion: our experience wih he sine funcion ells us ha ha and 4 5 and are also soluions We can represen muliples of he period wih he expression k where k is any ineger, ie, k, so we can represen all of he soluions ha are relaed o wih he expression k, k This expression represens infiniely many soluions, bu i sill doesn represen all of he soluions; see Figure Figure : The graph of y sin( ) and he line y The red dos represen poins wih horizonal coordinaes of he form k, k The oher insances where he blue graph inersecs he line soluions o he equaion sin( ) represened by k y are also bu hey are NOT
Haberman MTH Secion I: Chaper 7 Noice ha one of he soluions ha we are missing is jus as close o as our original soluion,, is o 0 Recall he ideniy sin( ) sin ha we firs noiced in Par of Chaper : his ideniy ells us ha he angles and always have he same sine value This means ha whenever we ve found a soluion,, o an equaion involving sine, we can find anoher soluion by compuing Now le s apply his observaion o find he res of he soluions o sin( ) : since we know ha is a soluion osin( ), we know ha 5 is anoher soluion And now we can again uilize he fac ha he period of he sine funcion is so we can express he res of he soluions wih 5 k, k ; in Figure, hese soluions are colored green So he complee soluion o he equaion sin( ) is: k or 5 k for all k Figure : The red dos represen poins wih horizonal coordinaes of he form k, k, while he green dos represen poins wih horizonal coordinaes of he form 5 k, k The red and green dos ogeher represen all of he soluions o sin( )
Haberman MTH Secion I: Chaper 7 EXAMPLE b: Solve he equaion sin( ) 0555 Unlike Example a where he equaion involved a friendly sine value, 0555 isn a friendly sine value: we don know wha inpu for he sine funcion is relaed o he oupu 0555, so we need o uilize he inverse sine funcion ha we consruced in Chaper in order o solve he equaion: sin( ) 0555 sin( ) sin 0555 sin sin 0555 0588 (apply he inverse sine funcion o boh sides of he equaion) (Noe ha you can uilize a calculaor o obain an approximaion for sin 0555 accessing a buon labeled sin ) by Alhough we've found one soluion o he equaion, we aren done ye! The inverse sine inverse only gives us one value, bu we know ha he periodic naure of he sine funcion suggess ha here are infiniely many soluions o an equaion like his; see Figure Figure : The graph of y sin( ) inersecing he line y 0555 many, many imes Each poin of inersecion represens a soluion o sin( ) 0555 We can find all of he soluions by using he soluion ha we found using he inverse sine funcion along wih he fac ha he sine funcion has period : since he sine funcion has period unis, we know ha he oupus repea every unis So if 0588 is a soluion, he values represened by 0588 k, k mus also be soluions This gives us LOTS of soluions, bu we are sill missing half of hem (Recall we had he same problem in Example a) In order o ge he res of he soluions, can use he sin( ) sin, and subrac our original soluion ( 0588 ) from : ideniy ( 0588) k, k Therefore, he complee soluion o he equaion is: 0588 k or 0588 k for all k
Haberman MTH Secion I: Chaper 7 4 EXAMPLE a: Solve he equaion cos( ) Like Example a, is a friendly cosine value so we can use our knowledge abou he cosine funcion, raher han he inverse cosine funcion, o solve he equaion Our experience wih he cosine funcion ells us ha ha 5 5 is a soluion o cos( ) cos, so we know ha We also know ha he cosine funcion is periodic wih period, so is values repea every unis, so angles like 5 7 and 5 4 9 and 5 7 are also soluions We can represen all of he soluions ha are relaed o 5 wih he expression 5 k, k This expression represens infiniely many soluions, bu i sill doesn represen all of he soluions; see Figure 4 y The red dos represen poins wih horizonal coordinaes of he form 5 k, k The oher insances where Figure 4: The graph of y cos( ) and he line he blue graph inersecs he line soluions o he equaion Recall he ideniy cos( ) cos represened by 5 k y are also cos( ) bu hey are NOT ha we noiced in Par of Chaper : his ideniy ells us ha he angles and always have he same cosine value This means ha whenever we ve found a soluion,, o an equaion involving cosine, we can find anoher soluion by compuing Now le s apply his observaion o find he res of he soluions o cos( ) : since we know ha 5 is a soluion o cos( ), we know ha 5 is anoher soluion Now we can again uilize he fac ha he period of cosine is
Haberman MTH Secion I: Chaper 7 5 so we can express he res of he soluions wih 5 k, k ; in Figure 5, hese soluions are colored green So he complee soluion o he equaion cos( ) is: k or 5 k for all k 5 Figure 5: The red dos represen poins wih horizonal coordinaes of he form 5 k, k, while he green dos represen poins wih horizonal coordinaes of he form 5 k, k The red and green dos ogeher represen all of he soluions o cos( ) EXAMPLE b: Solve he equaion cos( ) 04 Like in Example b, 04 isn a friendly cosine value so we need o uilize he inverse cosine funcion ha we consruced in Chaper in order o solve he equaion: cos cos( ) 04 cos( ) cos 04 cos 04 (apply he inverse cosine funcion o boh sides of he equaion) (Noe ha you can uilize a calculaor o obain an approximaion for accessing a buon labeled cos ) cos (04) by
Haberman MTH Secion I: Chaper 7 Alhough we have found a soluion o he given equaion, we aren done ye! The inverse cosine funcion only gives us one value bu we know ha he periodic naure of he cosine funcion suggess ha here are infiniely many soluions o an equaion like his; see Figure Figure : The graph of y cos( ) inersecing he line y 04 many, many imes Each poin of inersecion represens a soluion o cos( ) 04 We can find all of he soluions by using he soluion ha we found using he inverse cosine funcion along wih he fac ha he cosine funcion has period : since he cosine funcion has period unis, we know ha he oupus repea every unis So if is a soluion, he values represened by k, k mus also be soluions This gives us LOTS of soluions, bu we are sill missing half of hem (Recall we had he same problem in Example a) In order o ge he res of he soluions, we can cos( ) cos and ake he opposie of original soluion o find a use he ideniy second family of soluions: k, k Therefore, he complee soluion o he equaion cos( ) 04 is: k or k for all k
Haberman MTH Secion I: Chaper 7 7 EXAMPLE a: Solve he equaion cos( ) CLICK HERE o see a video of his example cos( ) cos( ) cos( ) cos cos (his sep is opional since cos or (since for k k k is a "friendly" cosine value) cos( ) cos( ) ) and [Since is a friendly cosine value, we didn need o use he inverse cosine funcion as we did in he hird sep he inverse rig funcions are available when solving equaions bu we don need o use hem if he values are friendly] EXAMPLE b: Solve he equaion 8cos( x) 9 0 8cos( x) 9 0 cos 8cos( x) cos( x) 8 cos( x) cos 8 or cos k cos k for k 445 k or 445 k for k (since cos 445) 8 [Since isn a friendly cosine value, we need o use he inverse cosine funcion as 8 we ve done in he fourh sep Also noe ha in he las sep we ve approximaed he soluions: his requires a calculaor, so i s no somehing ha we would need o do on no-calculaor exams]
Haberman MTH Secion I: Chaper 7 8 EXAMPLE 4a: Solve he equaion 4sin( ) CLICK HERE o see a video of his example 4sin( ) sin( ) sin 4sin( ) 4 sin( ) sin or k k k 4 k or k for k (his sep is opional since is a "friendly" sine value) sin for (since sin( ) sin( ) ) and [Since is a friendly value, we didn need o employ he inverse sine funcion as we did in he fourh sep he inverse rig funcions are always available when solving rig equaions bu we don need o use hem when he values are friendly] EXAMPLE 4b: Solve he equaion sin( ) sin( ) sin( ) sin( ) sin or sin sin k sin k for all k 048 k or 048 k for all k 048 k or k for all k (since sin 048) [Since isn a friendly sine value, we need o employ he inverse sine funcion as we ve done in he hird sep Also noe ha in he second-o-las sep we ve approximaed he soluions: his requires a calculaor, so i s no somehing ha we need o do on a no-calculaor aciviy]
Haberman MTH Secion I: Chaper 7 9 EXAMPLE 5a: Solve he equaion an( x) an an( x) an( x) an( x) an x k for all k (his sep is opional since is a "friendly" angen value) (since an ) [Noice ha we add k (raher han k ) o our soluions since, unlike sine and cosine, he period of angen is unis Also, here s only one family of soluions since angen only reaches each oupu value once in each period] EXAMPLE 5b: Solve he equaion 5an( ) 0 5an( ) 0 an 5an( ) 4 an( ) 4 5 an( ) an 4 5 an 4 k for all k 5 [Since 4 isn a friendly angen value, we need o use he inverse angen funcion 5 as we ve done in he fourh sep As menioned above in Example 5a, we add k (raher han k ) o our soluions since he period of angen is unis and here s only one family of soluions since angen only reaches each oupu value once in each period]
Haberman MTH Secion I: Chaper 7 0 EXAMPLE : a Find all of he soluions o he equaion sin( x) b Find he soluions o sin( x) ha are in he inerval [0, ] a Noice ha he rigonomeric funcion involved in he given equaion is sin( x ), and recall ha sin( x ) has period unis, ie, he values for sin( x ) repea every unis This means ha once we find a soluion o he given equaion we ll be able o add o i any ineger muliple of and obain anoher soluion Thus, we should expec he phrase k for all k o be involved in our soluions You ll see in he work below ha we add k o our soluions afer applying he inverse sine funcion since he period of he sine funcion is unis In he las sep, we finish solving for x and obain he desired period-shif of k unis sin( x) sin( x) s x sin sin in( ) x k or x k for all k 4 4 or x k x k for all k 4 4 x k or x k for all k b Now we need o subsiue specific values of k ino he soluions we found in par (a) and deermine which soluions are in he inerval [0, ] or x 7 4 k : x ( ) ( ) Boh of hese values are negaive so hey aren in he inerval [0, ] Smaller values of k will produce even smaller values of x so we don need o ry smaller values of k x k 0: x 0 or 0 Boh of hese values are in he inerval [0, ]
Haberman MTH Secion I: Chaper 7 x 9 k : x or Boh of hese values are in he inerval [0, ] x 7 9 Since boh of hese values are greaer han, hey aren in he inerval k : x or [0, ] Cerainly larger values of k will produce even larger values of x so we don need o ry larger values of k Therefore, he soluion se o he equaion sin( x) on he inerval [0, ] is,, 9, EXAMPLE 7: a Find all of he soluions o he equaion cos( ) b Find he soluions in he inerval [0, ] o he equaion cos( ) a Noice ha he rigonomeric funcion involved in he given equaion is cos( ), and recall ha cos( ) has period unis, ie, he values for cos( ) repea every unis This means ha once we find a soluion o he given equaion we ll be able o add o i any ineger muliple of and obain anoher soluion Therefore, we should expec he phrase k for all k o be involved in our soluions You ll see in he work below ha we add k o our soluions afer applying he inverse cosine funcion since he period of he cosine funcion is unis In he las sep, we finish solving for and obain he desired period-shif of k unis
Haberman MTH Secion I: Chaper 7 cos( ) cos( ) cos( ) cos cos k or k for all k or k k for all k k or k for all k 9 9 b Now we need o subsiue specific values of k ino he soluions we found in par (a) and deermine which soluions are in he inerval [0, ] k ( ) ( ) 9 9 4 8 9 9 : or Boh of hese values are negaive so hey aren in he inerval [0, ] Smaller values of k will produce even smaller values of so we don need o ry smaller values of k ( 0) ( ) 0 9 9 9 9 Only is in he inerval [0, ] since is negaive 9 9 k 0: or ( ) ( ) 9 9 8 4 9 9 k : or Boh of hese values are in he inerval [0, ] ( ) ( ) 9 9 4 0 9 9 k : or Boh of hese values are in he inerval [0, ]
Haberman MTH Secion I: Chaper 7 ( ) ( ) 9 9 0 9 9 k : or Only is in he inerval [0, ] since 9 0 is greaer han 9 ( 4) ( ) 4 9 9 9 9 Since boh of hese values are greaer han, hey aren in he inerval k 4: or [0, ] Cerainly larger values of k will produce even larger values of so we don need o ry larger values of k Therefore, he soluion se o he equaion sin( x) on he inerval [0, ] is, 4, 8, 0, 4, 9 9 9 9 9 9 EXAMPLE 8: Find he soluions o he equaion cos( x) 0 on he inerval [, ] cos( x) 0 cos( x) cos cos( x) cos or or or x cos k x cos k for all k x cos k x cos k for all k x cos k x cos k for all k Noice ha we were asked o find he soluions in he inerval [, ], so we need o find which of he infiniely many soluions we ve found are on he inerval I migh help if we approximae he values we found above:
Haberman MTH Secion I: Chaper 7 4 x cos k 04 k x cos k 04 k for all k or We know ha 4 so we need o find values ha saisfy he equaion as above and are beween 4 and 4 k : x 04 ( ) or x 04 ( ) 7 5 Only 7 is in he inerval [, ] k 0: x 04 0 or x 04 0 04 04 Boh of hese values are in he inerval [, ] k : x 04 or x 04 5 7 Only 7 is in he inerval [, ] k : x 04 or x 04 7 58 Neiher of hese values is in he inerval [, ] We could ry more values of k bu we can ell from he work we ve done hus far ha no oher values of k will give us soluions ha are on he inerval [, ] Thus, he soluion se o he equaion cos( x) 0 on he inerval [, ] is 7, 04, 04, 7