C H P T E R 1 toolo Ojectives To revise the properties of sine, cosine nd tngent To revise methods for solving right-ngled tringles To revise the sine rule nd cosine rule To revise sic tringle, prllel line nd circle geometr To revise rithmetic nd geometric sequences To revise rithmetic nd geometric series To revise infinite geometric series To revise crtesin equtions for circles To sketch grphs of ellipses from the generl crtesin reltion ( h) ( k) + = 1 To sketch grphs of hperols from the generl crtesin reltion ( h) ( k) = 1 To consider smptotic ehviour of hperols To work with prmetric equtions for circles, ellipses nd hperols The first si sections of this chpter revise res for which knowledge is required in this course, nd which re referred to in the Specilist Mthemtics Stud Design. The finl section introduces crtesin nd prmetric equtions for ellipses nd hperols. 1.1 Circulr functions Defining sine, cosine nd tngent The unit circle is circle of rdius one with centre t the origin. It is the grph of the reltion + = 1. 1 ( _ 1, ) (, 1) (, _ 1) (1, )
Essentil Specilist Mthemtics Sine nd cosine m e defined for n ngle through the unit circle. For the ngle of,point P on the unit circle is defined s illustrted opposite. The ngle is mesured in n nticlockwise direction from the positive direction of the is. cos( )isdefined s the -coordinte of the point P nd sin( )isdefined s the -coordinte of P. clcultor gives pproimte vlues for these coordintes where the ngle is given. P(cos(θ ), sin (θ )) θ ( _ ( (.8,.5).771,.771) _.17,.988) 15 1 sin =.5(ect vlue) sin 15 = 1.771 cos 1.17 cos =.8 cos 15 = 1.771 sin 1.988 tn( )is defined tn( ) = sin( ) cos( ). The vlue of tn( ) cn e illustrted geometricll through the unit circle. considering similr tringles OPP nd OTT, it cn e seen tht TT OT = PP OP i.e. TT = sin( ) cos( ) = tn( ) O P θ P' T(1, tn (θ )) T' sin (θ ) = PP' Forright-ngled tringle OC,similr tringle O C cn e constructed tht lies in the unit circle. the definition, OC = cos( ) nd C = sin( ). The scle fctor is the length O. Hence C = O sin( ) nd OC = O cos( ). This implies 1 ' C O = sin( ) nd OC O = cos( ) O θ C' C
Chpter 1 toolo This gives the rtio definition of sine nd cosine for right-ngled tringle. The nming of sides with respect to n ngle is s shown. sin = opp hp cos = dj hp tn = opp dj ( ) opposite hpotenuse ( ) djcent hpotenuse ( ) opposite djcent hpotenuse θ O djcent C opposite Definition of rdin In moving round the circle distnce of 1 unit from to P the ngle PO is defined. The mesure of this ngle is 1 rdin. One rdin (written 1 c )is the ngle sutended t the centre of the unit circle n rc of length 1 unit. _ 1 1 O P 1 c 1unit 1 Note: ngles formed moving nticlockwise round the circumference of the unit circle re defined s positive. Those formed moving in clockwise direction re sid to e negtive. _ 1 Degrees nd rdins The ngle, in rdins, swept out in one revolution of circle is c c = c = 18 1 c = 18 or 1 = c 18 Henceforth the c m e omitted. n ngle is ssumed to e mesured in rdins unless otherwise indicted. The following tle displs the conversions of some specil ngles from degrees to rdins. ngles in degrees 5 9 18 ngles in rdins
Essentil Specilist Mthemtics Some vlues for the trigonometric functions re given in the following tle. in rdins sin cos tn 1 1 1 1 1 undefined The grphs of sine nd cosine s sin = sin( + n), n Z, the function is periodic nd the period is. sketch of the grph of f : R R, f () = sin is shown opposite. The mplitude is 1. 1 f () = sin π π π π π π 1 sketch of the grph of f : R R, f () = cos is shown opposite. The period of the function is. The mplitude is 1. For = cos(n) nd = sin(n) >, n > Period = n, mplitude =, rnge = [, ] 1 f () = cos π π π π π π 1 Smmetr properties for sine nd cosine From the grph of the functions or from the unit circle definitions, the following results m e otined. sin( ) = sin cos( ) = cos sin( ) = sin cos( ) = cos sin( + ) = sin cos( + ) = cos sin( ) = sin cos( ) = cos sin( + n ) = sin cos( + n ) = cos for n Z ( ) sin = cos ( ) cos = sin
Chpter 1 toolo 5 Emple 1 c Chnge 15 into rdins. Chnge 1.5 c into degrees, correct to two deciml plces. Find the ect vlue of: ( ) 7 c i sin c ii cos ( ) 15 c ( 15 = = 18 ) c Note tht ngles in rdins which re epressed in terms of re left in tht form. ( ) 1.5 18 1.5 c = = 85.9, correct to two deciml plces. c i sin c = ( ) 7 c ( ) 7 c ii cos = cos = cos Emple ( ) c = cos c = Find the ect vlue of: sin 15 cos( 585 ) sin 15 = sin(18 15 ) = sin Emple =.5 cos( 585 ) = cos 585 = cos(585 ) = cos(5 ) = cos 5 = Find the ect vlue of: ( ) 11 sin ( ) 11 sin ( ) 5 cos ( = sin 11 ) = sin = 1 ( ) 5 cos = cos ( 7 1 ) ( = cos ) =
Essentil Specilist Mthemtics The Pthgoren identit Forn vlue of cos + sin = 1 Emple If sin =., < < 9, find: cos tn sin + cos = 1.9 + cos = 1 cos =.91 cos =±.91 s < < 9, cos =.91 = 91 91 1 = 1 tn = sin cos =..91 = 91 = 91 91 of equtions If trigonometric eqution hs solution, then it will hve corresponding solution in ech ccle of its domin. Such equtions re solved using the smmetr of the grphs to otin solutions within one ccle of the function. Other solutions m e otined dding multiples of the period to these solutions. Emple 5 The grph of = f ()where f: R R, f () = sin, [, ] is shown. Find the other vlue which hs the sme vlue s ech of the pronumerls mrked. For =, the vlue is. For =, the other vlue is. For = c, the other vlue is (c ) = c. For = d, the other vlue is + ( d) = d. 1 c d π π _ 1 Emple ( Solve the eqution sin + ) = 1 for [, ].
Chpter 1 toolo 7 Let = + Note + 1 1 Therefore solving the eqution sin solving the eqution sin( ) = 1 for Consider sin = 1 = or 5 The solutions 9 nd for. For 1 or + or + 5 = 5 + = 5 = = ( + ) = 1 for [, ] is chieved first 1 or + or + 5 or... re not required s the lie outside the restricted domin or or or or 1 1 11 11 1 or or or or 17 17 15 5 Using grphics clcultor grphics clcultor cn e( used to find numericl solution to the eqution sin + ) = 1 plotting ( the grphs of = sin + ) nd = 1 nd considering the points of intersection. It cn e seen from the grph for [, ] tht there re four solutions. The coordintes of the points of intersection re found through using 5:intersect from the CLC menu. It is sometimes possile to find the ect vlue clculting X/. The corresponding ect vlue is 11 1. or or or or 5 5 1
8 Essentil Specilist Mthemtics Using CS clcultor CS clcultor cn e used to solve this eqution. The snt is solve(sin( + /) = 1/, ) < = nd < =. The result is shown. Sketch grphs The grphs of functions defined rules of the form f () = sin(n + ε) + nd f () = cos(n + ε) + cn e otined from the grphs of sin nd cos trnsformtions. Emple 7 ( Sketch the grph of h: [, ] R, h() = cos + ) + 1. ( ( h() = cos + )) + 1 The trnsformtions from the grph of = cos re diltion from the is of fctor 1 diltion from the -is of fctor trnsltion of in the negtive direction of the is trnsltion of 1 in the positive direction of the is The grph with the diltions pplied is s shown elow. = cos () π π π π 5π π 7π π The trnsltion in the negtive direction of the is is then pplied.
Chpter 1 toolo 9 _ π π 1 _ π 7π 1 = cos + 5π 1π 1 π π 19π 1 11π 5π 1 π, The finl trnsltion is pplied nd the grph is given for the required domin. The -is intercepts re found solving the eqution. ( cos + ) + 1 = ( i.e. cos + ) = 1 5 _ π 5π π 11π π, 5 The grph of tn sketch of the grph of f : R\{(n + 1) ; n Z} R, f ( ) = tn is shown elow. _ π _ π π π 5π π π π θ Note: =,, nd 5 re smptotes. Oservtions from the grph The grph repets itself ever units, i.e. the period of tn is. Rnge of tn is R. The verticl smptotes hve eqution = (k + 1) where k Z.
1 Essentil Specilist Mthemtics Using grphics clcultor If the norml procedure is followed nd = tn is entered into the Y = menu nd then plotted pressing ZOOM 7,we get the grph s shown here. While it might pper tht the clcultor hs drwn in smptotes, it hsn t. Insted, it hs simpl joined up the lst two points plotted either side of the point where the smptote lies nd joined them up. Note tht, since tn = sin, the domin of tn must eclude cos ll rel numers where cos is zero, i.e. ll odd integrl multiples of. The grph (n + 1) shows verticl smptotes t =. This function hs period of s tn = tn( + n ), n Z. The concept of mplitude is not pplicle here. Smmetr properties for tn From the definition of tn, the following results re otined: tn( ) = tn tn( + ) = tn tn( ) = tn tn( ) = tn Emple 8 Find the ect vlues of: ( ) tn ( ) tn tn ( ) = tn( ) = tn () = ( ) tn ( = tn + ( ) = tn = s of equtions involving tn The procedure here is similr to tht used for solving equtions involving sin nd cos, ecept tht onl one solution needs to e selected then ll other solutions re or 18 prt. Emple 9 ) Solve the equtions: tn = 1 for [, ] tn( ) = for [, ] tn = 1 Now tn = 1.
Chpter 1 toolo 11 Therefore = or + or + or +. Hence = 7 11 15 or or or. tn( ) =. Therefore nd thus nd. In order to solve tn( ) = first solve tn =. = or or or = or or 5 or 8 nd s = = Therefore = nd = Eercise 1 or or or or or or 5 or 5. or 5 or 8 1 Chnge the following ngles from degrees to ect vlues in rdins: i 7 ii 5 iii 5 iv 15 v 1 vi 15 Chnge ( the ) following ngles from rdins to degrees: 5 c ( ) c ( ) 7 c i ii iii 1 ( ) 11 c ( ) 1 c ( ) 11 c iv v vi 9 1 Perform the correct conversion on ech of the following, giving the nswer correct to two deciml plces. Convert from degrees to rdins: i 7 ii 1 iii 5 iv 51 v vi 1 Convert from rdins to degrees: i 1.7 c ii.87 c iii.8 c iv.1 c v c vi 8.9 c Find the ect vlue of ech of the following: ( ) ( ) sin cos ( ) ( ) 5 9 d cos e cos ( ) ( ) 1 9 g cos h cos Find the ect vlue of ech of the following: c f i ( cos ) ( ) 11 sin ( sin ) sin(15 ) cos( ) c sin(8 ) d cos( ) e sin( 5 ) f sin( )
1 Essentil Specilist Mthemtics 5 If sin( ) =.5 nd 9 < < 18, find: cos( ) tn( ) If cos( ) =.7 nd 18 < < 7, find: sin( ) tn( ) 7 If sin() =.5 nd < <, find: cos() tn() 8 If sin() =. nd < <, find: cos() tn() 9 Solve ech of the following for [, ]: sin = sin () = ( c cos () = 1 d sin + ) = 1 ( ( e cos + )) ( = 1 f sin + ) = 1 Find the ect vlues of ech of the following: ( ) ( 5 tn tn ) ( c tn 9 ) d tn( ) 11 If tn = 1 nd, find the ect vlue of: sin cos c tn( ) d tn( ) 1 If tn = nd, find the ect vlue of: sin cos c tn( ) d tn( ) 1 Solve ech of the following for [, ]: tn = ( tn ) = ( ) ( ) c tn + = d tn + = 1 Sketch the grphs of ech of the following for the stted domin: f () = sin, [, ] f () = cos ( ( c f () = cos + ( e f () = sin ) +, [, ] ( + ) [ ],, )), [, ] d f () = sin() + 1, [, ] 15 Sketch the grphs of ech of the following for [, ], clerl lelling ll intercepts with the es nd ll smptotes: ( f () = tn() f () = tn ) ( c f () = tn + ) ( d f () = tn + )
Chpter 1 toolo 1 1. Solving right-ngled tringles Pthgors theorem This well-known theorem is pplicle to right-ngled tringles nd will e stted here without proof: Emple 1 (hp) = (opp) + (dj) In tringle C, C = 9 nd C =, = cm nd C = 5cm. Find: C the trigonometric rtios relted to c C 5 Pthgors theorem, C = 5 + = 1 C = 1 cm sin = 5 cos = tn = 5 1 1 c tn = 5 = 9.81 (correct to two deciml plces) Eercise 1 1 Find the trigonometric rtios tn, cos nd sin for ech of the following tringles: c 5 9 5 7 8 7 Find the ect vlue of in ech of the following tringles: c 1 5 5
1 Essentil Specilist Mthemtics Find the ect vlue of the pronumerls for ech of the following: 1 5 1 c 1 1 c h d 5 d 1 c 1 Find the vlues of,, z, w nd. Hence deduce ect vlues for sin(15 ), cos(15 ) nd tn(15 ). c Find the ect vlues for sin(75 ), cos(75 ) nd tn(75 ). z w 1 1. The sine nd cosine rules The sine rule In section 1., methods for finding unknown lengths nd ngles for right-ngled tringles were discussed. This section discusses methods for finding unknown quntities in non-right-ngled tringles. The sine rule is used to find unknown quntities in tringle when one of the following situtions rises: one side nd two ngles re given two sides nd non-included ngle re given In the first cse unique tringle is defined, ut for the second it is possile for two tringles to eist. Lelling convention The following convention is followed in the reminder of this chpter. Interior ngles re denoted upper-cse letters, nd the length of the side opposite n ngle is denoted the corresponding lower-cse letter. The mgnitude of ngle C is denoted. The length of side C is denoted. c 5 C
Chpter 1 toolo 15 The sine rule sttes tht for tringle C sin = sin = c sin C c C proof will onl e given for the cute-ngled tringle cse. The proof for otuse-ngled tringles is similr. Proof In tringle CD, sin = h C h = sin In tringle CD, sin = h h = sin h sin = sin i.e. sin = sin Similrl, strting with perpendiculr from to C would give sin = c sin C D Emple 11 Use the sine rule to find the length of. c sin 1 = 1 sin 7 1 sin 1 c = sin 7 c = 5.89... c 7 The length of is 5.8 cm correct to two deciml plces. 1 cm 1 C Emple 1 Use the sine rule to find the mgnitude of ngle XZY, given tht Y = 5, = 5 nd z =. X 5 cm cm Z 5 Y
1 Essentil Specilist Mthemtics 5 sin 5 = sin Z sin Z sin 5 = 5 sin 5 sin Z = 5 =.571... Z = sin 1 (.571...) Z 1 5 cm.7 X 5 cm Z =.7... or 18.7... 19.5 cm Z =.7 or Z = 19.5 correct to two deciml plces. Z 5 Y Rememer: sin(18 ) = sin There re two solutions for the eqution sin Z =.571... Note: When using the sine rule in the sitution where two sides nd non-included ngle re given, the possiilit of two such tringles eisting must e considered. Eistence cn e checked through the sum of the given ngle nd the clculted ngle not eceeding 18. The cosine rule The cosine rule is used to find unknown quntities in tringle when one of the following situtions rises: two sides nd n included ngle re given three sides re given The cosine rule sttes tht for tringle C = + c c cos or, equivlentl, cos = + c c c C The smmetricl results lso hold, i.e. = + c c cos c = + cos C The result will e proved for cute-ngled tringles. The proof for otuse-ngled tringles is similr. Proof In tringle CD C = + h (Pthgors theorem) cos = nd therefore = cos h In tringle CD = (c ) + h (Pthgors theorem) D c
Chpter 1 toolo 17 Epnding gives = c c + + h = c c + (s + h = ) = + c c cos (s = cos ) Emple 1 For tringle C, find the length of in centimetres correct to two deciml plces. c = 5 + 1 5 1 cos 7 = 85.98... c 7 5 cm C c 9.9 1 cm The length of is 9.7 cm correct to two deciml plces. Emple 1 Find the mgnitude of ngle C for tringle C correct to two deciml plces. i.e. cos = + c c = 1 + 15 1 =.15 = (18.99...) cm 15 cm 18.1 correct to two deciml plces. 1 cm C The mgnitude of ngle C is 18 1 (to the nerest second). Emple 15 In C, C = 8, C = 1 cm, = 15 cm. Find, correct to two deciml plces: C C 1 cm 8 15 cm C cm
18 Essentil Specilist Mthemtics C is found ppling the cosine rule: = + c c cos = 1 + 15 1 15 cos 8 = 1 + 5 cos 8 = 18.897... C = = 17.8 cm, correct to two deciml plces. C is found ppling the sine rule pir: sin = c sin C sin C = c sin 15 sin 8 = 17.8 C = 5.8, correct to two deciml plces. Note: 1.7 is lso solution to this eqution ut it is discrded s possile nswer s it is inconsistent with the informtion lred given. Eercise 1C 1 In tringle C, C = 7, C = 55 nd = 1 cm. Find, correct to two deciml plces: C C In tringle C, C = 58, =.5 cm nd C = 8 cm. Find, correct to two deciml plces: C C The djcent sides of prllelogrm re 9 cm nd 11 cm. One of its ngles is 7.Find the length of the longer digonl, correct to two deciml plces. In C, C =, C = 8.5 cm nd = 5. cm. Find, correct to two deciml plces: the two possile vlues of C (one cute nd one otuse) C in ech cse 5 In C, C = 5, = 1 cm nd C =.7 cm. Find, correct to two deciml plces: C C In C, C = 5, C = nd C = 1 cm. Find.
7 In PQR, QPR =, PQ = cm nd PR = cm. Find QR. Chpter 1 toolo 19 8 In C, C hs mgnitude, C = cm nd = 18 cm. Find the distnce C correct to deciml plces. 9 In C, C hs mgnitude, C = 1 cm nd = 8 cm. Find the distnce C using the cosine rule. 1 In C, = 5 cm, C = 1 cm nd C = 1 cm. Find: the mgnitude of C, correct to two deciml plces the mgnitude of C, correct to two deciml plces 1. Geometr prerequisites In the Specilist Mthemtics stud design it is stted tht students should e fmilir with severl geometric results nd e le to ppl them in emples. These results hve een proved in erlier ers stud. In this section the re listed. The sum of the interior ngles of tringle is 18. + + c = 18 c The sum of the eterior ngles of conve polgon is. e c d w z f c e d + + c + d + e = + + z + w = + + c + d + e + f = Corresponding ngles of lines cut trnsversl re equl if, nd onl if, the lines re prllel. = nd c = d nd re corresponding ngles c nd d re corresponding ngles d c
Essentil Specilist Mthemtics Opposite ngles of prllelogrm re equl, nd opposite sides re equl in length. c = DC D = C c = d nd = D d C The se ngles of n isosceles tringle re equl. ( = C) = X The line joining the verte to the midpoint of the se of n isosceles tringle is perpendiculr to the se. The perpendiculr isector of the se of n isosceles tringle psses through the opposite verte. The ngle sutended n rc t the centre of circle is twice the ngle sutended the sme rc t the circumference. C = The ngle in semicircle is right ngle. ngles in the sme segment of circle re equl. z = = z The sum of the opposite ngles of cclic qudrilterl is 18. w + = 18 z + w = 18 z = n eterior ngle of cclic qudrilterl nd the interior opposite ngle re equl.
Chpter 1 toolo 1 trngent to circle is perpendiculr to the rdius t the point of contct. The two tngents to circle from n eterior point re equl in length. X O X = X n ngle etween tngent to circle nd chord through the point of contct is equl to the ngle in the lternte segment. = Emple 1 Find the mgnitude of ech of the following ngles: C DC 9 c CD O d OCD 85 e D D C C = 9 (verticll opposite) DC = 87 (opposite ngle of cclic qudrilterl) c CO = 85 (verticll opposite) CD = [18 ( + 85)] = 5 (ngles of tringle, CO) d CD = 5 (ngle sutended the rc CD) DC = 87 (From ) OCD = [18 (87 + 5)] = 58 (ngles of tringle, CD) e D = [18 ( + 58)] = (opposite ngles of cclic qudrilterl) Eercise 1D 1 Find the vlue of,, z nd. 8 15 z
Essentil Specilist Mthemtics Find the mgnitude of ech of the following: R S c d RTW TSW TRS RWT T 15 7 W Find the vlue of, nd c. is tngent to the circle t C. 5 c C CD is squre nd X is n equilterl tringle. Find the mgnitude of: DXC XDC X D C 5 Find the vlues of,, c, d nd e. Find in terms of, nd c. X 9 7 Y W e d 15 c Z 7 Find the vlues of nd,given tht O is the center of the circle. c 8 Find the vlues of,, c nd d. O 7 5 c d 9 Find the vlues of nd. 1 O is the centre of the circle. Find the vlues of nd. X O 5 C
1.5 Sequences nd series The following re emples of sequences of numers: 1,, 5, 7,9....1,.11,.111,.1111... c Chpter 1 toolo 1, 1 9, 1 7, 1 81... d 1, 7,, 1,... e., 1.7,.8,.9... Note tht ech sequence is set of numers, with order eing importnt. For some sequences of numers rule cn e found connecting n numer to the preceding numer. For emple: for sequence, rule is: dd for sequence C, rule is: multipl 1 for sequence D, rule is: sutrct for sequence E, rule is: dd 1.1 The numers of sequence re clled terms. The nth term of sequence is denoted the smol t n.so the first term is t 1, the 1th term is t 1 nd so on. sequence cn e defined specifing rule which enles ech susequent term to e found using the previous term. In this cse, the rule specified is clled n itertive rule or difference eqution.for emple: sequence cn e defined t 1 = 1, t n = t n 1 + sequence C cn e defined t 1 = 1, t n = 1 t n 1 Emple 17 Use the difference eqution to find the first four terms of the sequence t 1 =, t n = t n 1 + 5 t 1 = t = t 1 + 5 = 8 t = t + 5 = 1 t = t + 5 = 18 The first four terms re, 8, 1, 18. Emple 18 Find the difference eqution for the following sequence. 9,, 1, 1... = 1 9 i.e. t = 1 t 1 1 = 1 i.e. t = 1 t t n = 1 t n 1, t 1 = 9
Essentil Specilist Mthemtics lterntivel, sequence cn e defined rule tht is stted in terms of n. For emple: t n = n defines the sequence t 1 =, t =, t =, t = 8... t n = n 1 defines the sequence t 1 = 1, t =, t =, t = 8... Emple 19 Find the first four terms of the sequence defined the rule t n = n +. t 1 = 1 + = 5 t = + = 7 t = + = 9 t = + = 11 The first four terms re 5, 7, 9, 11. rithmetic sequences sequence in which ech successive term is found dding constnt vlue to the previous term is clled n rithmetic sequence. For emple,, 5, 8, 11... is n rithmetic sequence. n rithmetic sequence cn e defined difference eqution of the form t n = t n 1 + d where d is constnt. If the first term of n rtithmetic sequence t 1 = then the nth term of the sequence cn lso e descried the rule: t n = + (n 1) d where = t 1 nd d = t n t n 1 d is the common difference. Emple Find the tenth term of the rithmetic sequence, 1,, 5... =, d =, n = 1 t n = + (n 1) d t 1 = + (1 1) = rithmetic series The sum of the terms in sequence is clled series. If the sequence in question is rithmetic, the series is clled n rithmetic series. The smol S n is used to denote the sum of n terms of sequence, i.e. S n = + ( + d) + ( + d) + +( + (n 1) d) If this sum is written in reverse order, then S n = ( + (n 1) d) + ( + (n ) d) + +( + d) +
dding these two epression together gives S n = n[ + (n 1) d] S n = n [ + (n 1) d] nd since the lst term l = t n = + (n 1) d S n = n ( + l) Chpter 1 toolo 5 Geometric sequences sequence in which ech successive term is found multipling the previous term fied vlue is clled geometric sequence. Foremple,,, 18, 5... is geometric sequence. geometric sequence cn e defined n itertive eqution of the form t n = rt n 1,where r is constnt. If the first term of geometric sequence t 1 =, then the nth term of the seqeunce cn lso e descried the rule t n = r n 1 where r = t n t n 1 r is clled the common rtio. Emple 1 Clculte the tenth term of the sequence,, 18,... =, r =, n = 1 t n = r n 1 t 1 = (1 1) = 9 Geometric series The sum of the terms in geometric sequence is clled geometric series. nepression for S n, the sum of n terms, of geometric sequence cn e found using similr method to tht used in the development of formul for n rithmetic series. Let S n = + r + r + +r n 1... 1 Then rs n = r + r + r + +r n... Sutrct 1 from rs n S n = r n S n (r 1) = (r n 1) nd S n = (r n 1) r 1
Essentil Specilist Mthemtics Forvlues of r such tht 1 < r < 1, it is often more convenient to use the lterntive formul S n = (1 r n ) 1 r which is otined sutrcting from 1 ove. Emple Find the sum of the first nine terms of the geometric sequence 1, 1 9, 1 7, 1 81... = 1, r = 1 (, n = 9 1 ( 1 ) ) 9 1 S 9 = = 1 1 ( 1 ( 1 ) ) 9 1 1 (.999 99).99 975 (to deciml plces) Infinite geometric series If the common rtio of geometric sequence hs mgnitude less thn 1, i.e. 1 < r < 1, then ech successive term of the sequence is closer to zero. e.g.,,, 1, 1, 1... When the terms of the sequence re dded, the corresponding series + r + r + +r n 1 will pproch limiting vlue, i.e. s n, S n limiting vlue. Such series is clled convergent. In emple ove, it ws found tht for the sequence 1, 1 9, 1 7, 1...the sum of the first 81 nine terms, S 9,ws.99975. For the sme sequence, S =.99 999 999 9.5 So even for reltivel smll vlue of n (), the sum pproches the limiting vlue of.5 ver quickl. Given tht S n = (1 r n ) 1 r S n = 1 r rn 1 r s n, r n nd hence rn 1 r It follows then tht the limit s n of S n is So S = 1 r 1 r This is lso referred to s the sum to infinit of the series.
Chpter 1 toolo 7 Emple Find the sum to infinit of the series 1 + 1 + 1 + 1 8 +... r = 1, = 1 S = 1 1 1 = Using grphics clcultor Emple Grph the terms of the geometric sequence defined : n = 51(.5) n 1 for n = 1,... Step 1 Set our clcultor to sequence mode. Enter the MODE menu; move cursor down to line nd cross to cover Seq. Press ENTER to select sequence mode. Step Enter epression defining sequence. Press Y= nd enter: set nmin = 1 u(n) = 51(.5) ˆ (n 1) (Note: n is otined pressing XT N ) c u(nmin) = 51 Step Set viewing window. Press WINDOW nd enter the vlues s shown. Step Plot. Press GRPH. Note tht terms of the sequence cn e viewed in the TLE window.
8 Essentil Specilist Mthemtics Emple 5 Generte the first 1 terms in the rithmetic sequence whose nth term is 8 + (n 1) nd store them in list L1. Find the sum of these first 1 terms. c This rithmetic sequence cn lso e generted the recursive form, t n = t n 1 +, with t 1 = 8. Generte the first 1 terms in this w Select Seq mode from the MODE menu. In the Y= window enter nmin = 1, u(n) = 8 + (n 1), nd u(nmin) = 8 Press STT nd select 1:Edit from the EDIT menu. Tke the cursor to the top of L (ctull over L ) nd press ENTER.Inthe entr line enter L = 8 + (L 1 1) nd press ENTER. The result is s shown. Tke the cursor to L nd enter L = cumsum(l ) nd press ENTER. The result is s shown. cumsum is otined selecting : cumsum from the OPS sumenu of LIST. c In the Y= window enter, nmin = 1, u(n) = u(n 1) + nd u(nmin) = 8. Note tht u is otined pressing ND 7 nd n is otined pressing XT N. The sequence cn e seen in the TLE screen. Using CS clcultor The method is ver similr to tht discussed ove. Press MODE nd use the right rrow from Grph to revel the Grph menu. Select :SEQUENCE nd press ENTER.
Chpter 1 toolo 9 In the Y= screen enter u1(n) = u1(n 1) + nd ui1 = 8. The sequence cn e seen in the TLE window. Eercise 1E 1 difference eqution hs rule t n+1 = t n 1, t 1 =. Find t nd t. Use grphics clcultor to find t 8. difference eqution hs rule n+1 = n +, 1 = 5. Find nd. Use grphics clcultor to find 1 nd to plot grph showing the first ten vlues. The Fioncci sequence is given the difference eqution t n+ = t n+1 + t n where t 1 = t = 1. Find the first ten terms of the Fioncci sequence. Find the sum of the first ten terms of n rithmetic sequence with first term nd common difference. 5 Find the sum to infinit of 1 1 + 1 9 1 7 +... The first, second nd third terms of geometric sequence re + 5, nd respectivel. Find: the common rtio c the difference etween the sum to infinit nd the sum of the first ten terms 7 Find the sum of the first eight terms of geometric sequence with first term nd common rtio. 8 Find the sum to infinit of the geometric sequence,,,...in terms of. 9 Consider the sum S n = 1 + + n 1 + + n 1 Clculte S 1 when = 1.5. i Find the possile vlues of for which the infinite sum eists. Denote this sum S. ii Find the vlues of for which S = S 1. 1 Find n epression for the infinite geometric sum 1 + sin + sin +... Find the vlues of for which the infinite geometric sum is. 1. Circles Forcircle with centre the origin nd rdius r, if the point with coordintes (, ) is on the circle then + = r.
Essentil Specilist Mthemtics The converse is lso true, i.e. point with coordintes (, ) such tht + = r lies on the circle with centre the origin nd rdius r. ppling Pthgors theorem to tringle OP ields P(, ) r = OP = + r O In generl, the following result holds: The circle with centre (h, k) nd rdius r is the grph of the eqution ( h) + ( k) = r This grph is otined from the grph of + = r the trnsltion defined (, ) ( + h, + k) Emple Sketch the grph of the circle with centre t (, 5) nd rdius, nd stte the crtesin eqution for this circle. The eqution is ( + ) + ( 5) = which m lso e written s + + 1 + 5 = 7 5 Note: The eqution + + 1 + 5 = cn e unsimplified completing the squre. + + 1 + 5 = implies + + + 1 + 5 + 5 = 9 i.e. ( + ) + ( 5) = This suggests generl form of the eqution of circle. + + D + E + F = Completing the squre gives i.e. + D + D + + E + E + F = D + E ( + D ) ( + + E ) = D + E F
Chpter 1 toolo 1 ( D If D + E F >, then the eqution represents circle with centre, E ) nd D + E rdius F. ( D If D + E F =, then the eqution represents one point, E ). If D + E F <, then the eqution hs no grphicl representtion in the crtesin plne. Emple 7 Sketch the grph of + + + 1 =. Stte the coordintes of the centre nd the rdius. + + + 1 = + + + + + 9 1 = 1 i.e. ( + ) + ( + ) = 5 The circle hs centre (, ) nd rdius 5. Emple 8 + 1 (, ) 1 Sketch grph of the region of the plne such tht + < 9 nd 1. = 1 required region Eercise 1F 1 Find the equtions of the circles with the following centres nd rdii: centre (, ); rdius 1 centre (, ); rdius 5 c centre (, 5); rdius 5 d centre (, ); rdius Find the rdii nd the coordintes of the centres of the circles with the following equtions: + + + 1 = + + 1 = c + = d + + 1 + 5 =
Essentil Specilist Mthemtics Sketch the grphs of ech of the following: + + + = + + = c + + 8 1 + 1 = d + 8 1 + 1 = e + 8 + 5 + 1 = f + + 9 = 1 Sketch the grphs of the regions of the plne specified the following: + 1 + 9 c ( ) + ( ) < d ( ) + ( + ) > 1 e + 1 nd f + 9 nd 1 5 The points (8, ) nd (, ) re the ends of dimeter of circle. Find the coordintes of the centre nd the rdius of the circle. Find the eqution of the circle, centre (, ), which touches the is. 7 Find the eqution of the circle which psses through (, 1), (8, ) nd (, ). 8 Find the rdii nd coordintes of the centre of the circles with equtions + 7 + 5 = nd + 1 1 + 9 =, nd find the coordintes of the points of intersection of the two curves. 9 Find the coordintes of the points of intersection of the circle with eqution + = 5 nd the line with eqution: = = 1.7 Ellipses nd hperols Ellipses The curve with eqution + = 1isn ellipse with centre the origin, -is intercepts (, ) nd (, ) nd -is intercepts (, ) nd (, ). If > the ellipse will pper s shown in the digrm on the left. If > the ellipse is s shown in the digrm on the right. If =, the eqution is tht of circle with centre the origin nd rdius. _ ' ' ' _ + = 1; > is the mjor is is the minor is + = 1; > is the minor is is the mjor is
Chpter 1 toolo The generl crtesin form is s given elow. The curve with eqution ( h) ( k) + = 1 is n ellipse with centre (h, k). It is otined trnsltion of the ellipse + = 1. The trnsltion is (, ) ( + h, + k). (h, k + ) (h _, k) (h, k) (h +, k) (h, k _ ) The ellipse with eqution + = 1 cn e otined ppling the following diltions to the circle with eqution + = 1: diltion of fctor from the is, i.e. (, ) (, ) diltion of fctor from the is, i.e. (, ) (, ) The result is the trnsformtion (, ) (, ). 1 1 1 (, ) (, ) (, ) (, ) 1 1 1 Emple 9 Sketch the grph of ech of the following. Give the es intercepts nd the coordintes of the centre. 9 + = 1 + 9 = 1 c ( ) + 9 ( ) 1 = 1 d + + + =
Essentil Specilist Mthemtics _ Centre (, ) es intercepts (±, ) nd (, ±) c Centre is t (, ) When = ( ) + = 1 9 1 ( ) = 5 1 9 ( ) = 1 5 9 = ± 5 Centre (, ) es intercepts (±, ) nd (, ±) When = ( ) + 9 9 1 = 1 ( ) = 7 9 1 ( ) = 9 7 1 = ± 7 (, 7) 5 + ( 1, ) (, ) (5, ) 5 _ 7 _ 7 + (, 1) d + + + = Completing the squre ields [ + 8 + 1] + + 8 = i.e. ( + ) + = 1 ( + ) Centre (, ) es intercepts (, ) nd (, ) + 1 = 1 ( _, ) ( _, ) ( _, ) ( _, ) ( _, _ )
Chpter 1 toolo 5 Defining n ellipse In the previous section circle ws defined s set of points which re ll constnt distnce from given point (the centre). n ellipse cn e defined in similr w. Consider the set of ll points P such tht PF 1 + PF is equl to constnt k with k > m, nd the coordintes of F 1 nd F re (m, ) nd ( m, ) respectivel. We cn show tht the eqution descriing this set of points is P 1 P + = 1where k =. m P 1 F 1 + P 1 F = P F 1 + P F = P F 1 + P F O F 1 F P This cn e pictured s string of length P 1 F 1 + P 1 F eing ttched nils to ord t F 1 nd F nd, considering the pth mpped out pencil, etending the string so tht it is tut, nd moving round the two points. Let the coordintes of P e (, ). nd ssume Then PF 1 = ( m) + nd PF = ( + m) + PF 1 + PF = k ( + m) + + ( + m) + = k Rerrnging nd squring gives ( + m) + = k k ( m) + + ( + m) + m = k k ( m) + Rerrnging nd squring gin gives Collecting like terms k ( m) + k = k 8k m + 1m Let = k, then (k m ) + k = k (k m ) k + k m = 1 + m = 1
Essentil Specilist Mthemtics The points F 1 nd F re clled the foci of the ellipse. The constnt k = is clled the focl sum. If = nd m =, the ellipse with eqution 9 + = 1isotined. 5 Forn ellipse with eqution + = 1 nd >, the foci re t (± ( ), ) Given n eqution of the form + + C + E + F =, where nd re oth positive (or oth negtive), the corresponding grph is n ellipse or point. If = the grph is tht of circle. In some cses, s for the circle, no pirs (, ) will stisf the eqution. Hperols The curve with eqution = 1ishperol with centre t the origin. The is intercepts re (, ) nd (, ). The hperol hs smptotes = nd =.ninforml rgument for this is s follows. The eqution = 1 cn e rerrnged: = 1 = ) (1 _ = = ut s ±, ( _, ) (, ) i.e. ± = 1 The generl eqution for hperol is formed suitle trnsltions. The curve with eqution ( h) ( k ) = 1 is hperol with centre (h, k). The smptotes re k =± ( h) This hperol is otined from the hperol with eqution = 1 the trnsltion defined (, ) ( + h, + k).
Chpter 1 toolo 7 Emple For ech of the following equtions, sketch the grph of the corresponding hperol, give the coordintes of the centre nd the es intercepts, nd the equtions of the smptotes. 9 = 1 9 = 1 c ( 1) ( + ) = 1 d 9 = 1 = (1 9 ) 9 Eqution of smptotes: ( 1) ( + ) 9 = 1 = _ = ( _, ) (, ) =± When =, = 9 nd therefore =±. es intercepts (, ) nd (, ), centre (, ). 9 = 1is the reflection of 9 = 1in the line =. smptotes re =± = _ = (, ) i.e. =± The -is intercepts re (, ) nd (, ). (, _ ) c ( 1) ( + ) = 1. The grph of = 1issketched first. The smptotes re = nd =. This hperol is clled rectngulr hperol s its smptotes re perpendiculr. The centre is (, ) nd the es intercepts re t (1, ) nd ( 1, ) = _ = ( _ 1, ) (1, )
8 Essentil Specilist Mthemtics trnsltion of (, ) ( + 1, ) is pplied. The new centre is (1, ) nd the smptotes hve equtions + =±( 1), i.e. = nd = 1. When =, = nd when = ( 1) = 5 = 1 ± 5 = 1 = 1 _ (1 _ 5, ) 1 (1, _ (1 + 5, ) ) (, _ ) (, _ ) _ d ( 1) ( + ) = 1is otined trnslting the hperol 9 9 = 1 through the trnsltion defined (, ) (, + 1). = _ = _ + _ 7 ( _, ) (, ) ( _ 1) ( + ) = 1 9 _ ( = 1 _, 1) 9, _ ( _, _ 1) ) = _ 1 _ = Note tht the smptotes for = 1 re the sme s for those of the hperol 9 9 = 1. The two hperols re clled conjugte hperols. Defining hperol Hperols cn e defined in mnner similr to the methods discussed erlier in this section for circles nd ellipses. Consider the set of ll points, P, such tht PF 1 PF = k where k is constnt nd F 1 nd F re points with coordintes (m, ) nd ( m, ) respectivel. Then the eqution of the curve defined in this w is m = 1 k =
Chpter 1 toolo 9 Eercise 1G 1 Sketch the grph of ech of the following. Lel the es intercepts. Stte the coordintes of the centre. 9 + 1 = 1 5 + 1 = c ( ) ( 1) + = 1 d ( ) + = 1 9 1 9 e 9 + 5 5 1 = f 9 + 5 = 5 g 5 + 9 + 18 1 = h 1 + 5 + 1 8 = i ( ) + ( ) 9 = 1 j ( ) + ( 1) = 1 Sketch the grphs of ech of the following. Lel the es intercepts nd give the equtions of the smptotes. 1 9 = 1 1 9 = 1 c = d = e 8 1 = f 9 5 9 + 15 = 5 g ( ) ( ) 9 = 1 h 8 + = i 9 1 18 + 151 = j 5 1 = Find the coordintes of the points of intersection of = 1 with: = 1 + = 1 Show tht there is no intersection point of the line = + 5 nd the ellipse + = 1. 5 Find the coordintes of the points of intersection of the curves 9 + + 9 = 1 nd = 1. Show tht the points of intersection re the vertices of squre. Find the coordintes of the points of intersection of eqution 5 =. 1 + 5 = 1 nd the line with 7 On the one set of es sketch the grphs of + = 9 nd = 9. 8 Sketch ech of the following regions: 1 c 1 d 9 + < 1
Essentil Specilist Mthemtics e 1 nd + f ( ) + 1 9 1 g nd 9 + 1 h > 1 nd + i ( ) + j 9 + 1 nd 1.8 Prmetric equtions of circles, ellipses nd hperols Circles It is sometimes useful to epress the rule of reltion in terms of third vrile, clled prmeter.wehve lred seen in the work on circulr functions tht the unit circle cn e epressed in crtesin form, i.e. {(, ): + = 1} or in the form {(, ): = cos t, = sin t, with t [, ]}. The ltter is clled the set of prmetric equtions of the unit circle which give the coordintes (, )of ll points on the unit circle. The restriction for the vlues of t is unnecessr in the representtion of the grph s {(, ): = cos t, = sin t, with t R}gives the sme points with repetitions since cos( + t) = cos t nd sin( + t) = sin t.if the set of vlues for t is the intervl [, ], onl the top hlf of the circle is otined. The set nottion is often omitted, nd in the following this will e done. The net three digrms illustrte the grphs resulting from the prmetric equtions = cos t nd = sin t for three different sets of vlues of t. t [, ] t [, ] t [, ] In generl, + =,where >, is the crtesin eqution of circle with centre t the origin nd rdius. The prmetric equtions re = cos t nd = sin t. The miniml intervl of t vlues to ield the entire circle is [, ]. The domin nd rnge of the corresponding crtesin reltion cn e determined the prmetric eqution determining the vlue nd the vlue respectivel. The rnge of the function with rule = cos t, t [, ] is [, ] nd hence the domin of the reltion + = is [, ]. The rnge of the function with rule = sin t, t [, ] is [, ] nd hence the rnge of the reltion + = is [, ].
Chpter 1 toolo 1 Emple 1 circle is defined the prmetric equtions = + cos nd = 1 + sin for [, ], Find the corresponding crtesin eqution of the circle nd stte the domin nd rnge of this reltion. Ellipses The rnge of the function with rule = + cos is [ 1, 5] nd hence the domin of the corresponding crtesin reltion is [ 1, 5]. The rnge of the function with rule = 1 + sin is [, ] nd hence the rnge of the corresponding crtesin reltion is [, ]. Rewrite the equtions s = cos nd 1 = sin. Squre oth sides of ech of these equtions nd dd: ( ) ( 1) + = cos + sin nd therefore 9 9 i.e. ( ) + ( 1) = 9 ( ) 9 + ( 1) 9 It hs een shown in the previous section tht + = 1, where nd re positive rel numers, is the crtesin eqution of n ellipse with centre t the origin, nd -is intercepts (±, ) nd -is intercepts (, ±). The prmetric equtions for such n ellipse re = cos t nd = sin t. The miniml intervl of t vlues to ield the entire ellipse is [, ]. The domin nd rnge of the corresponding crtesin reltion cn e determined the prmetric eqution determining the vlue nd the vlue respectivel. The rnge of the function with rule = cos t is [, ] nd hence the domin of the reltion + = 1 is [, ]. The rnge of the function with rule = sin t is [, ] nd hence the rnge of the reltion + = 1 is [, ]. The proof tht the two forms of eqution ield the sme grph uses the Pthgoren identit sin t + cos t = 1. Let = cos t nd = sin t. Therefore = cos t nd = sin t. Squring oth sides of ech of these equtions ields = cos t nd = sin t Now, since sin t + cos t = 1it follows tht Emple + = 1. = 1 Find the crtesin eqution of the curve with prmetric equtions = + sin t, = cos t with t R nd descrie the grph.
Essentil Specilist Mthemtics = + sin t nd = cos t, therefore = sin t nd Squre oth sides of ech eqution nd dd: Hence ut ( ) ( ) + = sin t + cos t 9 ( ) ( ) + = 1 9 ( ) = ( ) so this eqution is more netl written s = cos t. ( ) ( ) + = 1 9 Clerl this is n ellipse, with centre t (, ), nd es intercepts t (, ) nd (, ). Hperols The generl crtesin eqution for hperol with centre t the origin is = 1. The prmetric equtions re = sec t nd = tn t where sec t = 1 ( cos t nd t, ) gives ( the right-hnd rnch of the hperol. For the function with rule = sec t nd domin, ) the rnge is [, ). (The sec function is discussed further in Chpter.) = π = π ( The grph of = sec is shown for the intervl, ). ( For the function with rule = tn t nd domin, ) the rnge is R. The left rnch ( of the hperol cn e otined for t, ). The proof tht the two forms of eqution cn ield the sme grph uses form of the Pthgoren identit sin t + cos t = 1. Divide oth sides of this identit cos t. This ields tn t + 1 = sec t. Consider = sec t nd = tn t. Therefore = sec t nd = tn t. Squre oth sides of ech eqution to otin = sec t nd = tn t. Now, since tn t + 1 = sec t,itfollows tht + 1 =. Therefore = 1.
Chpter 1 toolo Emple Find the crtesin ( eqution of the curve with prmetric equtions = sec t, = tn t, where t, ) nd descrie the curve. Now = sec t nd = tn t. Therefore = sec t nd = tn t. Squre oth sides of ech eqution to otin 9 = sec t nd 1 = tn t. dd these two equtions to otin 1 + 1 = 9. So the crtesin form of the curve is 9 1 = 1. The rnge of the function with rule = sec t for t (, ) is (, ]. Hence the domin for the grph is (, ]. This is the left rnch of hperol, with centre t the origin, nd intercept t (,) nd with smptotes with equtions = nd =. Emple Give prmetric equtions for ech of the following: + = 9 1 + = 1 c ( 1) 9 c ( + 1) The prmetric equtions re = cos t nd = sin t or = sin t nd = cos t. There re infinitel mn pirs of equtions which determine the curve given the crtesin eqution + = 9. Others re = cos(t) nd = sin(t). For = cos t nd = sin t it is sufficient for t to e chosen for the intervl [, ]to otin the whole curve. For = cos(t) nd = sin(t)itissufficient for t to e chosen in the intervl [, ]. The ovious solution is = cos t nd = sin t The ovious solution is 1 = sec t nd + 1 = tn t = 1 Using grphics clcultor with prmetric equtions Press MODE nd select Pr from the fourth row of the menu. Ensure tht the clcultor is in rdin mode.
Essentil Specilist Mthemtics In the Y= screen enter X 1T = cos T nd Y 1T = sin T. The T is otined pressing the ke XT N. The period of oth functions is shown.. Press WINDOW nd complete the settings s From the ZOOM menu select Zsqure nd ctivte TRCE. The TRCE cn e used to see the order of plotting. Using CS clcultor with prmetric equtions Press MODE nd select PRMETRIC s shown. Enter the functions in the Y= screen. From WINDOW estlish the settings s shown. From the ZOOM menu choose ZoomSqure. ctivte TRCE. Eercise 1H 1 Find the crtesin eqution of the curve determined the prmetric equtions = cos t nd = sin t, nd determine the domin nd rnge of the corresponding reltion.
Chpter 1 toolo 5 Determine the corresponding crtesin eqution of the curve determined ech of the following prmetric equtions nd sketch the grph of ech of these. ( = sec t, = tn t, t, ) = cos t, = sin t [ c = cos t, = + sin t d = sin t, = cos t, t, ] ( e = sec t, = tn t, t, ) f = 1 sec (t), = 1 + tn(t), ( t, ) Give prmetric equtions corresponding to ech of the following: + = 1 9 = 1 c ( 1) + ( + ) = 9 d ( 1) 9 + ( + ) circle hs centre (1, ) nd rdius. If prmetric equtions for this circle re = + cos( t) nd = c + d sin( t), where,, c nd d re positive constnts, stte the vlues of,, c nd d. 5 n ellipse hs -is intercepts (, ) nd (, ) nd -is intercepts (, ) nd (, ). Stte possile pir of prmetric equtions for this ellipse. The grph of the circle with prmetric equtions = cos t nd = sin t is dilted fctor from the is. For the imge curve, stte: possile pir of prmetric equtions the crtesin eqution ( ) t 7 The grph of the ellipse with prmetric equtions = cos nd ( ) t = + sin is trnslted units in the negtive direction of the is nd units in the negtive direction of the is. For the imge curve stte: possile pir of prmetric equtions the crtesin eqution 8 Sketch the grph of the curve with prmetric equtions = + sin( t) nd = + cos( t) for: t [ ], 1 t [, 1 ] c t [ ], For ech of these, stte the domin nd rnge. = 9
Essentil Specilist Mthemtics Review Summr of circles, ellipses nd hperols Circles The circle with centre t the origin nd rdius is the grph of the eqution + =. The circle with centre (h, k) nd rdius is the grph of the eqution ( h) + ( k) =. In generl, + =,where >, is the crtesin eqution of circle with centre t the origin nd rdius. The prmetric equtions re = cos t nd = sin t. The miniml intervl of t vlues to ield the entire circle is [, ]. The circle with centre (h, k) nd rdius cn e descried through the prmetric equtions = h + cos t nd = k + sin t. Ellipses The curve with eqution + = 1isn ellipse with centre the origin, -is intercepts (, ) nd (, ) nd -is intercepts (, ) nd (, ). For > the ellipse will pper s shown in the digrm on the left. If > the ellipse is s shown in the digrm on the right. ' _ ' _ ' ' ( h) ( k) The curve with eqution + = 1isnellipse with centre (h, k). The ellipse with centre t the origin, nd -is intercepts (±, ) nd -is intercepts (, ±) hs prmetric equtions = cos t nd = sin t. The miniml intervl of t vlues to ield the entire ellipse is [, ]. The ellipse with centre (h, k) formed trnslting the ellipse with eqution + = 1 cn e descried through the prmetric equtions = h + cos t nd = k + sin t.
Chpter 1 toolo 7 Hperols The curve with eqution = 1 is hperol with centre t the origin. The is intercepts re (, ) nd (, ). The hperol hs smptotes = nd =. ( h) ( k) The curve with eqution = 1 is hperol with centre (h, k). The hperol hs _ = ( _, ) (, ) = smptotes h = ( k) nd h = ( k). The prmetric equtions for the hperol shown ove re = sec t nd = tn t where sec t = 1 cos t. The hperol with centre (h, k) formed trnslting the hperol with eqution = 1 cn e descried through the prmetric equtions = h + sec t nd = k + tn t. Review Multiple-choice questions 1 The rd term of geometric sequence is. If the 8th term is 18, then the 1st term is: 1 C D 5 E none of these If the numers 5, nd re in rithmetic sequence then: = + 5 = 5 C = + 5 D = 5 E none of these If cos =, the vlue of the cute ngle is: C 5 D 5 E 7.5 The eqution of the grph shown is: ( = sin ) 1 ( = cos + ) C = sin() π D = sin() ( 1 E = sin + ) ( ) ( ) ( ) 5 The ect vlue of the epression sin cos tn is: 1 1 C D E none of these π
8 Essentil Specilist Mthemtics Review In the digrm,,, C nd D re points on the circle. D = 5 nd X = 1. The mgnitude of XDC is: 5 C 5 D 5 E 55 5 1 X 7 In geometric sequence t = nd t = 5. The sum of the first 5 terms, if the common rtio is positive, is: 1 11 C 8 D 1.5 E 81 8 In tringle C, =, = 1 nd cos C = 51. The vlue of c to the nerest whole 5 numer is: 9 1 C 11 D 81 E 19 9 The coordintes of the centre of the circle with eqution 8 + = 8 re: ( 8, ) (8, ) C (, 1) D (, 1) E (1, ) 1 The eqution of the grph shown is: C D E ( + ) 7 ( ) 9 ( + ) 81 ( ) 81 ( + ) 9 18 = 1 = 1 = 1 = 1 = 1 7 D C 11 Short-nswer questions (technolog-free) 1 For the difference eqution f n = 5 f n 1, f = 1, find f n in terms of n. P nd P re tngents to the circle with centre O. If P = 1 cm, find OP. O α P
Chpter 1 toolo 9 Write down the eqution of the ellipse shown. ( _, 7) Find sin. θ 8 7 Review (, ) O 5 Find. cm 9 cm circle hs chord of length 1 cm situted cm from its centre. Find: the rdius length the ngle sutended the chord t the centre 7 Find the ect vlue of cos 15. Given tht tn = nd 18 < < 7, find n ect vlue of cos. c Find n ngle ( ) such tht sin = sin. 8 In the digrm, D is tngent to the circle with centre O, C intersects the circle t, nd D =. Find CD in terms of. If D = cm, = cm nd C = cm, epress in terms of nd. 9 C is horizontl right-ngled tringle with the right ngle t. P is point cm directl ove. The length of is 1 cm nd the length of C is 1 cm. Find, the ngle which the tringle CP mkes with the horizontl. 1 Solve cos( + ) 1 =,. Sketch the grph of = cos( + ) 1,, clerl lelling es intercepts. c Solve cos( + ) < 1,. C O D P C
5 Essentil Specilist Mthemtics Review 11 The tringulr se C of tetrhedron hs side lengths = 15 cm, C = 1 cm nd C = 9 cm. If the pe D is 9 cm verticll ove C, then find: the ngle C of the tringulr se the ngles tht the sloping edges mke with the horizontl 1 Two ships sil from port t the sme time. One sils nuticl miles due est in three hours, nd the other sils nuticl miles on ering of in the sme time. How fr prt re the ships three hours fter leving port? How fr prt would the e in five hours if the mintined the sme erings nd constnt speed? 1 Find. 18 cm cm 5 1 n irport is 8 km due est of irport. pilot flies on ering of 5 from to C nd then on ering of 15 from C to. Mke sketch of the sitution. Determine how fr the pilot flies from to C. c Determine the totl distnce the pilot flies. 15 Find the equtions of the smptotes for the hperol with rule ( ) = 15. 9 1 curve is defined the prmetric equtions = cos(t) + nd = sin(t). Give the crtesin eqution of the curve. 17 Find the vlue of. Find,, c nd d,given tht PR is tngent to the circle with centre O. P c O d R 18 curve is defined the prmetric equtions = cos( t) nd = sin( t) +. Give the crtesin eqution of the curve.
Chpter 1 toolo 51 ( 19 Sketch the grphs of = cos nd = cos ) on the sme set of es, for [, ]. ( Solve cos ) = for [, ]. c Solve cos for [, ]. Find ll ngles, such tht, where: sin = 1 cos = c tn = 1 1 circle hs centre (1, ) nd rdius. If prmetric equtions for this circle re = + cos( t) nd = c + d sin( t), where,, c nd d re positive constnts, stte the vlues of,, c nd d. O is the centre of circle with points, C, D nd E on the circle. Find: D EC E O c DC Review D C Find the centre nd rdius of the circle with eqution + 8 + 1 + =. Find the - nd -es intercepts of the grph of the ellipse 81 + 9 = 1. 5 The first term of n rithmetic sequence is (p + 5) where p is positive integer. The lst term is (17p + 17) nd the common difference is. Find in terms of p: i the numer of terms ii the sum of the sequence Show tht the sum of the sequence is divisile 1 onl when p is odd. sequence is formed using rising powers of :, 1,,... Find the nth term. Find the product of the first twent terms. Etended-response questions 1 hiker wlks from point on ering of 1 for 5 km nd then on ering of 75 for 7kmto rech point. Find the length of. Find the ering of from the strt point. second hiker trvels from point on ering of 8 for km to point P, nd then trvels in stright line to. c Find: i the totl distnce trvelled the second hiker ii the ering on which the hiker must trvel in order to rech from P.
5 Essentil Specilist Mthemtics Review third hiker lso trvels from point on ering of 8 nd continues on tht ering until he reches point C.He then turns nd wlks towrds. In doing so, the two legs of the journe re of equl length. d Find the distnce trvelled the third hiker to rech. n ellipse is defined the rule ( + ) + = 1. 5 Find: i the domin of the reltion ii the rnge of the reltion iii the centre of the ellipse. E is n ellipse given the rule ( h) + ( k) = 1. The domin of E is [ 1, ] nd its rnge is [ 1, 5]. Find the vlues of,, h nd k. The line = intersects the ellipse E t (1, 1) nd t P. c Find the coordintes of the point P. line perpendiculr to the line = is drwn t P. This line intersects the is t Q. d Find the coordintes of Q. e Find the eqution of the circle through, P nd Q. Show tht the circle with eqution + + =, touches oth the is nd the is. Show tht ever circle tht touches the is nd is hs n eqution of similr form. c Hence show tht there re ectl two circles pssing through the point (, ) nd just touching the is nd is nd give their equtions. d Stte the coordintes of the centres of these two circles nd give the rdius of ech of these circles. e For ech of the circles, find the grdient of the line which psses through the centre nd the point (, ). f Find n eqution to the tngent to ech circle t the point (, ). circle is defined the prmetric eqution = cos nd = sin. Let P e the point with coordintes ( cos, sin ). Find the eqution of the stright line which psses through the origin nd the point P. Stte the coordintes, in terms of, of the other point of intersection of the circle with the stright line through the origin nd P. c Find the eqution of the tngent to the circle t the point P. d Find the coordintes of the points of intersection nd of the tngent with the is nd is respectivel. e Find the re of tringle of O in terms of if < <.Find the vlue of for which the re of this tringle is minimum.