Trigonometric equations

Transcription

1 Trigonometric eqations In this nit we consider the soltion of trigonometric eqations. The strategy we adopt is to find one soltion sing knowledge of commonly occring angles, and then se the symmetries in the graphs of the trigonometric fnctions to dedce additional soltions. Familiarity with the graphs of these fnctions is essential. In order to master the techniqes eplained here it is vital that yo ndertake the practice eercises provided. After reading this tet, and/or viewing the video ttorial on this topic, yo shold be able to: find soltions of trigonometric eqations se trigonometric identities in the soltion of trigonometric eqations Contents. Introdction. Some special angles and their trigonometric ratios. Some simple trigonometric eqations 4. Using identities in the soltion of eqations 8 5. Some eamples where the interval is given in radians 0 c mathcentre Jne 4, 004

2 . Introdction This nit looks at the soltion of trigonometric eqations. In order to solve these eqations we shall make etensive se of the graphs of the fnctions sine, cosine and tangent. The symmetries which are apparent in these graphs, and their periodicities are particlarly important as we shall see.. Some special angles and their trigonometric ratios. In the eamples which follow a nmber of angles and their trigonometric ratios are sed freqently. We list these angles and their sines, cosines and tangents sin 0 cos tan 0 0. Some simple trigonometric eqations Sppose we wish to solve the eqation sin =0.5 and we look for all soltions lying in the interval This means we are looking for all the angles,, in this interval which have a sine of 0.5. We begin by sketching a graph of the fnction sin over the given interval. This is shown in Figre. sin o 90 o 50 o 80 o 70 o 60 o Figre. A graph of sin. We have drawn a dotted horizontal line on the graph indicating where sin =0.5. The soltions of the given eqation correspond to the points where this line crosses the crve. From the Table above we note that the first angle with a sine eqal to 0.5 is 0. This is indicated in Figre. Using the symmetries of the graph, we can dedce all the angles which have a sine of 0.5. These are: =0, 50 This is becase the second soltion, 50, is the same distance to the left of 80 that the first is to the right of 0. There are no more soltions within the given interval. c mathcentre Jne 4, 004

3 Sppose we wish to solve the eqation cos = 0.5 and we look for all soltions lying in the interval As before we start by looking at the graph of cos. This is shown in Figre. We have drawn a dotted horizontal line where cos = 0.5. The soltions of the eqation correspond to the points where this line intersects the crve. One fact we do know from the Table on page is that cos 60 =+0.5. This is indicated on the graph. We can then make se of the symmetry to dedce that the first angle with a cosine eqal to 0.5 is0. This is becase the angle mst be the same distance to the right of 90 that 60 is to the left. From the graph we see, from consideration of the symmetry, that the remaining soltion we seek is 40.Ths = 0, 40 cos o 40 o 60 o 90 o 80 o 70 o 60 o Figre. A graph of cos. Sppose we wish to solve sin = for Note that in this case we have the sine of a mltiple angle,. To enable s to cope with the mltiple angle we shall consider a new variable where =, so the problem becomes that of solving sin = for 0 70 We draw a graph of sin over this interval as shown in Figre. sin 0 60 o 0 o 80 o 60 o 40 o 480 o 540 o 70 o Figre. A graph of sin for lying between 0 and 70. c mathcentre Jne 4, 004

4 By referring to the Table on page we know that sin 60 =. This is indicated on the graph. From the graph we can dedce another angle which has a sine of. This is 0. Becase of the periodicity we can see there are two more angles, 40 and 480. We therefore know all the angles in the interval with sine eqal to, namely =60, 0, 40, 480 Bt = so that from which =60, 0, 40, 480 =0, 60, 0, 40 Sppose we wish to solve tan = for vales of in the interval Note that in this eample we have the tangent of a mltiple angle,. To enable s to cope with the mltiple angle we shall consider a new variable where =, so the problem becomes that of solving tan = for We draw a graph of tan over this interval as shown in Figre 4. tan Figre 4. A graph of tan. We know from the Table on page that an angle whose tangent is is 45, so sing the symmetry in the graph we can find the angles which have a tangent eqal to. The first will be the same distance to the right of 90 that 45 is to the left, that is 5. The other angles will each be 80 frther to the right becase of the periodicity of the tangent fnction. Conseqently the soltions of tan = are given by Bt = and so from which = 5, 5, 495, = 5, 5, 495, =45, 05, 65 c mathcentre Jne 4, 004 4

5 Sppose we wish to solve cos = for vales of in the interval In this we are dealing with the cosine of a mltiple angle,. To enable s to handle this we make a sbstittion = so that the eqation becomes cos = for 0 80 A graph of cos over this interval is shown in Figre 5. cos o 60 o 90 o 80 o Figre 5. A graph of cos. We know that the angle whose cosine is is 60. Using the symmetry in the graph we can find all the angles with a cosine eqal to. In the interval given there is only one angle with cosine eqal to and that is = 0 Bt = and so =. We conclde that there is a single soltion, = 40. Let s now look at some eamples over the interval Sppose we wish to solve sin = for From the graph of sin over this interval, shown in Figre 6, we see there is only one angle which has a sine eqal to, that is =90. sin -80 o -90 o 90 o 80 o Figre 6. A graph of the sine fnction 5 c mathcentre Jne 4, 004

6 Sppose we wish to solve cos = for In this we have a mltiple angle,. To handle this we let = and instead solve cos = for A graph of the cosine fnction over this interval is shown in Figre 7. cos o -70 o -80 o 90 o - 60 o 90 o 80 o 70 o 60 o - Figre 7. A graph of cos. The dotted line indicates where the cosine is eqal to. Remember we already know one angle which has cosine eqal to and this is 60. From the graph, and making se of symmetry, we can dedce all the other angles with cosine eqal to. These are = 00, 60, 60, 00 Then = so that = 00, 60, 60, 00 from which = 50, 0, 0, 50 Sppose we wish to solve tan = for We again have a mltiple angle,. We handle this by letting = so that the problem becomes that of solving tan = for c mathcentre Jne 4, 004 6

7 We plot a graph of tan between 60 and 60 as shown in Figre 8. tan Figre 8. A graph of tan. We know from the Table on page that one angle which has a tangent eqal to is =60. We can se the symmetry of the graph to dedce others. These are Bt = and so and so the reqired soltions are Eercise = 00, 0, 60, 40 = 00, 0, 60, 40 = 50, 60, 0, 0. Find all the soltions of each of the following eqations in the given range (a) sin = for 0 <<60 o (b) cos = for 0 <<60 o (c) tan = for 0 <<60 o (d) cos = for 0 <<60 o. Find all the soltions of each of the following eqations in the given range (a) tan = for 80 o <<80 o (b) tan = for 80 o <<80 o (c) cos = for -80o <<80 o (d) sin = for 80 o <<80 o 7 c mathcentre Jne 4, 004

8 . Find all the soltions of each of the following eqations in the given range (a) cos = for 80 o <<80 o (b) tan = for 90 o <<90 o (c) sin = for 80o <<0 (d) cos = for 80o <<80 o 4. Using identities in the soltion of eqations There are many trigonometric identities. Two commonly occring ones are sin + cos = sec = + tan We will now se these in the soltion of trigonometric eqations. (If necessary yo shold refer to the nit entitled Trigonometric Identities). Sppose we wish to solve the eqation cos + cos = sin for We can se the identity sin + cos =, rewriting it as sin = cos to write the given eqation entirely in terms of cosines. cos + cos = sin cos + cos = cos Rearranging, we can write cos + cos =0 This is a qadratic eqation in which the variable is cos. This can be factorised to ( cos )(cos +)=0 Hence cos = 0 or cos +=0 from which cos = or cos = We solve each of these eqations in trn. By referring to the graph of cos over the interval 0 80 which is shown in Figre 9, we see that there is only one soltion of the eqation cos = in this interval, and this is =60. From the same graph we can dedce the soltion of cos = tobe = 80. c mathcentre Jne 4, 004 8

9 So there are two soltions of the original eqation, 60 and 80. cos o 90 o 80 o - Figre 9. A graph of cos. Sppose we wish to solve the eqation tan = sec + for In this eample we will simplify the eqation sing the identity sec = + tan. tan = sec + tan = ( + tan )+ tan = + tan + Rearranging we can write tan = so that tan =+ or We solve each of these eqations separately. The soltions of tan = can be obtained by inspecting the graph in Figre 0. From the Table on page we know that one angle with a tangent of is60. There are no other soltions in the given interval. Using the symmetry of the graph we can dedce the soltion of the eqation tan =. This is = 0. So the given eqation has two soltions, =60 and = 0. tan Figre 0. A graph of tan. 9 c mathcentre Jne 4, 004

10 Eercise. Find all the soltions of each of the following eqations in the given range (a) cos = sin for 0 <<60 o (b) 5 cos =4 sin for -80 o <<80 o (c) tan = sec for -80 o <<80 o (d) cos + cos = sin for 0 <<60 o 5. Some eamples where the interval is given in radians In the previos eamples, we soght soltions of eqations where the angle reqired was measred in degrees. We now look at some eamples where the angle is measred in radians. In fact, it is advisable to work with angles in radians becase many trigonometric eqations only make sense when an angle is measred in this way. Sppose we wish to solve the eqation tan = for. A graph of the tangent fnction over this interval is shown in Figre. We know from the Table on page that one angle with a tangent eqal to is. Using the symmetry of the graph 4 we can dedce that the soltions of the eqation tan = are = 4, 4 tan Figre. A graph of tan. Sppose we wish to solve the eqation cos = for 0. c mathcentre Jne 4, 004 0

11 We handle the mltiple angle by letting = so that the problem becomes that of solving cos = for 0 4 So we have plotted a graph of cos over this interval in Figre. cos Figre. A graph of cos. We know from the Table on page that an angle which has cosine eqal to is 0, that is =. This is indicated on the graph. Using the symmetry of the graph we can dedce all the 6 angles which have cosine eqal to. These are Bt = and so and so = 6, 6, + 6, 4 6 = 6, 6, + 6, 4 6 = 6, 6, 6, 6 =,,, Sppose we wish to solve sin = for vales of in the interval. We handle the mltiple angle by letting = so that the problem becomes that of solving sin = for A graph of sin over the given interval is shown in Figre. sin 0 Figre. A graph of sin c mathcentre Jne 4, 004

12 We know from the Table on page that an angle which has sine eqal to is =60 or =. Using the symmetry of the graph we can dedce angles which have a sine eqal to. There is only one soltion in the given interval and this is =. Bt = and so Hence = = Sppose we wish to solve the eqation cos + sin = for 0. We shall se the identity sin + cos = to rewrite the eqation entirely in terms of sines. cos + sin = ( sin ) + sin = sin + sin = and rearranging sin sin =0 This is a qadratic eqation in sin which can be factorised to give ( sin + )(sin )=0 Hence from which sin + = 0 and sin =0 sin = and sin = We solve each of these separately. Graphs of sin are shown in Figre 4. From Figre 4(a) we can dedce the soltion of sin =tobe =. Soltions of sin = can be dedced from Figre 4(b). We know from the Table on page that an angle with sine eqal to is 0 or. Using the symmetry in the graph we can dedce the angles with sine eqal to to be 6 + and. Hence 6 6 = 7 6, 6 (a) (b) sin sin / / Figre 4. Graphs of the fnction sin c mathcentre Jne 4, 004

13 So, the fll set of soltions of the given eqation is =, 7 6, 6. So, we have seen a large nmber of eamples of the soltion of trigonometric eqations. The strategy is to obtain an initial soltion and then work with the graph and its symmetries to find additional soltions. Eercise. Find all the soltions of each of the following eqations in the given range (a) tan = for 0 << (b) sin = for << (c) cos = for << (d) tan = for <<0. Find all the soltions of each of the following eqations in the given range Answers (a) tan =5sec 9 for 0 << (b) cos 6 cos = sin for << Eercise. a) 45 o, 5 o b) 5 o, 5 o c) 0 o, 0 o d) 80 o. a) 0 o,60 o b) 60 o, 0 o c) 60 o,60 o d) 5 o, 45 o. a) 57.5 o,.5 o,.5 o, 57.5 o b) 45 o,5 o,75 o c) 65 o, 05 o d) No soltion Eercise. a) 0 o, 50 o, 0 o, 0 o b) 5 o, 45 o,45 o, 5 o c) 5 o, 45 o,45 o, 5 o d) 0 o, 80 o, 40 o Eercise. a), 4 b) 4, 4 c) 4, 7,,, 7, 4 d). a) 4, 4, 5 4, 7 4 b),0, c mathcentre Jne 4, 004