SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING. Self Study Course

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1 SHOOL OF MTHEMTIS MTHEMTIS FOR PRT I ENGINEERING Self Study ourse MODULE TRIGONOMETRY Module Topics 1 The measurement of angles and conversion etween degrees and radians The elementary trigonometric ratios and relations etween them 3 The solution of general triangles 4 Trigonometric ratios for various angles and graphs of the elementary trigonometric functions 5 Other trigonometric ratios and relationships etween them 6 Various useful formulae involving multiple angles 7 Trigonometric equations This module is again self-contained It is another long module ut it should e mainly revision for you Work Scheme Study the following sections, read carefully the worked Examples and do the stated Exercises Solutions to the Exercises are given towards the end of this Module, starting on p1 1 The measurement of angles ngles are usually measured either in degrees or in radians There are, of course, 360 degrees (360 ) in a complete revolution, 180 in a straight line angle and 90 in a right-angle y definition, if the angle θ in figure 1 is measured in radians, r θ L Figure 1 Length of arc then θ = = L radius r, a ratio which is independent of the units in which oth L and r are measured Since the length of the complete circumference of a circle of radius r is πr, in a complete revolution there are πr/r =πradians onversion from degrees to radians and vice-versa is straightforward since 360 is equivalent to π radians: To convert from degrees to radians, multiply y π 180 To convert from radians to degrees, multiply y 180 π 1

2 In trigonometry it is usually most convenient to work in radians and, therefore, you may assume that all angles are measured in radians unless stated otherwise If angles are measured in degrees then the usual notation will e used (eg 45 ) When using a pocket calculator, you must e very careful that it is operating in the correct ie radian mode unless you specifically intend to use degrees Rememer, though, that an accuracy of four decimal places in descriing an angle in radians is equivalent to an accuracy of only two or three places in degrees The answers to many prolems in this Module are given in oth radians and degrees These answers are (hopefully!) accurate to four places However, if you convert one answer directly to the other y using the appropriate scale factor, ecause of rounding errors in the calculation you may well find an apparent error in the last one or two decimal places Example 1 onvert 60 to radians 60 is 60 π 180 = π 3 radians Example onvert π radians to degrees 6 π 6 radians is π =30 π One or two frequently used conversions are worth committing to memory: 30 = π 6, 45 = π 4, 60 = π 3, 90 = π, 180 = π, 360 =π For other cases, use your pocket calculator (or tales) if necessary ***Do Exercise 1 Express the following in degrees: (i) π/4, (ii) 3π/, (iii) 7π/3 (N In all the Exercises in this module give numerical answers correct to 4 decimal places (ie 4 dpl), using degrees or radians as appropriate) ***Do Exercise Express the following in radians: (i) 18, (ii) 35, (iii) 350, (iv) 40 The three elementary trigonometric ratios In a right-angled triangle, the sine, cosine and tangent of an acute angle are defined to e: Sine of angle = side opposite hypotenuse, adjacent side osine of angle = hypotenuse side opposite Tangent of angle = adjacent side The notations used for the three ratios are respectively sin, cos and tan Thus, in the right-angled triangle shown in figure (the side laelled c, opposite the right-angle is, of course, the hypotenuse): sin θ = c, cos θ = a c, tan θ = a c θ a Figure

3 Once again, assuming of course that all lengths are measured in the same units, these ratios are independent of which units are used This means that, given a right-angled triangle, if the length of any side and the size of a second angle are known or if the lengths of any two sides are known then the triangle can e solved that is to say all other sides and angles of the triangle can e found Example 3 Solve the triangle shown in figure if a =cmandθ=π/6 Since tan θ = /a, cosθ=a/c, it follows that = a tan θ and c = a/ cos θ and hence, putting in the given values for a and θ, weseethat ( π =tan =11547 cm and c = 6) =3094 cm cos(π/6) [Note that we do not need trigonometry to calculate the third angle the sum of the angles of a triangle is π and so the value of the third angle is π/3 lso, having found we could have used Pythagoras theorem to give us c] Example 4 Solve the triangle shown in figure if a =cmand=1cm Firstly we use Pythagoras theorem to find c: c = a + = +1 = 5=361 cm Next, since tan θ = a = 5, the inverse tangent utton on your pocket calculator gives θ = 4636 or The remaining angle is then π ( 1 π ) = 1107 or 180 ( )= Important relations etween sin, cos and tan Referring again to figure you can see that, y Pythagoras theorem, c = a + However, a = c cos θ and = c sin θ and hence from which it follows that c = c cos θ + c sin θ, cos θ +sin θ=1 Notice the notation carefully We write, for example, cos θ to mean (cos θ) [The quantity cos θ has no meaning since it could represent cos(θ )or(cosθ) ] Example 5 Verify the formula cos θ +sin θ= 1 for the angle θ in Example 4 Since θ = 4636, your pocket calculators tell you that cos θ = 8, whilst sin θ = Hence cos θ + sin θ =1 nother relationship etween cosine and sine is ovious from figure The angles etween a and c and etween and c add to π/ since the third angle of the triangle is π/ and we know that the angles of a triangle add to π It follows immediately that ( π ) cos θ =sinθ, ( π ) sin θ =cosθ 3

4 In addition it is easily shown that since tan θ = sin θ cos θ, tan θ = a = /c a/c = sin θ cos θ You can check for yourself that the aove three formulae involving θ are true for the numers given in Example 4 The four major results in this section are identities that is to say, they hold for all values of θ They are important and you should e sure to memorise them They are repeated as numers 4, 5 and 7 of the useful results at the end of this module 4 The solution of general triangles If you are given any triangle together with any one of the following three pieces of information: (a) the lengths of all three sides () the lengths of two sides and the size of any angle (c) the length of one side and the size of any two angles you can still solve the triangle completely although in the second case, unless the given angle lies etween the two sides, the solution may not e unique To do this you can use two rules called respectively the sine rule and the cosine rule In the notation used in figure 3 (which is the conventional notation with a opposite etc), the sine rule is written and the cosine rule gives the three formulae a sin = sin = c sin a = + c c cos, = c + a ca cos, c = a + a cos c a Figure 3 The tale elow shows you which formula you need for which situation Facts known Rule to use 3 sides cosine sides and the angle etween sides and either of the other angles sine 1 side and any angles The sine and cosine rules appear on the Formula Sheet (and in the Data ook) ut it is useful to rememer them You do not need to know the proofs of the two rules ut they are included elow since they are very straightforward 4

5 c h a F Figure 4 In figure 4, you can see that the height, h, of the triangle is given equivalently y h = c sin and h = sin Hence c sin = sin and sin = c sin The remaining part of the sine rule is proved similarly or y symmetry To prove the cosine rule we see, again looking at figure 4, c = h +(a F) = sin +(a cos ) = sin + a + cos a cos However, cos +sin = 1, and hence c = a + a cos Once again, the other two results follow similarly or y symmetry Example 7 Solve the triangle if a = 1, =8,c= 10 Since all three sides ut no angles are known, we must use a cosine formula c = a + a cos tells us that cos = a + c a = 1 8 = =0565 Hence, using a calculator, = 9734, (= ) We could find the other angles in a similar way, ut we now know one angle and so a simpler alternative is to use the sine formula Since a sin = c sin, sin = a sin c = = Hence =14455, (= 8819 ) To find the remaining angle we could use the cosine or the sine formula, ut oviously the easiest way is to rememer that the angles of a triangle add to π, sothat=π =π = 77, (= ) Notice that these results have een otained retaining 10 digits throughout on a pocket calculator However, at each stage they have een stated to only four places If you use the four place answers (instead of the full 10 place ones) to perform your susequent calculations, you will otain an answer which is certainly not accurate to four places Try it and see! 5

6 Example 8 Solve the triangle if =cm,c=1cm,=0 Given two sides and non-included angle, we must use the sine formula In some situations, and this is one of them, the given information does not produce a unique triangle (that is one solution only) and so it is always advisale to draw a reasonaly accurate diagram efore proceeding further It is easily seen that in this example two solutions are possile Figure 5 (a) 0 Figure 5 () 0 sin = c sin = 1 sin 0 = 6840 Hence =43160 (= 7533), see figure 5(), or = = (= 3883), as shown on figure 5(a) It follows that = 180 = = ,(=039) or =3160,(=404) Finally we may find a from the cosine formula a = + c c cos = +1 1 cos =60883 cm or +1 1 cos 3160 =11499 cm 5 The area of a triangle The area of a triangle is half that of a rectangle on the same ase that is to say, it is one half of the length of its ase times its height Hence, using figure 4 and rememering that h = sin, rea of = 1 ase height = 1 a h = 1 a sin In a similar way it can also e shown that rea of = 1 c sin = 1 ca sin lthough these formulae could e rememered and used it is usually easier to proceed from first principles y calculating the height directly (see example elow) Example 9 Find the area of the triangle if =1cm,c=cm,=0 This is the same triangle as in Example 8 ut with the lengths of the given sides interchanged ut this time figure 6 confirms the answer is unique Figure 6 0 sin = c sin = 1 sin 0 =

7 Hence =98466 It follows that = 180 = = This enales us to find a from the cosine formula: a = + c c cos = cos Using a calculator we deduce a =910 cm The height h is easily found from h 1 =sin0, and hence the area of the triangle is given y rea of = 1 a h = sin 0 = 4977 cm ***Do Exercise 3 Solve the right-angled triangle shown elow in each of the following cases: c a (i) a =1m,=π/3; (ii) a =m,=10 ; (iii) a =3m,=1m ***Do Exercise 4 Solve the triangle shown elow in the following cases: (i) c =51m, =47,=70 ; (ii) a =11m, =17m, c=184m; (iii) =7m,c=3m,=10 c a ***Do Exercise 5 Find the area of each of the triangles in Exercise 4 6 Trigonometric ratios for special angles up to π/ The angles 0,30,45,60 and 90 (ie 0, π/6, π/4, π/3andπ/) occur frequently in trigonometric work and it is helpful to memorise the corresponding values of sine, cosine and tangent for each of these angles lternatively, you should e ale to sketch the appropriate triangle and deduce the value of the ratios (see elow), or use your calculator! (a) ngle 0 θ c a Figure 7 7

8 In this case the triangle looks like figure 7 as θ approaches 0 learly approaches 0 and as this happens a and c ecome equal However sin θ = c, cos θ = a c, tan θ =, and so we conclude that a sin 0 = 0, cos 0 = 1, tan 0 = 0 () ngle π/6 or π/3 When θ = π/3 the forms half of an equilateral triangle see figure 8 Hence, if = units, = 1 unit and Pythagoras theorem tells us that = 1 = 3 units π/3 π/ Figure 8 It follows that cos π 3 = 1 = 5, sin π 3 3 = = 8660, tan π 3 = 3=1731 From the same triangle we can also read off the results for an angle of π/6: cos π 6 = 3 = 8660, sin π 6 = 1 = 5, tan π 6 = 1 3 = 5774 (c) ngle π/4 This time the picture is as in figure 9, with a = =1andc= π/4 1 Figure 9 1 It follows that: cos π 4 = 1 = 7071, sin π 4 = 1 = 7071, tan π 4 =1 (d) ngle π/ This case is much the same as (a), ut with the triangle the other way round, as shown in figure 10 This time, however, it is the angle at which approaches 0 so that a approaches zero length and and c eventually ecome equal Thus cos π =0, sin π =1 8

9 s θ ecomes nearer and nearer to π/ youcanalsoseethattan(π/) gets larger and larger We write tan θ as θ π If you calculate tan(π/), your calculator will show an error c θ a Figure 10 7 ngles etween π/ and π cos θ, sinθand tan θ have so far only een defined when θ is an acute angle, although in using your calculator to evaluate some of your results you have used larger angles If θ lies etween π/ andπ, figure 11 shows what we must do α F θ Figure 11 x hoose to e along the positive x-axis of a coordinate system with at the origin We define the three trigonometrical ratios cos θ, sinθand tan θ to have the same numerical values respectively as those of cos α, sinαand tan α ut, since F is in the negative direction of the x-axis (shown in the figure along the direction of ), we put a minus sign in front of each of the ratios which involve this length Notice that other distances are still treated as positive Thus: cos θ = F = cos α, F sin θ = =+sinα, F tan θ = = tan α F Since α + θ = π, it follows that α = π θ and the aove formulae can e written cos θ = cos(π θ), sin θ = sin(π θ), tan θ = tan(π θ) 9

10 8 ngles etween π and π and > π or < 0 F α θ Figure 1 x θ α Figure 13 F x To find the trigonometric ratios for these angles, we simply extend what we did in section 7 Let us again assume that is along the positive x-axis of a coordinate system with at the origin Then the magnitudes of the trigonometric ratios for the angles ˆ shown in figures 1 and 13 are the same as for the acute angles that makes with this x-axis, ut an appropriate sign must also e attached s in section 7, this sign is calculated y looking at the signs attached to the lines F and F (note is always assumed to have a positive sign) If F is to the right, it is positive; if it is to the left, it is negative If F is up, it is positive; if it is down, it is negative Thus in figure 1 F is to the left and F is down: hence cosines will e negative, sines negative and tangents (negative over negative) positive whilst, in figure 13, F is to the right and F down: hence cosines will e positive, sines negative and tangents negative Finally, if the angle is in excess of π we have gone full circle (at least once!) and we simply knock off as many whole numer multiples of π (complete revolutions) as we need in order to otain an angle etween 0andπand find the appropriate trigonometric ratio for this Thus, for example, sin 9π ( =sin π+ π ) =sin π =1 Likewise, if the angle is less than 0, we have to add as many whole numer multiples of π as is necessary to put the angle into the range 0 to π Thus cos(θ +nπ) =cosθ, sin(θ +nπ) =sinθ, tan(θ +nπ) =tanθ, if n is an integer In fact tan(θ + nπ) =tanθ, ut that you will see later! Most people find it easiest to rememer which signs are positive y drawing the following diagram (figure 14) in which stands for ll, S stands for Sine, T stands for Tangent and stands for osine ll other signs are negative π/ S π 0, π, T 3π/ Figure 14 10

11 Notice and rememer the very useful results that: cos( θ) = cos θ, sin( θ) = sin θ, tan( θ) = tan θ Example 10 Evaluate sin(π/3) π/3 lies in the range from π/ toπ Figure 14 therefore shows that sin(π/3) is positive and so sin(π/3) = + sin(π π/3) = sin(π/3) Using the results otained with figure 8 it follows that sin(π/3) = sin(π/3) = 3/ =8660 heck that your calculator gives the latter value for oth sin(π/3) and sin(π/3) Example 11 Evaluate cos is larger than π/ (=15708) and smaller than π (= 31416) Hence it follows from figure 14 that cos is the negative of the cosine of π (=11416) Using your calculator verify that the values for cos and cos(π ) are oth equal to 4161 Example 1 Evaluate cos(190 ) 190 (= 33161) is etween π and 3π/ Hence, looking at figure 14, its cosine is the negative of the cosine of π heck the values given y your calculator for oth cos(33161) and cos(33161 π) You ll find that they are oth the same, namely 9848 Example 13 Evaluate tan 5 5 is igger than 3π/ (=4713) and smaller than π (= 6831) Hence, from figure 14, tan 5 = tan(π 5) gain, your calculator should give the same answer for oth tan 5 and tan(π 5), namely It may e that you are more used to using degrees than radians in that case you may prefer to check the range in which 5 lies y converting it to degrees ( ) and noting that this lies etween 70 and 360 Example 14 Evaluate sin(33π/4) The angle is outside the range 0 to π and so we have to sutract whole numer multiples of π to ring it into range 33π/4 =4 π+ 1 4 π Hence sin(33π/4) = sin(π/4) = 1/ heck that the answer (7071) given y your calculator agrees whichever way you do the calculation that is to say, if your calculator allows you do this type of calculation directly! Example 15 Evaluate tan( π/4) tan( π/4) = tan(π/4) = 1 Example 16 Evaluate sin( 4π) 4π is equivalent to 0 for our purposes two negative revolutions takes us ack to where we started and so sin( 4π) = sin0 = 0 11

12 9 Graphs of trigonometric functions With a graph-plotting calculator you can very easily see what the graphs of cos θ, sinθand tan θ look like for θ in the range ( π, π) If you don t have such a calculator, the previous sections give sufficient values of each function for you to e ale to make a good plot The three graphs are: θ Figure 15: Graph of cos θ 1 +1 π 3π/ π π/ π/ π 3π/ π θ Figure 16: Graph of sin θ 1 +1 π 3π/ π π/ π/ π 3π/ π θ Figure 17: Graph of tan θ π 3π/ π π/ π/ π 3π/ π 1 1 s we saw in previous sections, outside the range (0, π) each curve repeats itself every π, the extensions joining on smoothly to the curves shown in each case We say that these curves have period π In fact, the tangent curve repeats itself every π has period π a fact which means that we have to take particular 1

13 care if we are asked to find the angle in the range (0, π) which has a particular value for its tangent: there are two possile answers! We shall discuss this in more detail later Notice that, if the cosine curve is displaced to the right y π/, the sine curve is otained whereas if the sine curve is displaced to the right y π/ then the negative of the cosine curve is derived This means that ( π ) sin + θ =cosθ, ( π ) cos + θ = sin θ, which are two useful results Notice also that the tangent curve ecomes infinite (either in the positive or negative direction) at the points θ =3π/, π/,π/,3π/ ***Do Exercise 6 (i) Use your calculator to find the sine, cosine and tangent of each of the following: (a) 140, () 7, (c) 700 (ii) Use your calculator to find (a) cos( 0 ), () sin( 50 ), (c) tan( 450 ), (d) sin( 10) (iii) Show y direct calculation that (a) cos 10 +sin 10 =1, () cos 410 sin 410 = cos 80, (c) sin15cos15 = sin( 3) (iv) Given that π/ α π and sinα = 4/5, find cosα and tanα without using your calculator (check your answers with the calculator) 10 Other trigonometric ratios There are three other ratios that you will meet These are the reciprocals of (ie one over) the cosine, sine and tangent and are called respectively the secant, cosecant and cotangent In the standard notation: sec θ 1 cos θ, cosec θ 1 sin θ, cot θ 1 tan θ e very careful of your notation cos 1 θ is used to mean the angle whose cosine is θ not 1/ cos θ, and similarly for sin 1 θ and tan 1 θ Thus, for example, sec θ =(cosθ) 1,not cos 1 θ cos n θ means (cos θ) n whenever n 1 ut not when n = 1 wkward, isn t it? It is worth pointing out that the aove difficulty with notation is the main reason why the angle whose cosine is θ is often denoted y arc cos θ, instead of cos 1 θ Sketch for yourselves the graphs of the functions secθ, cosecθ and cotθ, noticing in particular that, whilst 1 cos θ 1and 1 sin θ 1 for all values of θ, the functions sec θ and cosec θ are always either 1 or 1 lso notice that, unlike tan θ, cotθis infinite at 0 and π ut finite at π/ and3π/ Example 17 Evaluate sec 314, cosec 314 and tan 314 to 6 dpl sec 314 = 1/ cos 314 = 1/ = cosec 314 = 1/ sin 314 = 1/ = cot 314 = 1/ tan 314 = 1/ = Notice how nearly numerically equal the values of sin 314 and tan 314 (and therefore of cosec314 and cot 314) are for these values of θ so close to π 11 Two further identities You will rememer that we proved in section 3 that cos θ +sin θ =1and that tan θ =sinθ/ cos θ for all values of θ If we divide the first equation y cos θ we find cos θ cos θ + sin θ cos θ = 1 cos θ, 13

14 and making use of the second equation then gives 1+tan θ=sec θ In a similar way (dividing the first equation y sin θ instead of cos θ), we find cot θ +1=cosec θ These two formulae are useful and, if possile, should e rememered, although they do appear on the Formula Sheet (and in the Data ook) The important point is to know that formulae exist connecting tan θ to sec θ (and cot θ to cosec θ) 1 Trigonometric ratios for sums and differences of angles It is often necessary to e ale to express cos θ in terms of cos θ, a formula which is a special case of that which gives cos( + ) intermsof the cosines and sines of and alone From figure 18 we can easily show that cos( + ) =coscos sin sin In the figure, the perpendicular, F, from onto has F as its foot and the extension of is of length e In addition, the angle ĈF is of size + since it is an exterior angle of the triangle In triangle F, However, from triangle F, where x = a cos and y = cos It follows that y x + a e F l Figure 18 cos( + ) = e = l a l =(x+y)cos, cos( + ) = (x+y)cos a (acos + cos )cos a = = a(cos 1) +coscos = a sin +coscos and, since the sine rule tells us that sin = sin, a we can replace a sin y sin and conclude that cos( + ) =coscos sin sin 14 h

15 as required If in this formula we replace y (π/), we otain a corresponding formula for sin( ) and if in the two formulae we now have we replace y, we otain two more formulae for cos( ) and sin(+) From your point of view, the proofs are relatively unimportant What matters is that you know the following formulae are on the Formula Sheet (and in the Data ook) and you are ale to use them cos( + ) =coscos sin sin sin( + ) =sincos +cossin cos( ) =coscos +sinsin sin( ) =sincos cos sin Notice the somewhat unexpected arrangement of the signs in the aove equations The corresponding formulae for tan( ± ) are less important since they can e otained, if needed, from those for cos( ± ) and sin( ± ) y division For example tan( + ) = sin( + ) cos( + ), and the expressions for sin( + ) andcos(+) can then e used These formulae for sums and differences of angles are very useful in determining a numer of the formulae quoted earlier For example, using the formula for cos( ), we otain cos(π θ) =cosπcos θ +sinπsin θ = 1 cos θ +0 sin θ = cos θ 13 Doule angle formulae These are the formulae mentioned at the eginning of the previous note If we put oth and equal to θ in the formulae for cos( + ) and sin( + ), we otain cos θ =cos θ sin θ, sin θ =sinθcos θ These results are highly important and must e committed to memory The former may e written (using sin θ +cos θ= 1) in the alternative forms: cos θ =cos θ sin θ =cos θ 1=1 sin θ These alternative identities are also very important and you must either learn them or e ale to derive them quickly Example 18 Prove that sin( D) + sin( + D) =sincos D sin( D) =sincos D cos sin D and sin( + D) =sincos D +cossin D y adding these two equations we otain the required result 15

16 14 Sum and difference formulae If in Example 18 we put = 1 ( + ) andd= 1 ( ), we otain sin +sin=sin 1 (+)cos1 ( ) This is one of a set of four formulae which are useful in some integration prolems in calculus, ut not worth rememering just know where to find them (on the Formula Sheet and in the Data ook) or how to deduce them if you need them! The other three are: sin sin =cos 1 (+)sin1 ( ), cos +cos=cos 1 (+)cos1 ( ), cos cos = sin 1 (+)sin1 ( ), Notice the negative sign in the last of these Example 19 Express the following as products: (a) sin 6θ + sin4θ, () cos18θ cosθ (a) sin 6θ +sin4θ=sin 1 (6θ +4θ)cos1 (6θ 4θ) =sin5θcos θ () cos 18θ cos θ = sin 1 (18θ +θ)sin1 (18θ θ) = sin10θsin 8θ Example 0 Express cos 7θ cos 3θ as a sum or difference of two cosines Here we use the formula cos +cos=cos 1 (+)cos1 ( ) We put 7θ = 1 ( + ) and3θ=1 ( ) so that, after multiplying oth equations y, we otain + =14θand =6θ dding these two equations and dividing y two we see that =10θ It follows that =4θThus cos 7θ cos 3θ = 1 (cos 10θ +cos4θ) ***Do Exercise 7 Prove the following identities: (i) (sin θ +cosθ) =1+sinθ, (ii) tan( + ) = tan +tan tanθ, (iii) tan θ = 1 tan tan 1 tan θ, tan θ +cotθ (iv) =cosecθ sec θ ***Do Exercise 8 α) ***Do Exercise 9 and If sin α = 5/13, where α is acute, find tan α, cot α and cosecα (without calculating If tan θ = a/, whereθis acute, find expressions for sin θ and sec θ in terms of a ***Do Exercise 10 y writing 3θ =θ+θ, and then using the identities for sums of angles and doule angles, show that (i) sin 3θ =3sinθ 4sin 3 θ, (ii) cos 3θ =4cos 3 θ 3cosθ ***Do Exercise 11 ( π ) (i) sin + θ =cosθ, Using the formulae for sums and differences of angles show that ( π ) ( ) 3π (ii) cos + θ = sin θ, (iii) sin (π θ) =sinθ, (iv) cos +θ =sinθ 16

17 ***Do Exercise 1 Express as products of sines and/or cosines: (i) sin 4θ +sinθ, (ii) sin 8θ sin 6θ, (iii) cos 1θ +cos 10θ, (iv) cosθ cos θ, (v) sin(π/3)+sin (π/4) ***Do Exercise 13 Express as sums or differences of trigonometric functions: (i) sin 6θ cos θ, (ii) cos 5θ cos 3θ, (iii) cos 4θ sin θ, (iv) sin7θsin 5θ 15 Simple trigonometric equations It is important to realise that, since sin θ and cos θ repeat themselves every π, even a simple equation like cos θ = 1 is satisfied y an infinite numer of values of θ y referring ack to figure 15, or using your graphics calculator, you will see that not only is 1 3 π a solution, ut so are π π,4π+1 3 π, π+1 3πand so on In fact it s worse than this even etween θ =0andθ=πthere are two different solutions, as you can again see y looking at the graph of cos θ in figure 15 (or on your graphics calculator) or y using the ST diagram in figure 14 Either method shows you that π 1 3 π = 5 3π is also a solution We can therefore write the general solution of this equation as θ = 1 3 π +nπ, or θ = 5 3 π +nπ, where n is any whole numer (positive, negative or zero) It makes things simpler, to define a unique inverse to each of the trigonometric functions and then calculate all other solutions to equations like the one aove from it Pocket calculators have the same prolem They cannot come up with several different inverses at the same time and, instead, have to choose one and leave you to calculate the rest alculators use the following convention, and you should do the same: cos 1 x is defined to e that solution of cos θ = x which lies in the range 0 θ π, sin 1 xis that solution of sin θ = x which lies in the range 1 π θ 1 π and tan 1 x is that solution of tan θ = x which lies in the range 1 π θ 1 π Notice the ranges are not the same for each function: this is necessary in order to ensure that each possile value is covered once and once only as can e seen from figures Example 1 Find the general solution of cos θ = 3 Your calculator will tell you that cos 1 ( 3) is This is the solution of the equation which lies in the range [0,π] in fact etween 1 π and π The graph then shows you that π = is also a solution, giving you a second answer etween 0 and π The general solution is then for any integer n, positive, negative or zero θ = nπ, or θ = nπ Example Find the general solution of sin θ = 1 sin 1 ( 1 ) = 1 4 π = 7854 since this is the appropriate value lying in the range [ 1 π, 1 π] Your calculator will give you the same answer s you can see from the graph, the two solutions lying etween 0 and π are θ = π π and π 1 4 π, ie θ = 5 4 π and θ = 7 4π The general solution is then for integer n θ = 5π 4 +nπ, or θ = 7π 4 +nπ 17

18 Example 3 Find the solution of tan θ =16 which lies etween π and π Your calculator gives you tan 1 16 =101 This is the value which lies in [ 1 π, 1 π] The tangent function (see the graph) repeats itself every π The general solution of the equation is tan 1 θ = nπ and the solution we require, otained y taking n =1,isthus41538 Some trigonometric equations can readily e solved using the doule angle formulae (section 13) and simple algera Example 4 Find the solutions to the equation 6 cos θ +5cosθ+ 4 = 0 which lie in the range 0 θ π The doule angle formulae tell us that cos θ =cos θ 1 and hence the given equation can e written 6( cos θ 1) + 5 cos θ +4=0 or 1 cos θ +5cosθ =0, which is a quadratic equation in cos θ This equation factorises to give (3 cos θ + )(4 cos θ 1) = 0 and hence either 3 cos θ + = 0 or 4 cos θ 1 = 0 It follows that cos θ = /3 orcosθ=1/4 (results which could also have een found from the quadratic equation using the formula) ccording to the calculator, these solutions give ( θ =cos 1 ) ( ) 1 =3005 or θ =cos 1 =13181, 3 4 and from figure 14, or figure 15, it is clear that each of the aove solutions leads to a corresponding second solution etween 0 and π Hence there are four solutions etween 0 and π: θ =3005, θ =π 3005 = 3987, θ =13181, θ =π = Slightly more difficult trigonometric equations Equations of the form sin ± sin = 0 or cos ± cos = 0 can e solved y using the appropriate sum or difference formula Example 5 Find the roots of the equation sin 7θ =sinθfor 0 θ 1 π Here we use the formula for the difference of two sines: sin 7θ sin θ =cos 1 (7θ + θ)sin1 (7θ θ) =cos4θsin 3θ Hence, equating this to zero, we see that either cos 4θ =0orsin3θ=0 It follows from the first of these equations (look at the graphs!) that 4θ = 1 π + nπ and from the second that 3θ = nπ Hence θ = 1 8 π nπ or 1 3nπ and, picking out the solutions which are in the required range, we see (n =0,1 in the first equation) that θ = 1 8 π or 3 8π or (n =0,1 in the second equation) θ =0or 1 3π Thus there are four solutions in the required range Finally, equations of the form a cos θ ± sin θ = 0 can e solved y expressing the left-hand side in one of the forms sin(θ ± α) orcos(θ ± α) for suitaly chosen and α This is an extremely important form for the Engineer since it is expressed in terms of phase (α) and amplitude () 18

19 Example 6 First notice that Solve the equation 3 sin θ +4cosθ=3for0 θ π sin(θ + α) =cos α sin θ + sin α cos θ Hence, if this is to e identically equal to the left-hand side of the given equation, cos α =3 and sin α =4 Squaring and adding, we see that (sin α +cos α)=3 +4 so that = 5 It follows that cos α = 3 5 and sin α = 4 5, so that, y calculator, α = 973 Notice carefully that the signs of cos α and sin α force the solution for α to lie in a single quadrant, and hence there is only one solution to this pair of equations in [0, π] If we tried to use just one of the equations instead of oth, we would get two solutions, one wrong since it would not satisfy the second equation! Our equation now simplifies to 5sin(θ+973) = 3 or sin(θ + 973) = 3/5 Once more using our calculator and looking at the sine graph, we find that θ = 6435 or π (θ + 973) = 6435 That is to say θ = 838 or θ =+15708, with general solutions θ = nπ or θ = nπ Finally, choice of n = 1 in the first equation or n = 0 in the second gives the required solutions: θ =59994 or ie π/ ***Do Exercise 14 Find the values of θ in the range 0 θ π which satisfy the following equations: (i) sin θ = 87, (ii) tan θ =139, (iii) cos θ = ***Do Exercise 15 Find all values of θ which satisfy the following equations: (i) sin θ = 81, (ii) cos 3θ = 49, (iii) tan 1 θ = 63 ***Do Exercise 16 Find all values of θ in the range 0 θ 180 which satisfy the following equations: (i) sin θ =sin60, (ii) cos 3θ = cos 10, (iii) tan θ = 1 3 ***Do Exercise 17 Find all solutions of the following equations which lie in the range 0 θ π: (i) sin θ +sinθ 1 = 0, (ii) 16 tan θ 4 tan θ + 9 = 0, (iii) sin θ =sinθ ***Do Exercise 18 Find all solutions of the following equations which lie in the range 0 θ π: (i) cos θ =cos4θ, (ii) sin 3θ +sinθ= 0, (iii) 7 sin θ 8cosθ =9, (iv)5cosθ+1sinθ=4 19

20 You have now worked through the Module, so please attend a testing session as soon as possile The following questions are similar to ones that you will e asked in the Test (solutions can e found on p8) Specimen Test 1 Express π 6 radians in degrees In the triangle find (correct to 4 decimal places) If sin θ = 03 find (i) cosec θ, (ii) cos θ 4 In the triangle find (correct to 4 decimal places) In the triangle PQR find β Q 6 1 β P 10 R 6 Sketch the graph of sin x for π <x<π 7 Express sin(π + θ) in terms of sinθ 8 Find all values of α which satisfy cos α = 05 9 Express cos 4θ +cosθ as a product of trigonometric functions 0

21 Worked solutions to Exercises Note carefully that, whilst alternative answers have een given in degrees and radians, these have een calculated independently Owing to rounding errors, the last decimal place(s) may not agree if you try to convert directly from an answer given in degrees to the corresponding answer in radians or vice versa 1 (i) π 4 = ( 180 π (iii) 7π 3 = ( 180 π ) = 180 =45, (ii) 4 ) 7π = 40 3 π 4 3π = ( 180 π ) 3π = 70, (i) 18 = π = π 10 = 314, (ii) 35 = π 35 = 6109, 180 (iii) 350 = π = 61087, (iv) 40 = π 40 = (i) c = a cos = 1 cos(π/3) = 1 5 =m, =atan =tan(π/3) = 1731 m, ( ) 1 = π π + = π 1 π 1 3 π = 1 6 π (ii) c = a cos = cos(10 ) = =0309 m, 9848 = a tan =tan(10 )=357 m, = 180 (90 + ) = =80 (iii) c = a + = 10 m = 3163 m, tan = a = 1 3 Hence, since <1 π,=tan = 318 (= ) ( ) 1 = π π + = 1 π 318 = 1490, (= ) 4 (i) Firstly, + + = 180 Hence, = =63 Now we can use the sine rule to give a us a and : sin = sin = c implies that sin a = c sin sin = =53787 m, = c sin sin = =4186 m (ii) Here we must use the cosine formula: a = + c c cos tells us that cos = + c a = = 7710 c and so =395594,(=6904) In the same way, the angle cos is given y cos = c + a = = 447 ca and so =648655,(=1131) Finally, = 180 = =755751,(=13190) (iii) Given two sides and non-included angle, we must use the sine formula and the answer is not unique (draw a diagram to confirm this)! sin = c sin = 7 3 sin 10 = 405 1

22 Hence =3903,(=417) or = = ,(=744) It follows that = 180 = = ,(=5499) or =13903, (= 46) Finally we may find a from the cosine formula a = + c c cos = cos = = m or cos = = m 5 The area in each case is given y, for example 1 ase height = 1 a sin = 1 a sin Thusthe answers are (i) sin 63 = m, (ii) sin = m, (iii) sin 10 =58567 m or sin 10 =58 m 6 (i) (a) 648, 7660, 8391, () 6570, 7539, 8714, (c) 340, 9397, 3640 (ii) (a) 9397, () 9397, (c), (d)5440 (iii) (a) cos 10 +sin 10 = = 1, () cos 410 sin 410 = = 1736 = cos 80 (c) sin15cos15= = 1411 = sin( 3) (iv) cos α +sin α= 1 Hence cos α = ± 1 sin α = ± 1 (4/5) = ±7/5 However, α is otuse and so lies etween 90 and 180 It follows that cos α is negative and so cos α = 7/5 tan α = sin α = 4/7 The sign is correct for an otuse angle cos α 7 (i) (sin θ +cosθ) =(sin θ+cos θ)+sinθcos θ =1+sinθcos θ =1+sinθ (ii) sin( + ) sin cos +cossin tan( + ) = = cos( + ) cos cos sin sin On dividing top and ottom of the right hand side y cos cos we otain tan( + ) = tan +tan 1 tan tan (iii) Putting = = θ in part (ii), the result follows immediately (iv) tan θ +cotθ (sin θ/ cos θ)+(cosθ/ sin θ) = sec θ (1/ cos θ) = cos θ sin θ( (sin θ/ cos θ)+(cosθ/ sin θ) ) cos θ sin θ(1/ cos θ) = sin θ +cos θ sin θ = 1 sin θ =cosecθ 8 Given that sin α = 5/13,where α is acute, the lengths of the side opposite α and the hypotenuse are 5 and 13 respectively Using Pyhthagoras theorem the length of the remaining side is 13 5 = 144 = 1 Hence tan α =5/1, cot α =1/tan α =1/5andcosecα=a/ sin α =13/5

23 9 Since tan θ = a/ and θ is acute, a and are respectively the side opposite to and the side y θ in a right-angled triangle whose hypotenuse is of length a + It follows that sin θ = a/ a + In addition 1 + tan θ =sec θand hence sec θ =(a/) (i) (ii) sin 3θ =sin(θ+θ) =sinθcos θ +cosθsin θ, using the formula for sin( + ) =sinθcos θ +(1 sin θ)sinθ, using the formulae for sin θ and cos θ =sinθ(1 sin θ)+(1 sin θ)sinθ =3sinθ 4sin 3 θ cos 3θ =cos(θ+θ) =cosθcos θ sin θ sin θ, using the formula for cos( + ) =(cos θ 1) cos θ sin θcos θ, using the formulae for sin θ and cos θ =(cos θ 1) cos θ (1 cos θ)cosθ =4cos 3 θ 3cosθ ( π ) 11 (i) sin + θ =sin π cos θ +cosπ sin θ =1 cos θ +0 sin θ =cosθ ( π ) (ii) cos + θ =cos π cos θ sin π sin θ =0 cos θ 1 sin θ = sin θ (iii) sin (π θ) = sinπcosθ cosπ sin θ = 0 cosθ ( 1) sinθ = sinθ ( ) 3π (iv) cos + θ =cos 3π cos θ sin 3π sin θ =0 cos θ ( 1) sin θ =sinθ 1 (i) sin 4θ +sinθ=sin 5 θcos 3 θ (ii) sin 8θ sin 6θ =cos7θsin θ (iii) cos 1θ + cos 10θ =cos11θcos θ (iv) cos θ cos θ = sin 3 θsin( 1 θ)=sin3 θsin 1 θ (v) sin 1 3 π +sin1 4 π=sin 7 4 π cos 1 4 π 13 (i) sin 6θ cos θ =sin8θ+sin4θ (ii) cos 5θ cos 3θ =cos8θ+cosθ (iii) cos 4θ sin θ =sin5θ sin 3θ (iv) sin 7θ sin 5θ = cos 1θ +cosθ 14 (i) θ =1055, 0864 (= π 1055), (= , ) (ii) θ = 947, (= π + 947), (= 54678, ) (iii) θ =13490, 4934 (= π 13490, (= 77910, ) 15 (i) θ = nπ, nπ Hence θ = nπ, nπ, (= n, n ) 3

24 (ii) 3θ =089 + nπ, nπ Hence θ = nπ, nπ, (= n, n ) (iii) 1 θ = 56 + nπ Hence θ = nπ, (= n ) 16 (i) sin θ sin 60 =cos(θ+30 )sin(θ 30 )=0 It follows that θ+30 = n so that θ = n,orθ 30 = 180n so that θ = n Hence the values of θ in the required range are θ =30,60,(=536, 1047) (ii) cos 3θ cos 10 = sin( 3 θ+60 ) sin( 3 θ 60 ) It follows that 3 θ+60 = 180n so that θ = n,or 3 θ 60 = 180n,sothatθ=40 +10n Hence the values of θ in the required range are θ =40,160, 80,(=6981, 795, ) (iii) tan θ = ± 1 3 Hence the values of θ in the required range are θ =30,150,(= 1 6 π, 5 6 π) 17 (i) The equation factorises to give ( sin θ 1)(sin θ + 1) = 0, so that either sin θ = 1 Hence the only solutions in the required range are θ = 1 6 π, 5 6 π,(=30,150 ) (ii) The equation factorises to give (4 tan θ 3) =0,sothattanθ= 3 4 Hence there is just one solution in the range, namely θ = 6435, (= ) (iii) Here, sin θ sin θ =sinθ( cos θ 1) = 0, and so either sin θ =0orcosθ= 1 Hence the solutions in the required range are θ =0,π, 1 3 π,(=0,180, 60 ) or sin θ = 1 18 (i) cos θ cos 4θ = sin 5 θsin 3 θ =0 Thus 5 θ = nπ so that θ = 5 nπ, or 3 θ=nπ so that θ = 3 nπ It follows that the solutions in the required range are θ =0, 5 π, 4 5 π, 6 5 π, 8 5 π, 3 π, 4 3 π, π,(=0,7, 144, 16, 88, 10, 40, 360 ) (ii) sin 3θ +sinθ=sinθcos θ =0 It follows that θ = nπ so that θ = 1 nπ, orθ=1 (n +1)π Hence the solutions in the required range are θ =0, 1 π, π, 3 π, π,(=0,90, 180, 70, 360 ) (iii) Dividing y 7 +8 = 113 and writing cos α =7/ 113, sin α =8/ 113, so that α = 850, we find sin(θ α) = Hence θ α = nπ or θ α = nπ It follows that θ = nπ or θ = nπ The values in the required range are thus θ =18617, 9839, (= , ) (iv) Dividing y 5 +1 = 13, and writing sin α =5/13, cos α =1/13, so that α = 3948, the equation ecomes sin(θ + α) =4/13 Hence θ + α = nπ or θ + α =888 + nπ and θ = nπ or θ = nπ It follows that the values in the required range are θ =601, 4340, (= , ) 4

25 Useful results (a) ommit the following to memory (Some of the results can easily e worked out from graphs, the Formula Sheet, calculators etc ut it is often necessary to have the results at your fingertips) 1 To convert from degrees to radians, multiply y π 180 To convert from radians to degrees, multiply y 180 π Some useful values: 30 = π 6, 45 = π 4, 60 = π 3, 90 = π, 180 = π, 360 =π 3 In the right-angled triangle shown sine, cosine and tangent are defined as follows: sin θ = side opposite hypotenuse adjacent side, cos θ = c hypotenuse a side opposite, tan θ = c adjacent side a c θ a 4 cos θ +sin θ=1 5 ( π ) cos θ =sinθ, ( π ) sin θ =cosθ 6 7 cos( θ) = cos θ, sin( θ) = sin θ, tan( θ) = tan θ tan θ = sin θ cos θ 8 The area of a general triangle is given y rea of = 1 ase height = 1 a sin or, of course, y either of the other two similar expressions 9 The values of sine, cosine and tangents of certain important angles etween 0 and π/ are: θ 0 π/6 π/4 π/3 π/ sin θ 0 1/ 1/ 3/ 1 cos θ 1 3/ 1/ 1/ 0 tan θ 0 1/

26 10 The sine and cosine functions have period π, the tangent function has period π: if n is an integer cos(θ +nπ) =cos(θ), sin(θ +nπ) =sinθ, tan(θ + nπ) =tanθ, 11 The other three trigonometric ratios are defined as follows: sec θ 1 cos θ, cosec θ 1 sin θ, cot θ 1 tan θ 1 cos θ =cos θ 1, sin θ =sinθcos θ 13 The diagram to help you rememer the signs of the trigonometric functions of θ for all real values of θ is: π/ π S T 0, π, 3π/ () The following appear on the Mathematics Formula Sheet ut it is very useful to know them! 1 The sine and cosine formulae for a general triangle are a sin = sin = c sin, a = + c c cos, with similar results giving in terms of c, a, cos and c in terms of a,, cos The graphs of the three elementary trigonometric functions etween π and π are: +1 π 3π/ π π/ π/ π 3π/ π θ 1 Graph of cos θ 6

27 θ Graph of sin θ 1 +1 π 3π/ π π/ π/ π 3π/ π θ Graph of tan θ π 3π/ π π/ π/ π 3π/ π tan θ=sec θ 4 1+cot θ=cosec θ 5 cos( + ) =coscos sin sin, sin( + ) =sincos +cossin cos( ) =coscos +sinsin, sin( ) =sincos cos sin (c) lso on the Mathematics Formula Sheet are: 1 sin +sin=sin 1 (+)cos 1 ( ), sin sin =cos 1 (+)sin1 ( ), cos +cos=cos 1 (+)cos1 ( ), cos cos = sin 1 (+)sin1 ( ), 7