Module 2 - Trigonometry (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 2

TRIGONOMETRY

Module Topics 1. The measurement of angles and conversion between degrees and radians 2. The elementary trigonometric ratios and relations between 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 between them 6. Various useful formulae involving multiple angles 7. Trigonometric equations This module is again self-contained. It is another long module but it should be 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 p.21. 1. The measurement of angles Angles 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. By definition, if the angle θ in figure 1 is measured in radians, ......................... .......... ......... ....... ....... ...... ......... ...... .... ... ..... . . . ....... .. ... ... ...... ... . . . ... ... . ... . . . . ... .. .... ... .. .... ... . ... ......... . ... ... ... . .. ... . ... .. . ... . .. ... . . . ... . ........................................................ ......... ... . ... .... ... . . . ... . . . ... ... ... .. ... ... .... ... .... . . . . ..... ..... ...... ...... ....... ............ ........ .............................

r

L

θ

Figure 1 L Length of arc = , radius r a ratio which is independent of the units in which both L and r are measured. then

θ=

Since the length of the complete circumference of a circle of radius r is 2πr, in a complete revolution there are 2πr/r = 2π radians. Conversion from degrees to radians and vice-versa is straightforward since 360◦ is equivalent to 2π radians: π . To convert from degrees to radians, multiply by 180 180 . To convert from radians to degrees, multiply by π –1–

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 be used (e.g. 45◦ ). When using a pocket calculator, you must be very careful that it is operating in the correct—i.e. radian— mode unless you specifically intend to use degrees. Remember, though, that an accuracy of four decimal places in describing an angle in radians is equivalent to an accuracy of only two or three places in degrees. The answers to many problems in this Module are given in both radians and degrees. These answers are (hopefully!) accurate to four places. However, if you convert one answer directly to the other by using the appropriate scale factor, because of rounding errors in the calculation you may well find an apparent error in the last one or two decimal places. Example 1. Convert 60◦ to radians. π π = radians. 60◦ is 60 × 180 3 π Example 2. Convert radians to degrees. 6 π 180 ◦ π radians is × = 30◦ . 6 6 π One or two frequently used conversions are worth committing to memory: 30◦ =

π , 6

45◦ =

π , 4

60◦ =

π , 3

π , 2

90◦ =

180◦ = π ,

360◦ = 2π .

For other cases, use your pocket calculator (or tables) if necessary. ***Do Exercise 1.

Express the following in degrees:

(i) π/4,

(ii) 3π/2,

(iii) 7π/3.

(N.B. In all the Exercises in this module give numerical answers correct to 4 decimal places (i.e. 4 d.pl.), using degrees or radians as appropriate.) ***Do Exercise 2.

Express the following in radians: (i) 18◦ ,

2. The three elementary trigonometric ratios gent of an acute angle are defined to be: Sine of angle =

side opposite , hypotenuse

Tangent of angle =

(ii) 35◦ ,

(iii) 350◦,

(iv) 420◦ .

In a right-angled triangle, the sine, cosine and tan-

Cosine of angle =

adjacent side hypotenuse

side opposite . adjacent side

The notations used for the three ratios are respectively sin, cos and tan. Thus, in the right-angled triangle shown in figure 2 (the side labelled c, opposite the right-angle is, of course, the hypotenuse): b sin θ = , c

cos θ =

a , c

tan θ =

. ...... ..... .. ..... .... ..... .. ..... . . . . .... . ..... .. ..... . . . ... . .... . . . ... . ... . . . ... . .... . . . . .... .... . . . .. . .... . . . ... . ... . . . . ... .... . . . . ... .... . . . ... . ...... . . . ... . ... .... . . . . . . . . . . . . . . . .. . .... .... . . . . . . . .. . . .................................................................................................................

c

θ

a Figure 2

–2–

b

b . a

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 be ‘solved’—that is to say all other sides and angles of the triangle can be found. Example 3.

Solve the triangle shown in figure 2 if a = 2 cm and θ = π/6.

Since tan θ = b/a, cos θ = a/c, it follows that b = a tan θ and c = a/ cos θ and hence, putting in the given values for a and θ, we see that b = 2 tan

π  6

= 1.1547 cm and c =

2 = 2.3094 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. Also, having found b we could have used Pythagoras’ theorem to give us c.] Example 4.

Solve the triangle shown in figure 2 if a = 2 cm and b = 1 cm.

Firstly we use Pythagoras’ theorem to find c: c=

p p √ a2 + b2 = 22 + 12 = 5 = 2.2361 cm .

Next, since

b = .5 , a the inverse tangent button on your pocket calculator gives θ = .4636 or 26.5651◦. The remaining angle is then π − ( 12 π + .4636) = 1.1072 or 180◦ − (90◦ + 26.5651◦) = 63.4349◦. tan θ =

3. Important relations between sin, cos and tan Referring again to figure 2 you can see that, by Pythagoras’ theorem, c2 = a 2 + b 2 . However, a = c cos θ and b = c sin θ and hence c2 = c2 cos2 θ + c2 sin2 θ , from which it follows that

cos2 θ + sin2 θ = 1 .

Notice the notation carefully. We write, for example, cos2 θ to mean (cos θ)2 . [The quantity cos θ2 has no meaning since it could represent cos(θ2 ) or (cos θ)2 ]. Example 5.

Verify the formula cos2 θ + sin2 θ = 1 for the angle θ in Example 4.

Since θ = .4636, your pocket calculators tell you that cos2 θ = .8, whilst sin2 θ = .2. Hence cos2 θ + sin2 θ = 1. Another relationship between cosine and sine is obvious from figure 2. The angles between a and c and between b and c add to π/2 since the third angle of the triangle is π/2 and we know that the angles of a triangle add to π. It follows immediately that  π − θ = sin θ , cos 2 π  sin − θ = cos θ . 2 –3–

In addition it is easily shown that tan θ = since tan θ =

sin θ , cos θ

b/c sin θ b = = . a a/c cos θ

You can check for yourself that the above three formulae involving θ are true for the numbers 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 be sure to memorise them. They are repeated as numbers 4, 5 and 7 of the ‘useful results’ at the end of this module. 4. The solution of general triangles lowing three pieces of information:

If you are given any triangle together with any one of the fol-

(a) the lengths of all three sides (b) 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 between the two sides, the solution may not be 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 A etc.), the ‘sine’ rule is written b c a = = sin A sin B sin C and the ‘cosine’ rule gives the three formulae a2 = b2 + c2 − 2bc cos A,

b2 = c2 + a2 − 2ca cos B,

c2 = a2 + b2 − 2ab cos C.

A

...... ... ................. ......... ... ......... .... . ......... . . . ......... .... ......... ... ......... . . ......... .. . . ......... .. ......... . . .. ......... . . . ......... ... ......... . . ......... .. . . ......... .. . ......... . ........ ... . . .............................................................................................................................................................................................................................

c

B

b

a

C

Figure 3 The table below shows you which formula you need for which situation. Facts known

Rule to use

3 sides 2 sides and the angle between

cosine

2 sides and either of the other angles sine 1 side and any 2 angles

The sine and cosine rules appear on the Formula Sheet (and in the Data Book) but it is useful to remember them. You do not need to know the proofs of the two rules but they are included below since they are very straightforward. –4–

A

...... ... ... .......... ..... ... ..... ... ... . . ..... ..... ... ... ..... ... . ..... . . .. ..... .. . . ..... . .. . . ..... . . ..... ... . . . ..... .. .. ..... . . ..... .. .. . . . ..... .. ..... . . . ..... .. .. . ..... . ..... .. ... . . ..... . . . ..... . . ..... .. .. . . ..... ... .. ..... . . . ..... .. . . . ......................................................................................................................................................................................

c

B

h

b

a F Figure 4

C

In figure 4, you can see that the height, h, of the triangle is given equivalently by h = c sin B and h = b sin C. Hence c sin B = b sin C and c b = . sin B sin C The remaining part of the sine rule is proved similarly or by symmetry. To prove the cosine rule we see, again looking at figure 4, c2 = h2 + (a − CF )2 = b2 sin2 C + (a − b cos C)2 = b2 sin2 C + a2 + b2 cos2 C − 2ab cos C . However, cos2 C + sin2 C = 1, and hence c2 = a2 + b2 − 2ab cos C. Once again, the other two results follow similarly or by symmetry. Example 7.

Solve the triangle ABC if a = 12, b = 8, c = 10.

Since all three sides but no angles are known, we must use a cosine formula. c2 = a2 + b2 − 2ab cos C tells us that a2 + b2 − c2 2ab 144 + 64 − 100 = 2 × 12 × 8 108 = 0.5625 . = 192

cos C =

Hence, using a calculator, C = .9734, (= 55.7711◦). We could find the other angles in a similar way, but we now know one angle and so a simpler alternative is to use the sine formula. Since c a = , sin A sin C a sin C sin A = c 12 × .8268 = .9922 . = 10 Hence A = 1.4455, (= 82.8192◦). To find the remaining angle we could use the cosine or the sine formula, but obviously the easiest way is to remember that the angles of a triangle add to π, so that B = π −A−C = π −1.4455−.9734 = .7227, (= 41.4097◦). Notice that these results have been obtained retaining 10 digits throughout on a pocket calculator. However, at each stage they have been stated to only four places. If you use the four place answers (instead of the full 10 place ones) to perform your subsequent calculations, you will obtain an answer which is certainly not accurate to four places. Try it and see! –5–

Example 8.

Solve the triangle ABC if b = 2 cm, c = 1 cm, C = 20◦ .

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 advisable to draw a reasonably accurate diagram before proceeding further. It is easily seen that in this example two solutions are possible. A

A

........... ................ .... ............ ..... .......... .......... ..... ......... .... .......... ..... .......... .... ......... ..... ............... ..... .......... ... .... ......... ..... .. ........ .. .... . . ..........................................................................................

...................... .......... ..... .......... .... ......... ..... . . .......... . . .. .......... . . . . ......... .... . .......... . . ........... ... . . . ......... .... .......... . . ... . .......... ... .. . . . ........ .. .. . . . . . . ...............................................................................................................................................................................................

20

20

B Figure 5 (a)

B

C

C Figure 5 (b)

2 b sin C = sin 20◦ = .6840 c 1 Hence B = 43.1602◦ (= .7533), see figure 5(b), or B = 180◦ − 43.1602◦ = 136.8398◦ (= 2.3883), as shown on figure 5(a). It follows that sin B =

A = 180◦ − B − C = 180◦ − 43.1602◦ − 20◦ = 116.8398◦, (= 2.0392) or 180◦ − 136.8398◦ − 20◦ = 23.1602◦, (= .4042). Finally we may find a from the cosine formula. √ √ a = b2 + c2 − 2bc cos A = 22 + 12 − 2 × 2 × 1 × cos 116.8398◦ = 2.60883 cm √ or 22 + 12 − 2 × 2 × 1 × cos 23.1602◦ = 1.1499 cm. 5. The area of a triangle The area of a triangle is half that of a rectangle on the same base—that is to say, it is one half of the length of its base times its height. Hence, using figure 4 and remembering that h = b sin C, Area of 4ABC =

1 1 1 × base × height = × a × h = ab sin C . 2 2 2

In a similar way it can also be shown that Area of 4ABC =

1 1 bc sin A = ca sin B . 2 2

Although these formulae could be remembered and used it is usually easier to proceed from first principles by calculating the height directly (see example below). Example 9.

Find the area of the triangle ABC if b = 1 cm, c = 2 cm, C = 20◦ .

This is the same triangle as in Example 8 but with the lengths of the given sides interchanged. But this time figure 6 confirms the answer is unique. A

................................. ................... ................. .................. .......... ................... ... .......... .................. . . . . . . . . . .. . . . . . . . ......... . . .. ............. . . . . . . . . . . . . . . . . . . . . ...... .............................................................................................................................................................................................................

20

B

C Figure 6

sin B =

1 b sin C = sin 20◦ = .1710 c 2 –6–

Hence B = 9.8466◦. It follows that A = 180◦ − B − C = 180◦ − 9.8466◦ − 20◦ = 150.1534◦. This enables us to find a from the cosine formula: √ √ a = b2 + c2 − 2bc cos A = 12 + 22 − 2 × 1 × 2 × cos 150.1534◦. Using a calculator we deduce a = 2.9102 cm. The height h is easily found from

h = sin 20◦ , and hence the area of the triangle is given by 1 1 ×a×h 2 1 = × 2.9102 × 1 × sin 20◦ 2 = .4977 cm2 .

Area of 4ABC =

***Do Exercise 3.

Solve the right-angled triangle shown below in each of the following cases: A

.... ..... .. ..... .. ..... ... ..... . . . . .... ..... ... ..... ..... ... ..... . . . ... . ... . . . ... . .... . . . . .... .... . . . .. . .... . . . . ... ... . . . . ... .... . . . . ... .... . . . . . ............ .... . . . . . . . .... .... . . . . . . .................................................................................................

c

a

B (i) a = 1 m, B = π/3; ***Do Exercise 4.

◦

b

C

(iii) a = 3 m, b = 1 m.

(ii) a = 2 m, B = 10 ;

Solve the triangle shown below in the following cases:

(i) c = 5.1 m, B = 47◦ , A = 70◦ ;

(ii) a = 12.1 m, b = 17.2 m, c = 18.4 m;

(iii) b = 7 m, c = 3 m, C = 10◦ .

A

.... ... ................. ... ......... ......... ... ......... .... . . ......... ... ......... ......... ... . . . ......... .. . ......... . . ......... .. . . . ......... ... ......... . . ......... .. . . ......... .. . ......... . . . . .........................................................................................................................................................................................................

c

B ***Do Exercise 5.

b

a

C

Find the area of each of the triangles in Exercise 4.

6. Trigonometric ratios for special angles up to π/2 The angles 0◦ , 30◦ , 45◦ , 60◦ and 90◦ (i.e. 0, π/6, π/4, π/3 and π/2) occur frequently in trigonometric work and it is helpful to memorise the corresponding values of sine, cosine and tangent for each of these angles. Alternatively, you should be able to sketch the appropriate triangle and deduce the value of the ratios (see below), or use your calculator! (a) Angle 0 A

.................. ................................ ... ............................... ................................ ... ............................... . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . ....................... . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . ................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................................................................................................................................................................................................................................................................

c

B

θ

a Figure 7

–7–

b

C

In this case the triangle looks like figure 7 as θ approaches 0. Clearly b approaches 0 and as this happens a a b b and c become equal. However sin θ = , cos θ = , tan θ = , and so we conclude that c c a sin 0 = 0,

cos 0 = 1,

tan 0 = 0.

(b) Angle π/6 or π/3 When θ = π/3 the 4ABC forms half of an equilateral√ triangle—see √ figure 8. Hence, if AB = 2 units, BC = 1 unit and Pythagoras’ theorem tells us that AC = 22 − 12 = 3 units. A

.. ...... .. ... .. . ... .. .. .. .... ... . . . .. ... ... ... ... .... ... .. ... ... ... . . . . ... ... .... ......... . . . . ... ... ....................... . . ... . . . . . . ... ... .... . ... . ... . . . . . . . ... .. ... . . . . ... . . . . . ... ... .... . ... . . ... . . . . . . . ... .. . . . . ... ... .... .. . . . . ... ........ .. . . ... . ..... . . . . . ... . ... ... ... .... ... . ... ... . . ... . . . . . . ... .. .. ... . . . ............................................................................................. .... .... .... .... .... .... .... .... .... .... ....

π/6

√ 3

2

2

π/3

B It follows that

1 π cos = = .5, 3 2

C Figure 8

1

1

√ π 3 sin = = .8660, 3 2

tan

π √ = 3 = 1.7321. 3

From the same triangle we can also read off the results for an angle of π/6: cos

(c) Angle π/4

√ π 3 = = .8660, 6 2

sin

π 1 = = .5, 6 2

tan

π 1 = √ = .5774. 6 3

This time the picture is as in figure 9, with a = b = 1 and c =

√ 2.

A

. ...... .... .. ... ... .... . . ..... . .. .. ... .... ... .... . . ... .... . ... . ... . ... . . . . ... . ... . ... . ... . ... . .... . ... . .. . . ... . ........ . ... . ... ... . ... . ... .. . . ... . . .. .. . . . ... . ... .. . . . . ...................................................................................................

√

2

1

π/4

B It follows that: cos

1 π = √ = .7071, 4 2

1 Figure 9

sin

C

1 π = √ = .7071, 4 2

tan

π = 1. 4

(d) Angle π/2 This case is much the same as (a), but with the triangle the other way round, as shown in figure 10. This time, however, it is the angle at A which approaches 0 so that a approaches zero length and b and c eventually become equal. Thus cos

π = 0, 2 –8–

sin

π = 1. 2

As θ becomes nearer and nearer to π/2 you can also see that tan(π/2) gets larger and larger. We write tan θ → ∞ as θ →

π . 2

If you calculate tan(π/2), your calculator will show an error. A ..

.... ..... ....... .... ... .. ... ... .. ... ... . ... .... . .. ... .. .. ... ... .. ... ... . ... .. ... ... .. ... .... ... ... ... ... ... ... ... ... .... .. .... .. ... ... ... ... ... ... .... .. ... ... ......... ... ..... ... ... ... .. ...................................

c

b

θ B a C Figure 10 7. Angles between π/2 and π cos θ, sin θ and tan θ have so far only been 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 between π/2 and π, figure 11 shows what we must do. A...

. ... ...... .. ..... ... .... ... .... .. ... .... .... ... .... .. ... .... .... ... .... ... ... .... .... ... .... ... ... ... ... ....... .................... ... ... ........ ... ... .. ... . . . . ....... .... .... .... .... .... .... ...... .... .... .....................................................................................................

F

α θ B Figure 11

C

x

Choose BC to be along the positive x-axis of a coordinate system with B at the origin. We define the three trigonometrical ratios cos θ, sin θ and tan θ to have the same numerical values respectively as those of cos α, sin α and tan α but, since BF is in the negative direction of the x-axis (shown in the figure along the direction of BC), 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 = −

BF = − cos α , BA

sin θ =

FA = + sin α , BA

tan θ = −

FA = − tan α . BF

Since α + θ = π, it follows that α = π − θ and the above formulae can be written cos θ = − cos(π − θ),

sin θ = sin(π − θ),

–9–

tan θ = − tan(π − θ).

8. Angles between π and 2π and > 2π or < 0 .......................... ....... ..... ..... .... ... ... ... ... . . .. .... . . . ...... .... .... .... .... .......... ....... .... ............................................................................................ ... .. . ... ... .... ..... . . . . . . . ... .. ... ....... . ... .... ... ... .. . . ... .. . . ... ... ... ... ... ... ... ... . .... .. .. ... ... ..... . . . .. ... ...

θ

F

C

α B

..................... ....... ..... .... ... ... ... .... .. ... ...................................................................................................................................................................... ... ...... ...... .... ... ... ....... ... ... . . ..... . . ... ......... ............. .......... ....... ...... ... ...... ...... .. ...... . ...... .. ...... . ...... ... ...... ...... ... ...... ...... ... ...... ...... ...... .. ...... ...... .. ...... ...... ... ...... . ..

x

θ

A

B

C

α

F

x

A

Figure 13

Figure 12

To find the trigonometric ratios for these angles, we simply extend what we did in section 7. Let us again assume that BC is along the positive x-axis of a coordinate system with B at the origin. Then the magnitudes ˆ shown in figures 12 and 13 are the same as for the acute of the trigonometric ratios for the angles ABC angles that AB makes with this x-axis, but an appropriate sign must also be attached. As in section 7, this sign is calculated by looking at the signs attached to the lines BF and AF (note AB is always assumed to have a positive sign). If BF is to the right, it is positive; if it is to the left, it is negative. If AF is up, it is positive; if it is down, it is negative. Thus in figure 12 BF is to the left and AF is down: hence cosines will be negative, sines negative and tangents (negative over negative) positive whilst, in figure 13, BF is to the right and AF down: hence cosines will be positive, sines negative and tangents negative. Finally, if the angle is in excess of 2π we have gone full circle (at least once!) and we simply knock off as many whole number multiples of 2π (complete revolutions) as we need in order to obtain an angle between 0 and 2π and find the appropriate trigonometric ratio for this. Thus, for example, sin

 π π 9π = sin 2 × 2π + = sin = 1 . 2 2 2

Likewise, if the angle is less than 0, we have to add as many whole number multiples of 2π as is necessary to put the angle into the range 0 to 2π. Thus cos(θ + 2nπ) = cos θ,

sin(θ + 2nπ) = sin θ,

tan(θ + 2nπ) = tan θ,

if n is an integer. In fact tan(θ + nπ) = tan θ, but that you will see later! Most people find it easiest to remember which signs are positive by drawing the following diagram (figure 14) in which A stands for ‘All’, S stands for ‘Sine’, T stands for ‘Tangent’ and C stands for ‘Cosine’. All other signs are negative. π/2

..................... ........... ..... ................. ...... ....... .. ..... ...... .... ..... ... ... .... . . ... ... . ... . . ... .. .. ... . . .. .. ... . . . . ... . . . . ... .... .... ... ... ... .................................................................................................................. ... . . ... ... .. .. . . ... .. .... ... .. . . . . ... . ... ..... ... ... .. ... ... .... ... .... ... . ..... . . . . . ...... ... .. ........ ...... ............. .... .................. ...............

S

π

T

A

C

3π/2 Figure 14 – 10 –

0, 2π,...

Notice and remember the very useful results that: cos(−θ) = cos θ, Example 10.

sin(−θ) = − sin θ,

tan(−θ) = − tan θ.

Evaluate sin(2π/3).

2π/3 lies in the range from π/2 to π. Figure 14 therefore shows that sin(2π/3) is positive and so sin(2π/3) = + sin(π − 2π/3) = sin(π/3). Using the results obtained with figure 8 it follows that √ sin(2π/3) = sin(π/3) = 3/2 = .8660. Check that your calculator gives the latter value for both sin(2π/3) and sin(π/3). Example 11.

Evaluate cos 2.

2 is larger than π/2 (= 1.5708) and smaller than π (= 3.1416). Hence it follows from figure 14 that cos 2 is the negative of the cosine of π − 2 (= 1.1416). Using your calculator verify that the values for cos 2 and − cos(π − 2) are both equal to −.4161 Example 12.

Evaluate cos(190◦ ).

190◦ (= 3.3161) is between π and 3π/2. Hence, looking at figure 14, its cosine is the negative of the cosine of 3.3161−π. Check the values given by your calculator for both cos(3.3161) and − cos(3.3161− π). You’ll find that they are both the same, namely −.9848. Example 13.

Evaluate tan 5.

5 is bigger than 3π/2 (= 4.7123) and smaller than 2π (= 6.2831). Hence, from figure 14, tan 5 = − tan(2π − 5). Again, your calculator should give the same answer for both tan 5 and − tan(2π − 5), namely −3.3805. It may be that you are more used to using degrees than radians—in that case you may prefer to check the range in which 5 lies by converting it to degrees (286.4789◦) and noting that this lies between 270◦ and 360◦ . Example 14.

Evaluate sin(33π/4).

The angle is outside the range 0 to 2π and so we have to subtract whole number multiples of 2π to √ bring it into range. 33π/4 = 4 × 2π + 14 π. Hence sin(33π/4) = sin(π/4) = 1/ 2. Check that the answer (.7071) given by 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 back to where we started— and so sin(−4π) = sin 0 = 0.

– 11 –

9. Graphs of trigonometric functions With a graph-plotting calculator you can very easily see what the graphs of cos θ, sin θ and tan θ look like for θ in the range (−2π, 2π). If you don’t have such a calculator, the previous sections give sufficient values of each function for you to be able to make a good plot. The three graphs are: ... ..... ........ ... ... ... .......... . ... . . . . . ........ ... ... ............. ..... . . . ...... . . .... .. ... ... .... ... . ... . .... ... . ... ... . . . . . . . ... ... .. ... ... ... .... ... .. ... ... ... ... ... ... ... .. ... ... ... ... ... .. . . . . ... . ... ... ... ... ... .... ... ... ... ... ... ... ... .. .. ... .... .. . .. ... . . . . . . . . . . . . . . . . ..................................................................................................................................................................................................................................................................................................................................................................................................................... . ... ... . .. .. ... .. ... .. ... ... .. ... ... . . . . . . ... ... . .... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... ... ... ... . . . . . . ... . ... .. ... ... ..... ... ... ... ... ... .. ... ... .... ... .... ..... ... ..... ..... . ........ ............ . . . . . . . . . . . ........... ..... ......... ... ... .

+1

θ

−2π

−3π/2

−π

−π/2

π/2

π

3π/2

2π

−1

Figure 15: Graph of cos θ ... ...... ......... .... ... . ....... ................. . .................... . . . .. . ..... .... ...... .... . . ... .... .... . .... .. ... ... . ... . . ... ... . ... . ... . . . . ... . ... . .. . . . . . . . ... ... . .. .. . . . ... . . . . ... . .. ... . ... . . . . ... ... .. ... .. . . . . .. ... ... . .. . . . ... ... . .. ... . . ... . . . ... . . .. . . .. . . . . . . ................................................................................................................................................................................................................................................................................................................................................................................................................................... . ... .... ... . . ... ... ... . ... . . . ... . . ... . ... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... ... ... . .. . . . . ... . . . ... . .. ... ... .... ... ... ... ... ... ... ... ... ... .. ... .... ... .... ... ... ..... . . . . . . . . . . . . . ....... ....... ........ ........ ......... ...... ........ ... ... ...

+1

θ

−2π

−3π/2 −π

−π/2

π/2

π

3π/2

2π

−1

Figure 16: Graph of sin θ . . .. .. .. ... ... ... ... ........ .. . .. . .. . .. ......... .. .. ... ... .. .. .. .. ... .. ... .. .... .. . .. .... .. ... . .... ... ... .. .. . .. ... ... .. .. .. ... ... ... .. ....... .. .. .. ... .. .. . . ... ... .. .. .... .. .. ... .. ... .... ... ... ... ... .. . . .. ... ... .. .. .. .. .. ... . . . ... . . . . .... . . . .. .. . . ... . .. . . . . ... . .. . .. .. .... ... .. .. .... .... .... . . . . . . ..... ... . . .. .. ... .. .. .... ... . ... . . . . . . . . .. .. ... ... .. ... .. .. .. . . . . . . . . . . . . . ... . . . .. .. .. . .. ... .. ... . . . . ... . . . . .. .. .. .... ... .. ... .. .. .. .. . . . . . .. . ... . . . . . . .. .. .. .. .. .. .. . . . ... . . . . . . .. .. .. ... ... ..... ... .. .. .. . . . . .. . ... . . . . . .. .. .. .. .. .. ... .. . ... . . . . . . .. .. .. .... .... ... .. .. .. .. .. . . . . . . ... . . . . . . .................................................................................................................................................................................................................................................................................................................................................................................................................................. . . .. . . .. .. .. ... .. ..... ... ... .. .. . . . .. ... .. . . . . . . . . ... ... .. ... ... .. ... ... ... ... ... ... ... ... .. .. ... . ... ... .... ... ... ... . .. .. .. .. .. .. . . ... . . . . . . . ... .... .. .. ... ... ... ... . ... ... ... ... ... ... . .. .. ... .. ... ... ... ... .. .. ........ . . .. ... . .... .... . . .. ... .... .. .. ... ... ... ... .. ... ... ... ... ... .... .. .. .. .. .. .. ... ... .. .. .. .. .. . . . . . . . . . . . . ... ... .. ... ... ... ... .... ... ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... .... .... .... .... ... ... .. ... . ... .. .. ... .. .. .. .. ....... . ... ... ... ... ... .... .... ... .... ... .... .. .... ... ... ... ... ... .. ... .. ... . .. ... .. .. .. . ... .... ... ... .... ... .... .. ... .. .. ... ... .. ... .. .. .. .

2

1

−2π

−3π/2 −π

−π/2

π/2

π

θ

3π/2

2π

−1

−2

Figure 17: Graph of tan θ As we saw in previous sections, outside the range (0, 2π) each curve repeats itself every 2π, the extensions joining on smoothly to the curves shown in each case. We say that these curves ‘have period 2π’. In fact, the tangent curve repeats itself every π—has period π—a fact which means that we have to take particular – 12 –

care if we are asked to find the angle in the range (0, 2π) which has a particular value for its tangent: there are two possible answers! We shall discuss this in more detail later. Notice that, if the cosine curve is displaced to the right by π/2, the sine curve is obtained whereas if the sine curve is displaced to the right by π/2 then the negative of the cosine curve is derived. This means that sin

π 2

 + θ = cos θ ,

cos

π 2

 + θ = − sin θ ,

which are two useful results. Notice also that the tangent curve becomes infinite (either in the positive or negative direction) at the points θ = 3π/2, −π/2 , π/2 , 3π/2. ***Do Exercise 6. (i) Use your calculator to find the sine, cosine and tangent of each of the following: (a) 140◦, (b) 7, (c) 700◦ . (ii) Use your calculator to find (a) cos(−20◦ ), (b) sin(−250◦ ), (c) tan(−450◦), (d) sin(−10). (iii) Show by direct calculation that (a) cos2 120◦ + sin2 120◦ = 1, (b) cos2 410◦ − sin2 410◦ = cos 820◦ , (c) −2 sin 1.5 cos 1.5 = sin(−3). (iv) Given that π/2 ≤ α ≤ π and sin α = 24/25, find cos α and tan α 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 (i.e. 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

Be 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 = (cos θ)−1 , not cos−1 θ. cosn θ means (cos θ)n whenever n 6= −1 but not when n = −1. Awkward, isn’t it? It is worth pointing out that the above difficulty with notation is the main reason why the angle whose cosine is θ is often denoted by arc cos θ, instead of cos−1 θ. Sketch for yourselves the graphs of the functions sec θ, cosec θ and cot θ, noticing in particular that, whilst −1 ≤ cos θ ≤ 1 and −1 ≤ sin θ ≤ 1 for all values of θ, the functions sec θ and cosec θ are always either ≥ 1 or ≤ −1. Also notice that, unlike tan θ, cot θ is infinite at 0 and π but finite at π/2 and 3π/2. Example 17.

Evaluate sec 3.14, cosec 3.14 and tan 3.14 to 6 d.pl.

sec 3.14 = 1/ cos 3.14 = 1/ − .999999 = −1.000001. cosec 3.14 = 1/ sin 3.14 = 1/.001593 = 627.883190. cot 3.14 = 1/ tan 3.14 = 1/ − .001593 = −627.882394. Notice how nearly numerically equal the values of sin 3.14 and tan 3.14 (and therefore of cosec 3.14 and cot 3.14) are for these values of θ so close to π. 11. Two further identities You will remember that we proved in section 3 that cos2 θ + sin2 θ = 1 and that tan θ = sin θ/ cos θ for all values of θ. If we divide the first equation by cos2 θ we find sin2 θ 1 cos2 θ + = , cos2 θ cos2 θ cos2 θ – 13 –

and making use of the second equation then gives 1 + tan2 θ = sec2 θ . In a similar way (dividing the first equation by sin2 θ instead of cos2 θ), we find cot2 θ + 1 = cosec2 θ . These two formulae are useful and, if possible, should be remembered, although they do appear on the Formula Sheet (and in the Data Book). The important point is to know that formulae exist connecting tan2 θ to sec2 θ (and cot2 θ to cosec2 θ). 12. Trigonometric ratios for sums and differences of angles It is often necessary to be able to express cos 2θ in terms of cos θ, a formula which is a special case of that which gives cos(A + B) in terms of the cosines and sines of A and B alone. From figure 18 we can easily show that cos(A + B) = cos A cos B − sin A sin B . In the figure, the perpendicular, AF , from A onto BC has F as its foot and the extension of BC is of length ˆ is of size A + B since it is an exterior angle of the triangle ABC. e. In addition, the angle ACF A

... .......... ............ . ....... ..... ... ....... ..... .. ....... ......... . . . . . . .. ..... ....... ... ....... ..... ......... . . . . . . ... . ........ ....... ....... ..... . . . . . . . . . ... . . ..... .... . . . . . . . . . ... . ........ ..... ....... ..... . . . . . . . . ... . . . ..... .... . . . . . . . . ... . . .. . ....... ..... ..... ....... . . . . . . ... . . . . ...... ...... . . . . . . . . ... . . . . ..... ..... ........ .. .... ....... ....... ... ..... . . . . . . . . ... . . . . ... . ... . ...... . . . . . . . . ... . . ... ... . ..... .......... .. ... ......... ....... .... . . . ... . . . . .. ... .. ..... . . . . . . . . . . . . . ...........................................................................................................................................................................................................................

A

y

b

h

x

B

B

a

A+B

C

.. ..................................................................................................... .

e

F

. .................................................................................................... .

l Figure 18

In triangle ACF , cos(A + B) = However, from triangle ABF ,

l−a e = . b b

l = (x + y) cos B ,

where x = a cos B and y = b cos A. It follows that (a cos B + b cos A) cos B − a (x + y) cos B − a = b b a a(cos2 B − 1) + cos A cos B = − sin2 B + cos A cos B = b b

cos(A + B) =

and, since the sine rule tells us that

we can replace

sin A sin B = , b a

a sin B by sin A and conclude that b

cos(A + B) = cos A cos B − sin A sin B – 14 –

as required. If in this formula we replace A by (π/2) − A, we obtain a corresponding formula for sin(A − B) and if in the two formulae we now have we replace B by −B, we obtain two more formulae for cos(A − B) and sin(A + B). 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 Book) and you are able to use them. cos(A + B) = cos A cos B − sin A sin B sin(A + B) = sin A cos B + cos A sin B cos(A − B) = cos A cos B + sin A sin B sin(A − B) = sin A cos B − cos A sin B . Notice the somewhat unexpected arrangement of the signs in the above equations. The corresponding formulae for tan(A ± B) are less important since they can be obtained, if needed, from those for cos(A ± B) and sin(A ± B) by division. For example tan(A + B) =

sin(A + B) , cos(A + B)

and the expressions for sin(A + B) and cos(A + B) can then be used. These formulae for sums and differences of angles are very useful in determining a number of the formulae quoted earlier. For example, using the formula for cos(A − B), we obtain cos(π − θ) = cos π cos θ + sin π sin θ = −1 × cos θ + 0 × sin θ = − cos θ .

13. Double angle formulae These are the formulae mentioned at the beginning of the previous note. If we put both A and B equal to θ in the formulae for cos(A + B) and sin(A + B), we obtain cos 2θ = cos2 θ − sin2 θ ,

sin 2θ = 2 sin θ cos θ .

These results are highly important and must be committed to memory. The former may be written (using sin2 θ + cos2 θ = 1) in the alternative forms: cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ . These alternative identities are also very important and you must either learn them or be able to derive them quickly. Example 18.

Prove that sin(C − D) + sin(C + D) = 2 sin C cos D.

sin(C − D) = sin C cos D − cos C sin D and sin(C + D) = sin C cos D + cos C sin D . By adding these two equations we obtain the required result.

– 15 –

14. Sum and difference formulae If in Example 18 we put C = 12 (A + B) and D = obtain 1 1 sin A + sin B = 2 sin (A + B) cos (A − B) . 2 2

1 2 (A

− B), we

This is one of a set of four formulae which are useful in some integration problems in calculus, but not worth remembering—just know where to find them (on the Formula Sheet and in the Data Book) or how to deduce them if you need them! The other three are: 1 1 (A + B) sin (A − B) , 2 2 1 1 cos A + cos B = 2 cos (A + B) cos (A − B) , 2 2 1 1 cos A − cos B = −2 sin (A + B) sin (A − B) , 2 2 sin A − sin B = 2 cos

Notice the negative sign in the last of these. Example 19. (a) (b)

Express the following as products: (a) sin 6θ + sin 4θ,

(b) cos 18θ − cos 2θ.

1 1 (6θ + 4θ) cos (6θ − 4θ) = 2 sin 5θ cos θ . 2 2 1 1 cos 18θ − cos 2θ = −2 sin (18θ + 2θ) sin (18θ − 2θ) = −2 sin 10θ sin 8θ . 2 2

sin 6θ + sin 4θ = 2 sin

Example 20.

Express cos 7θ cos 3θ as a sum or difference of two cosines.

Here we use the formula

cos A + cos B = 2 cos

1 1 (A + B) cos (A − B) . 2 2

We put 7θ = 12 (A + B) and 3θ = 12 (A − B) so that, after multiplying both equations by 2, we obtain A + B = 14θ and A − B = 6θ. Adding these two equations and dividing by two we see that A = 10θ. It follows that B = 4θ. Thus cos 7θ cos 3θ =

***Do Exercise 7. (ii) tan(A + B) =

1 (cos 10θ + cos 4θ) . 2

Prove the following identities:

tan A + tan B , 1 − tan A tan B

(iii) tan 2θ =

(i) (sin θ + cos θ)2 = 1 + sin 2θ,

2 tan θ , 1 − tan2 θ

(iv)

tan θ + cot θ = cosec θ. sec θ

***Do Exercise 8. α).

If sin α = 5/13, where α is acute, find tan α, cot α and cosec α (without calculating

***Do Exercise 9. and b.

If tan θ = a/b, where θ is acute, find expressions for sin θ and sec2 θ in terms of a

***Do Exercise 10. By writing 3θ = 2θ + θ, and then using the identities for sums of angles and double angles, show that (i) sin 3θ = 3 sin θ − 4 sin3 θ, (ii) cos 3θ = 4 cos3 θ − 3 cos θ. ***Do Exercise 11. Using the formulae for sums and differences of angles show that   π  π 3π + θ = cos θ, (ii) cos + θ = − sin θ, (iii) sin (π − θ) = sin θ, (iv) cos + θ = sin θ. (i) sin 2 2 2

– 16 –

***Do Exercise 12. (i) sin 4θ + sin θ,

Express as products of sines and/or cosines:

(ii) sin 8θ − sin 6θ,

***Do Exercise 13. (i) 2 sin 6θ cos 2θ,

(iii) cos 12θ + cos 10θ,

(iv) cos θ − cos 2θ,

(v) sin (π/3) + sin (π/4).

Express as sums or differences of trigonometric functions:

(ii) 2 cos 5θ cos 3θ,

(iii) 2 cos 4θ sin θ,

(iv) 2 sin 7θ sin 5θ.

15. Simple trigonometric equations It is important to realise that, since sin θ and cos θ repeat themselves every 2π, even a simple equation like 1 cos θ = 2 is satisfied by an infinite number of values of θ. By referring back to figure 15, or using your graphics calculator, you will see that not only is 13 π a solution, but so are 2π + 13 π, 4π + 13 π, −2π + 13 π and so on. In fact it’s worse than this—even between θ = 0 and θ = 2π there are two different solutions, as you can again see by looking at the graph of cos θ in figure 15 (or on your graphics calculator) or by using the ‘SATC’ diagram in figure 14. Either method shows you that 2π − 13 π = 53 π is also a solution. We can therefore write the general solution of this equation as θ=

1 π + 2nπ , 3

or θ =

5 π + 2nπ , 3

where n is any whole number (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 above from it. Pocket calculators have the same problem. 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. Calculators use the following convention, and you should do the same: cos−1 x is defined to be that solution of cos θ = x which lies in the range 0 ≤ θ ≤ π, sin−1 x is that solution of sin θ = x which lies in the range − 12 π ≤ θ ≤ 12 π and tan−1 x is that solution of tan θ = x which lies in the range − 21 π ≤ θ ≤ 12 π. Notice the ranges are not the same for each function: this is necessary in order to ensure that each possible value is covered once and once only as can be seen from figures 15-17. Example 21.

Find the general solution of cos θ = −.3.

Your calculator will tell you that cos−1 (−.3) is 1.8755. This is the solution of the equation which lies in the range [0, π]—in fact between 12 π and π. The graph then shows you that 2π − 1.8755 = 4.4077 is also a solution, giving you a second answer between 0 and 2π. The general solution is then θ = 1.8755 + 2nπ ,

or

θ = 4.4077 + 2nπ

for any integer n, positive, negative or zero.

√ Example 22. Find the general solution of sin θ = − 12 2. √ sin−1 (− 12 2) = − 41 π = −.7854 since this is the appropriate value lying in the range [− 12 π, 12 π]. Your calculator will give you the same answer. As you can see from the graph, the two solutions lying between 0 and 2π are θ = π + 14 π and 2π − 14 π, i.e. θ = 54 π and θ = 74 π. The general solution is then θ=

5π + 2nπ , 4

or

for integer n.

– 17 –

θ=

7π + 2nπ 4

Example 23.

Find the solution of tan θ = 1.6 which lies between π and 2π.

Your calculator gives you tan−1 1.6 = 1.0122. This is the value which lies in [− 12 π, 12 π]. The tangent function (see the graph) repeats itself every π. The general solution of the equation is tan−1 θ = 1.0122 + nπ and the solution we require, obtained by taking n = 1, is thus 4.1538. Some trigonometric equations can readily be solved using the double angle formulae (section 13) and simple algebra. Example 24. 0 ≤ θ ≤ 2π.

Find the solutions to the equation 6 cos 2θ + 5 cos θ + 4 = 0 which lie in the range

The double angle formulae tell us that cos 2θ = 2 cos2 θ − 1 and hence the given equation can be written 6(2 cos2 θ − 1) + 5 cos θ + 4 = 0 or

12 cos2 θ + 5 cos θ − 2 = 0 ,

which is a quadratic equation in cos θ. This equation factorises to give (3 cos θ + 2)(4 cos θ − 1) = 0 and hence either 3 cos θ + 2 = 0 or 4 cos θ − 1 = 0. It follows that cos θ = −2/3 or cos θ = 1/4 (results which could also have been found from the quadratic equation using the formula). According to the calculator, these solutions give     2 1 = 2.3005 or θ = cos−1 = 1.3181 , θ = cos−1 − 3 4 and from figure 14, or figure 15, it is clear that each of the above solutions leads to a corresponding second solution between 0 and 2π. Hence there are four solutions between 0 and 2π: θ = 2.3005 ,

θ = 2π − 2.3005 = 3.9827 ,

θ = 1.3181 ,

θ = 2π − 1.3181 = 4.9651 .

16. Slightly more difficult trigonometric equations Equations of the form sin A ± sin B = 0 or cos A ± cos B = 0 can be solved by using the appropriate sum or difference formula. Example 25.

Find the roots of the equation sin 7θ = sin θ for 0 ≤ θ ≤ 12 π.

Here we use the formula for the difference of two sines: 1 1 sin 7θ − sin θ = 2 cos (7θ + θ) sin (7θ − θ) = 2 cos 4θ sin 3θ . 2 2 Hence, equating this to zero, we see that either cos 4θ = 0 or sin 3θ = 0. It follows from the first of these equations (look at the graphs!) that 4θ = 12 π + nπ and from the second that 3θ = nπ. Hence θ = 18 π + 14 nπ or 13 nπ and, picking out the solutions which are in the required range, we see (n = 0, 1 in the first equation) that θ = 18 π or 38 π or (n = 0, 1 in the second equation) θ = 0 or 13 π. Thus there are four solutions in the required range. Finally, equations of the form a cos θ ± b sin θ = 0 can be solved by expressing the left-hand side in one of the forms A sin(θ ± α) or A cos(θ ± α) for suitably chosen A and α. This is an extremely important form for the Engineer since it is expressed in terms of ‘phase’ (α) and ‘amplitude’ (A).

– 18 –

Solve the equation 3 sin θ + 4 cos θ = 3 for 0 ≤ θ ≤ 2π.

Example 26. First notice that

A sin(θ + α) = A cos α sin θ + A sin α cos θ . Hence, if this is to be identically equal to the left-hand side of the given equation, A cos α = 3

and

A sin α = 4 .

Squaring and adding, we see that A2 (sin2 α + cos2 α) = 32 + 42 so that A = 5. It follows that cos α = 35 and sin α = 45 , so that, by calculator, α = .9273. 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, 2π]. If we tried to use just one of the equations instead of both, we would get two solutions, one wrong since it would not satisfy the second equation! Our equation now simplifies to 5 sin(θ + .9273) = 3 or sin(θ + .9273) = 3/5 . Once more using our calculator and looking at the sine graph, we find that θ + .9273 = .6435 or π − (θ + .9273) = .6435. That is to say θ = −.2838 or θ = +1.5708, with general solutions θ = −.2838 + 2nπ or θ = 1.5708 + 2nπ. Finally, choice of n = 1 in the first equation or n = 0 in the second gives the required solutions: θ = 5.9994 or 1.5708—i.e. π/2. Find the values of θ in the range 0 ≤ θ ≤ 2π which satisfy the following equations:

***Do Exercise 14. (i) sin θ = .87,

(ii) tan θ = 1.39,

***Do Exercise 15. (i) sin 2θ = −.81,

Find all values of θ which satisfy the following equations:

(ii) cos 3θ = −.49,

(ii) cos 3θ = cos 120◦ ,

***Do Exercise 17. 2

***Do Exercise 18.

(iii) tan2 θ = 13 .

Find all solutions of the following equations which lie in the range 0 ≤ θ ≤ π:

(i) 2 sin θ + sin θ − 1 = 0,

(i) cos θ = cos 4θ,

(iii) tan 12 θ = .63

Find all values of θ in the range 0 ≤ θ ≤ 180◦ which satisfy the following equations:

***Do Exercise 16. (i) sin 2θ = sin 60◦ ,

(iii) cos θ = .22

(ii) 16 tan2 θ − 24 tan θ + 9 = 0,

(iii) sin 2θ = sin θ.

Find all solutions of the following equations which lie in the range 0 ≤ θ ≤ 2π:

(ii) sin 3θ + sin θ = 0,

(iii) 7 sin θ − 8 cos θ = 9,

– 19 –

(iv) 5 cos θ + 12 sin θ = 4.

You have now worked through the Module, so please attend a testing session as soon as possible. The following questions are similar to ones that you will be asked in the Test (solutions can be found on p.28). Specimen Test 2 π radians in degrees 6

1.

Express

2.

In the triangle ABC find AB (correct to 4 decimal places) ....... ...... . ...... .... ...... .. ...... . . . . . .... . ...... .. ...... . . . . ... . ..... . . . . . .... .... . . . . .. . ..... . . . . ... . .... . . . . . ... ..... . . . . ... . .... . . . . . ... ..... . . . . ... . .... . . . . . ... ..... . . . . ... . .... . . ◦ . . . ... ..... . . . . ... . .... . . . . . ..............................................................................................................................................

C

8

43

A

B

3.

If sin θ = 0.3 find

(i) cosec θ,

(ii) cos θ

4.

In the triangle ABC find AC (correct to 4 decimal places) C

.... ..... ......... ..... ..... ...... ..... ..... .... . . . . . ..... ...... ..... ...... ..... . . . . ..... .... . . . . ..... .... . . ..... . . ... . ..... . . . . .... ..... . . ..... . . ... . ..... . . . . .. ..... . . . . ◦ ◦ ............ .... . . . . ..... ... . . . . . .. . ........................................................................................................................................................................................

8

35

40

A

5.

B

In the triangle P QR find β Q

............ ... ............ .......... ... .......... ... .......... ... .......... .......... ... .......... ... .......... ... .......... .......... ... .......... ... .......... ... .......... ... .......... .......... ... .......... ... .......... ... .......... ... .......... ...... ... .................................................................................................................................................................................................

12

6

β

P

10

6.

Sketch the graph of sin x for −2π < x < 2π

7.

Express sin(π + θ) in terms of sin θ

8.

Find all values of α which satisfy cos α = −0.5

9.

Express cos 4θ + cos 2θ as a product of trigonometric functions – 20 –

R

Worked solutions to Exercises Note carefully that, whilst alternative answers have been given in degrees and radians, these have been 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. ◦ 180 ◦ 180 π = = 45◦ , π 4 4  ◦ 7π 180 7π = = 420◦. (iii) 3 π 3

π 1. (i) = 4



π π 18 = = .3142, 180 10 π (iii) 350◦ = 350 = 6.1087, 180

2. (i) 18◦ =

(ii)

(ii) (iv)

3π = 2



180 3π π 2

◦

= 270◦ ,

π 35 = .6109, 180 π 420◦ = 420 = 7.3304. 180 35◦ =

a 1 1 = = = 2 m, cos B cos(π/3) .5 b = a tan B = tan(π/3) = 1.7321 m,   1 1 1 1 π + B = π − π − π = π. A=π− 2 2 3 6

3. (i) c =

2 2 a = = = 2.0309 m, cos B cos(10◦ ) .9848 b = a tan B = 2 tan(10◦ ) = .3527 m, A = 180◦ − (90◦ + B) = 180◦ − 90◦ − 10◦ = 80◦ . p √ (iii) c = a2 + b2 = 10 m = 3.1623 m, 1 b tan B = = . Hence, since B < 12 π, B = tan−1 .3333 = .3218 (= 18.4349◦). a 3  1 1 π + B = π − .3218 = 1.2490, (= 71.5651◦). A=π− 2 2 (ii) c =

4. (i) Firstly, A + B + C = 180◦ . Hence, C = 180◦ − 47◦ − 70◦ = 63◦ . Now we can use the sine rule to give b c a = = implies that us a and b: sin A sin B sin C c sin A 5.1 × .9397 c sin B 5.1 × .7314 a= = = 5.3787 m, b = = = 4.1862 m. sin C .8910 sin C .8910 (ii) Here we must use the cosine formula: a2 = b2 + c2 − 2bc cos A tells us that 17.22 + 18.42 − 12.12 b 2 + c2 − a 2 = = .7710 cos A = 2bc 2 × 17.2 × 18.4 and so A = 39.5594◦, (= .6904). In the same way, the angle cos B is given by 18.42 + 12.12 − 17.22 c2 + a 2 − b 2 = = .4247 cos B = 2ca 2 × 18.4 × 12.1 ◦ and so B = 64.8655 , (= 1.1321). Finally, C = 180◦ − A − B = 180◦ − 39.5594◦ − 64.8655◦ = 75.5751◦, (= 1.3190). (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)! 7 b sin B = sin C = sin 10◦ = .4052 c 3 – 21 –

Hence B = 23.9023◦, (= .4172) or B = 180◦ − 23.9023◦ = 156.0977◦, (= 2.7244). It follows that A = 180◦ −B −C = 180◦ −23.9023◦ −10◦ = 146.0977◦, (= 2.5499) or 180◦ −156.0977◦ −10◦ = 13.9023◦, (= .2426). Finally we may find a from the cosine formula. √ √ √ a = b2 + c2 − 2bc cos A = 72 + 32 − 2 × 7 × 3 × cos 146.0977◦ = 92.8596 = 9.6364 m √ √ or 72 + 32 − 2 × 7 × 3 × cos 13.9023◦ = 17.2303 = 4.1509 m. 5.

The area in each case is given by, for example answers are (i) (ii)

(iii)

1 2 1 2 1 2

1 2

× base × height =

1 2

× a × b sin C = 12 ab sin C. Thus the

× 5.3787 × 4.1862 × sin 63◦ = 10.0311 m2 , × 12.1 × 17.2 × sin 75.5751◦ = 100.7795 m2 , × 9.6364 × 7 × sin 10◦ = 5.8567 m2 or

1 2

× 4.1509 × 7 × sin 10◦ = 2.5228 m2 .

6. (i) (a) .6428, −.7660, −.8391, (b) .6570, .7539, .8714, (c) −.3420, .9397, −.3640. (ii) (a) .9397, (b) .9397, (c) ∞, (d) .5440. (iii) (a) cos2 120◦ + sin2 120◦ = .25 + .75 = 1, (b) cos2 410◦ − sin2 410◦ = .4132 − .5868 = −.1736 = cos 820◦ (c) −2 sin 1.5 cos 1.5 = −2 × .9975 × .0707 = −.1411 = sin(−3). p p (iv) cos2 α + sin2 α = 1. Hence cos α = ± 1 − sin2 α = ± 1 − (24/25)2 = ±7/25. However, α is obtuse and so lies between 90◦ and 180◦. It follows that cos α is negative and so cos α = −7/25. sin α = −24/7. The sign is correct for an obtuse angle. tan α = cos α 7. (i) (sin θ + cos θ)2 = (sin2 θ + cos2 θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ = 1 + sin 2θ . (ii) tan(A + B) =

sin A cos B + cos A sin B sin(A + B) = cos(A + B) cos A cos B − sin A sin B

On dividing top and bottom of the right hand side by cos A cos B we obtain tan(A + B) =

tan A + tan B . 1 − tan A tan B

(iii) Putting A = B = θ in part (ii), the result follows immediately. (iv)

(sin θ/ cos θ) + (cos θ/ sin θ) tan θ + cot θ = sec θ (1/ cos θ)


cos θ sin θ (sin θ/ cos θ) + (cos θ/ sin θ) = cos θ sin θ(1/ cos θ)

sin2 θ + cos2 θ sin θ 1 = cosec θ . = sin θ =

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 132 − 52 = 144 = 12. Hence tan α = 5/12, cot α = 1/ tan α = 12/5 and cosec α = a/ sin α = 13/5. – 22 –

Since tan θ = a/b and θ is acute, a and b are respectively the side opposite to and the side √ by θ in a √ right-angled triangle whose hypotenuse is of length a2 + b2 . It follows that sin θ = a/ a2 + b2 . In addition 1 + tan2 θ = sec2 θ and hence sec2 θ = (a/b)2 + 1.

9.

10. (i) sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ, 2

using the formula for sin(A + B)

2

= 2 sin θ cos θ + (1 − 2 sin θ) sin θ, 2

using the formulae for sin 2θ and cos 2θ

2

= 2 sin θ(1 − sin θ) + (1 − 2 sin θ) sin θ = 3 sin θ − 4 sin3 θ . (ii) cos 3θ = cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ, 2

using the formula for cos(A + B) 2

= (2 cos θ − 1) cos θ − 2 sin θ cos θ, 2

using the formulae for sin 2θ and cos 2θ

2

= (2 cos θ − 1) cos θ − 2(1 − cos θ) cos θ = 4 cos3 θ − 3 cos θ .  π π + θ = sin cos θ + cos sin θ = 1 × cos θ + 0 × sin θ = cos θ . 2 2 2 π  π π (ii) cos + θ = cos cos θ − sin sin θ = 0 × cos θ − 1 × sin θ = − sin θ . 2 2 2 (iii) sin (π − θ) = sin π cos θ − cos π sin θ = 0 × cos θ − (−1) × sin θ = sin θ .   3π 3π 3π + θ = cos cos θ − sin sin θ = 0 × cos θ − (−1) × sin θ = sin θ . (iv) cos 2 2 2

11. (i) sin

π

12. (i) sin 4θ + sin θ = 2 sin 52 θ cos 32 θ. (ii) sin 8θ − sin 6θ = 2 cos 7θ sin θ. (iii) cos 12θ + cos 10θ = 2 cos 11θ cos θ. (iv) cos θ − cos 2θ = −2 sin 32 θ sin(− 12 θ) = 2 sin 32 θ sin 12 θ. 7 1 π cos 24 π. (v) sin 13 π + sin 14 π = 2 sin 24

13. (i) 2 sin 6θ cos 2θ = sin 8θ + sin 4θ. (ii) 2 cos 5θ cos 3θ = cos 8θ + cos 2θ. (iii) 2 cos 4θ sin θ = sin 5θ − sin 3θ. (iv) 2 sin 7θ sin 5θ = − cos 12θ + cos 2θ. 14. (i) θ = 1.0552 , 2.0864 (= π − 1.0552), (= 60.4586◦ , 119.5414◦). (ii) θ = .9472 , 4.0887 (= π + .9472), (= 54.2678◦ , 234.2678◦). (iii) θ = 1.3490 , 4.9342 (= 2π − 1.3490, (= 77.2910◦ , 282.7090◦). 15. (i) 2θ = −.9442 + 2nπ , 4.0858 + 2nπ. Hence θ = −.4721 + nπ , 2.0429 + nπ, (= −27.0480◦ + 180n◦ , 117.0480◦ + 180n◦ ). – 23 –

(ii) 3θ = 2.0829 + 2nπ , 4.2003◦ + 2nπ. Hence θ = .6943 + 23 nπ , 1.4001 + 23 nπ, (= 39.7802◦ + 120n◦ , 80.2198◦ + 120n◦). (iii)

1 2θ

= .5622 + nπ.

Hence θ = 1.1244 + 2nπ, (= 64.4219◦ + 360n◦). 16. (i) sin 2θ − sin 60◦ = 2 cos(θ + 30◦ ) sin(θ − 30◦ ) = 0. It follows that θ +30◦ = 90◦ +180n◦ so that θ = 60◦ +180n◦, or θ −30◦ = 180n◦ so that θ = 30◦ +180n◦. Hence the values of θ in the required range are θ = 30◦ , 60◦ , (= .5236 , 1.0472). (ii) cos 3θ − cos 120◦ = −2 sin( 32 θ + 60◦ ) sin( 32 θ − 60◦ ). It follows that 32 θ + 60◦ = 180n◦ so that θ = −40◦ + 120n◦, or 32 θ − 60◦ = 180n◦, so that θ = 40◦ + 120n◦. Hence the values of θ in the required range are θ = 40◦ , 160◦ , 80◦ , (= .6981 , 2.7925 , 1.3963◦).

(iii) tan θ = ± √13 .

Hence the values of θ in the required range are θ = 30◦ , 150◦ , (= 16 π , 56 π).

17. (i) The equation factorises to give (2 sin θ − 1)(sin θ + 1) = 0, so that either sin θ = Hence the only solutions in the required range are θ = 16 π ,

5 6 π,

1 2

or sin θ = −1.

(= 30◦ , 150◦ ).

(ii) The equation factorises to give (4 tan θ − 3)2 = 0, so that tan θ = 34 .

Hence there is just one solution in the range, namely θ = .6435, (= 36.8699◦).

(iii) Here, sin 2θ − sin θ = sin θ(2 cos θ − 1) = 0, and so either sin θ = 0 or cos θ = 12 . Hence the solutions in the required range are θ = 0 , π ,

1 3 π,

(= 0◦ , 180◦ , 60◦ ).

18. (i) cos θ − cos 4θ = −2 sin 52 θ sin 32 θ = 0. Thus 52 θ = nπ so that θ = 25 nπ, or 32 θ = nπ so that θ = 23 nπ. It follows that the solutions in the required range are θ = 0,

2 4 6 8 2 4 5π , 5π , 5π , 5π , 3π , 3π ,

2π, (= 0◦ , 72◦ , 144◦ , 216◦ , 288◦ , 120◦ , 240◦ , 360◦ ).

(ii) sin 3θ + sin θ = 2 sin 2θ cos θ = 0. It follows that 2θ = nπ so that θ = 12 nπ, or θ = 12 (2n + 1)π.

Hence the solutions in the required range are θ = 0 , 12 π , π , 32 π , 2π, (= 0 , 90◦ , 180◦ , 270◦ , 360◦ ). √ √ √ √ (iii) Dividing by 72 + 82 = 113 and writing cos α = 7/ 113, sin α = 8/ 113, so that α = .8520, we find sin(θ − α) = √

9 . 113

Hence θ − α = 1.0097 + 2nπ or θ − α = 2.1319 + 2nπ. It follows that θ = 1.8617 + 2nπ or θ = 2.9839 + 2nπ. The values in the required range are thus θ = 1.8617 , 2.9839, (= 106.6648◦ , 170.9668◦). √ (iv) Dividing by 52 + 122 = 13, and writing sin α = 5/13, cos α = 12/13, so that α = .3948, the equation becomes sin(θ + α) = 4/13 . Hence θ + α = .3128 + 2nπ or θ + α = 2.8288 + 2nπ and θ = −.0820 + 2nπ or θ = 2.4340 + 2nπ. It follows that the values in the required range are θ = 6.2012 , 2.4340, (= 355.3003◦ , 139.4599◦). – 24 –

Useful results (a) Commit the following to memory. (Some of the results can easily be worked out from graphs, the Formula Sheet, calculators etc. but it is often necessary to have the results at your fingertips.) π 180 180 . To convert from radians to degrees, multiply by π 2. Some useful values: 1. To convert from degrees to radians, multiply by

30◦ =

π , 6

45◦ =

π , 4

π , 3

60◦ =

π , 2

90◦ =

180◦ = π ,

360◦ = 2π .

3. In the right-angled triangle shown sine, cosine and tangent are defined as follows: sin θ =

b side opposite ≡ , hypotenuse c

cos θ =

adjacent side a ≡ , hypotenuse c

. ........ ...... .. ...... .... ..... . . . . ... . ..... ... ..... ...... ... ...... . . . .... . .... . . . .. . ... . . . . ... . ..... . . . ... . ... . . . . . ... ..... . . . ... . .... . . . . . ... . ..... .... . . . . . . . . . . . . . . . .. . .... ... . . . . . . . . ................................................................................................................

c

θ

4.

tan θ =

side opposite b ≡ . adjacent side a

b

a

cos2 θ + sin2 θ = 1 .

5. cos

π 2

 − θ = sin θ ,

sin

π 2

 − θ = cos θ .

6. cos(−θ) = cos θ ,

sin(−θ) = − sin θ ,

7. tan θ =

tan(−θ) = − tan θ .

sin θ . cos θ

8. The area of a general triangle ABC is given by Area of 4ABC =

1 1 × base × height = ab sin C 2 2

or, of course, by either of the other two similar expressions. 9. The values of sine, cosine and tangents of certain important angles between 0 and π/2 are: θ

0

π/6

sin θ

0

cos θ

1

tan θ

0

1/2 √ 3/2 √ 1/ 3

– 25 –

π/4 √ 1/ 2 √ 1/ 2 1

π/3 √ 3/2

π/2

1/2 √ 3

0

1 ∞

10. The sine and cosine functions have period 2π, the tangent function has period π: cos(θ + 2nπ) = cos(θ) ,

sin(θ + 2nπ) = sin θ ,

tan(θ + nπ) = tan θ ,

if n is an integer. 11. The other three trigonometric ratios are defined as follows: sec θ ≡ 12.

1 , cos θ

cosec θ ≡

1 , sin θ

cos 2θ = 2 cos2 θ − 1 ,

cot θ ≡

1 . tan θ

sin 2θ = 2 sin θ cos θ .

13. The diagram to help you remember the signs of the trigonometric functions of θ for all real values of θ is: π/2

π

......... ............ .... .................. ...... ...... ...... .... .... .... ... . . ... . ... ... ... ... . .. . . ... .. .... ..... .. ... .. ....................................................................................... ... ... .. . ... . .. . . ... ... .... ... .. ... . ... .. .... ... .... ..... .. ...... ..... ......... ... .............. .......................

S

A

T

0, 2π,...

C

3π/2

(b) The following appear on the Mathematics Formula Sheet but it is very useful to know them! 1. The ‘sine’ and ‘cosine’ formulae for a general triangle ABC are b c a = = , sin A sin B sin C a2 = b2 + c2 − 2bc cos A , with similar results giving b2 in terms of c, a, cos B and c2 in terms of a, b, cos C. 2. The graphs of the three elementary trigonometric functions between −2π and 2π are: ... ..... .......... .. .... .......... ...... ............ ......... ..... ...... .... ... ........ .... . .... . ... .. ... ... . .... . ... . ... ... . . . . . . . . ... ... . .. ... ... .... ... ... ... ... ... ... ... ... .. ... .. ... ... ... ... ... . . . . . . . . . . ........................................................................................................................................................................................................................................................................................................................................................................ . ... ... . .. ... ... .. ... ... ... ... ... ... .. . . . . ... . . . ... . .. ... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... .... ... . . . . . . . . . ..... . ..... ...................... ..................... ........ ... ... ..

+1

−2π −3π/2 −π −π/2

π/2

−1

Graph of cos θ

– 26 –

π

θ

3π/2

2π

. ....... .......... ... ........ ..... ......... ........ ............ .. ....... .......... .... .... .... .... . ... . ... ... ... ... ... ... ... ... . ... . . . . . . ... ... . . . .. . . . ... . . . . ... ... .. .... ... ... . . ... ... .. .... .. . . ... . .. . ... . . .. . . . . .................................................................................................................................................................................................................................................................................................................................................................... ... ... . . . ... . . ... ... .. ... . ... . ... ... .. . ... ... ... . . . ... . . .. ... ... ... .... ... ... ... ... .. ... ... ... ... ... ... ... .... .. .... ... . . . . . . . . . ..... . ..... . .. ..................... .................... ....... .. .... .

+1

−2π −3π/2 −π

−π/2

π/2

π

θ

3π/2

2π

−1

Graph of sin θ . . . . . . .. . .... .. . .. . ...... .. .. ... . .. .. .. . .. .. .......... .. .. ... .. .. .. .. .. . ... ... ..... .. ..... .. ... ... .. .... ... ..... . . . .. . .. .. . .. . .. . .. . .. . ..... ... .. .. ... .. ... .. . .. .. .. . .. .. ... . . .. . . ... . ... . . . . . .. .. .. ... .... ... .. .. .. ... .. .. .. ... ... ... ... ... ... ... .... .... ... ... ... . . . . .. . ... . ... . . ... .. . ....... .. ... .. ... .. .. . .. .. .. .. .. ... . ... . . . . . . .. .. .. .. .. .. . .. . . . . . . . . . ... ... ... .... .... .. .. . .. . .. .. .. ... . ... . . . . . .. . . . . . . . . . . .. . . . . . . . . . . . .. .. .. .. .... .... ... . .. .. .. . ... ... . . . . . . . . . . . . . . . . .................................................................................................................................................................................................................................................................................................................................................... . . . . .. . . . . . . . ... ... ... ...... .. .. .. . .. .. ... .. ... ... . . . . . . . ... .. ... ... ... ... ... ... .. ... ... ... ... .... .. . ... ... ... ... ... ... ... ... .. .. .. .. . .. . . . . . . . . . . ... . . .. ... ... ... ... .. ....... . . ... ... ... ... ... ... .. .. .. ... ... ... ... ... .. ... .. .. . . . . .... .... .. ... ... . ... ... ... ... ... .. .. ... ... .. ... ... ... ... ... ... .. .. ... .. .... ... .... .. .... .. .... . .. .. . . . . . ...... ... .... ... .... .. .... ... .... .. ... ... .. .. ... ... . .. ... ... ... .... .. ... ... .... .. .... ... ... .... ... ... . .. .. ... .... ... ... . . . . .. ... .

2 1

−2π −3π/2 −π

−π/2

π/2

π

3π/2

θ

2π

−1 −2

Graph of tan θ 3.

4.

5.

1 + tan2 θ = sec2 θ . 1 + cot2 θ = cosec2 θ . cos(A + B) = cos A cos B − sin A sin B , sin(A + B) = sin A cos B + cos A sin B . cos(A − B) = cos A cos B + sin A sin B , sin(A − B) = sin A cos B − cos A sin B .

(c) Also on the Mathematics Formula Sheet are: 1.

1 1 sin A + sin B = 2 sin (A + B) cos (A − B) , 2 2 1 1 sin A − sin B = 2 cos (A + B) sin (A − B) , 2 2 1 1 cos A + cos B = 2 cos (A + B) cos (A − B) , 2 2 1 1 cos A − cos B = −2 sin (A + B) sin (A − B) , 2 2

– 27 –