Maths in Focus - Margaret Grove - ch6

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TERMINOLOGY

6

Trigonometry

Angle of depression: The angle between the horizontal

and the line of sight when looking down to an object below

Angle of elevation: The angle between the horizontal and the line of sight when looking up to an object above Angles of any magnitude: Angles can be measured around a circle at the centre to fi nd the trigonometric ratios of angles of any size from 0c to 360c and beyond

Bearing: The direction relative to north. Bearings may be written as true bearings (clockwise from North) or as compass bearings (using N, S, E and W)

Complementary angles: Two or more angles that add up to 90c

Cosecant: The reciprocal ratio of sine (sin). It is the hypotenuse over the opposite side in a right triangle Cotangent: The reciprocal ratio of tangent (tan). It is the adjacent over the opposite side in a right triangle Secant: The reciprocal ratio of cosine (cos). It is the hypotenuse over the adjacent side in a right triangle Trigonometric identities: A statement that is true for all trigonometric values in the domain. Relationships between trigonometric ratios

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INTRODUCTION

TRIGONOMETRY IS USED IN many ﬁ elds, such as building, surveying and navigating. Wave theory also uses trigonometry.

This chapter revises basic right-angled triangle problems and applies them to real-life situations. Some properties of trigonometric ratios, angles greater than 90c and trigonometric equations are introduced. You will also study trigonometry in non-right-angled triangles.

Ptolemy (Claudius Ptolemaeus), in the second century, wrote He¯ mathe¯ matike¯ syntaxis (or

Almagest as it is now known) on astronomy. This is considered to be the fi rst treatise on trigonometry, but was based on circles and spheres rather than on triangles. The notation ‘chord of an angle’ was used rather than sin, cos or tan.

Ptolemy constructed a table of sines from 0c to 90c in steps of a quarter of a degree. He also calculated a value of r to 5 decimal places, and established the relationship for sin (X !Y) and cos (X !Y) .

DID YOU KNOW?

Trigonometric Ratios

In similar triangles, pairs of corresponding angles are equal and sides are in proportion. For example:

the

• hypotenuse is the longest side, and is always opposite

the right angle the

• opposite side is opposite the angle marked in the

triangle the

• adjacent side is next to the angle marked

In any triangle containing an angle of 30c, the ratio of AB AC: =1 2: . Similarly, the ratios of other corresponding sides will be equal. These ratios of sides form the basis of the trigonometric ratios.

In order to refer to these ratios, we name the sides in relation to the angle being studied:

You studied similar triangles in Geometry 1 in Chapter 4.

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The opposite and adjacent sides vary according to where the angle is marked. For example:

The trigonometric ratios are

You can learn these by their initials SOH , CAH , TOA .

What about S ome O ld H ags C an’t A lways H ide T heir O ld A ge?

DID YOU KNOW?

Trigonometry, or triangle measurement , progressed from the study of geometry in ancient Greece. Trigonometry was seen as applied mathematics. It gave a tool for the measurement of planets and their motion. It was also used extensively in navigation, surveying and mapping, and it is still used in these fi elds today.

Trigonometry was crucial in the setting up of an accurate calendar, since this involved measuring the distances between the Earth, sun and moon.

sin cos tan hypotenuse opposite hypotenuse adjacent adjacent opposite Sine Cosine Tangent i i i = = =

As well as these ratios, there are three inverse ratios,

cosec sin sec cos cot tan 1 1 1 opposite hypotenuse adjacent hypotenuse opposite adjacent Cosecant Secant Cotangent i i i _i i i = = = = = = f f f p p p

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EXAMPLES

1. Find sina, tana and seca.

Solution

sin tan sec _cos AB BC AC 5 3 4 5 3 4 3 1 4 5 hypotenuse opposite side adjacent side hypotenuse opposite adjacent opposite adjacent hypotenuse a a a _a = = = = = = = = = = = = = 2. If 7 2 ,

sini = ﬁ nd the exact ratios of cosi, tani and coti.

Solution

By Pythagoras’ theorem: 7 2 49 4 45 c a b a a a a 45 2 2 2 2 2 2 2 2 ` = + = + = + = = CONTINUED

To fi nd the other ratios you need to fi nd the adjacent side.

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cos tan cot tan 7 45 45 2 1 2 45 hypotenuse adjacent adjacent opposite i i i _i = = = = = =

Complementary angles

ABC, B , A 90 InD if+ =i then+ = c -i sin cos tan sec cosec cot c b c a a b a c bc b a i i i i i i = = = = = = (angle sum of a Δ) ( ) ( ) ( ) ( ) ( ) ( ) sin cos tan sec cosec cot c a c b b a bc ac a b 90 90 90 90 90 90 c c c c c c i i i i i i - = - = - = - = - = - =

From these ratios come the results.

( ) ( ) ( ) ( ) ( ) ° ° ° ° ° sin cos cos sin sec cosec cosec sec tan cot 90 90 90 90 90 i i i i i i i i i i = -= -= -= -=

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1. Write down the ratios of

,

cosi sini and tani.

2. Find sinb,cotb and secb.

3. Find the exact ratios of

,

sinb tanb and cosb.

4. Find exact values for cosx,tanx and cosecx.

EXAMPLES

1. Simplify tan50c-cot40 .c

Solution

tan cot

cot

tan cot tan tan

50 90 50 40 50 40 50 50 0 ` c c c c c c c c = -= - = -= ] g

2. Find the value of m if sec55c=cosec]2m-15gc.

Solution

90 55 35 sec cosec cosec m m m 55 2 15 35 2 50 25 ` c c c c = -= - = = = ] g

Check this by substituting m into the equation.

6.1 Exercises

Check this answer on your calculator.

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5. If tan , 3 4

i = ﬁ nd cosi and sini.

6. If

3 2_,

cosi = ﬁ nd exact values for

,

tani seci and sini.

7. If

6 1 ,

sini = ﬁ nd the exact ratios of cosi and tani.

8. If cosi =0.7, ﬁ nd exact values for tani and sini.

9. DABC is a right-angled isosceles triangle with +ABC=90c and

1.

AB=BC=

Find the exact length of

(a) AC .

Find (b) +BAC.

From the triangle, write down (c)

the exact ratios of sin45 ,c cos45c and tan 45c.

10.

Using Pythagoras’ theorem, (a)

ﬁ nd the exact length of AC . Write down the exact ratios of (b)

30 , 30

sin c cos c and tan30 .c

Write down the exact ratios of (c)

60 , 60

sin c cos c and tan60 .c

11. Show sin67c=cos23c.

12. Show sec82c=cosec8c.

13. Show tan48c=cot42c.

14. Simplify

(a) cos61c+sin29c

(b) seci-cosec]90c-ig

(c) tan70c+cot20c-2tan70c (d) 35 55 cos sin c c (e) 25 25 65 cot cot tan c c+ c

15. Find the value of x if

.

sin80c=cos]90-xgc

16. Find the value of y if

.

tan22c=cot^90-yhc

17. Find the value of p if

.

cos49c=sin^p+10hc

18. Find the value of b if

.

sin35c=cos]b+30gc

19. Find the value of t if

.

cot]2t+5gc=tan]3t-15gc

20. Find the value of k if

.

tan]15-kgc=cot]2k+60gc Hint: Change 0.7 to a fraction.

Trigonometric ratios and the calculator

Angles are usually given in degrees and minutes. In this section you will practise rounding off angles and ﬁ nding trigonometric ratios on the calculator.

Angles are usually given in degrees and minutes in this course. The calculator uses degrees, minutes and seconds, so you need to round off.

utes ree onds ute ( ) 60 1 (60 1 ) min deg sec min 60 =1 60 =1c = lm= l

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EXAMPLES

Round off to the nearest minute. 1. 23 12 22c l m

Solution

23 12 22c l m=23 12c l 2. 59 34 41c l m

Solution

59 34 41c l m=59 35c l 3. 16 54 30c l m

Solution

16 54 30c l m=16 55c l

_%, ,,

_KEY

This key changes decimal angles into degrees, minutes and seconds and vice versa.

Some calculators have deg or dms keys.

EXAMPLES

1. Change 58 19c l into a decimal.

Solution

58 %, ,,19 %, ,, %, ,,

Press =

So 58 19c l=58 31666667.

2. Change 45.236c into degrees and minutes.

Solution

45 236. %, ,,

Press = SHIFT

So 45 236. c=45 14c l

If your calculator does not give these answers, check the instructions for its use. Because 30 seconds is half a minute, we round up to the next minute.

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In order to use trigonometry in right-angled triangle problems, you need to ﬁ nd the ratios of angles on your calculator.

EXAMPLES

1. Find cos 58 19c l, correct to 3 decimal places.

Solution

58 %, ,,19 %, ,,

Press COS =

So cos 58 19c l=0 525.

2. Find sin 38 14c l, correct to 3 decimal places.

Solution

38 %, ,,14 %, ,,

Press SIN =

So sin 38 14c l=0 619.

3. If tani =0.348, ﬁ nd i in degrees and minutes.

Solution

This is the reverse of ﬁ nding trigonometric ratios.

To ﬁ nd the angle, given the ratio, use the inverse key _^tan-1_h.

TAN 0 348. %, ,,

Press SHIFT -1 ₌ SHIFT

. ( . ) tan tan 0 348 0 348 19 11 1 c i i = = = -l

4. Find i in degrees and minutes if cosi =0 675. .

Solution

0 675. %, ,,

Press SHIFT COS-1 ₌ SHIFT

. ( . ) cos cos 0 675 0 675 47 33 1 c i i== = -l

6.2 Exercises

1. Round off to the nearest degree. (a) 47 13 12° l m (b) 81 45 43° l m (c) 19 25 34° l m (d) 76 37 19° l m

2. Round off to the nearest minute. (a) 47 13 12° l m (b) 81 45 43° l m (c) 19 25 34° l m (d) 76 37 19° l m If your calculator

doesn't give this answer, check that it is in degree mode.

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3. Change to a decimal. (a) 77 45c l (b) 65 30c l (c) 24 51c l (d) 68 21c l (e) 82 31c l

4. Change into degrees and minutes. (a) 59.53c (b) 72.231c (c) 85.887c (d) 46.9c (e) 73.213c

5. Find correct to 3 decimal places. (a) sin39 25c l (b) cos 45 51c l (c) tan18 43c l (d) sin68 06c l (e) tan54 20c l

6. Find i in degrees and minutes if (a) sini =0 298. (b) tani =0 683. (c) cosi =0 827. (d) tani =1 056. (e) cosi =0 188.

Right-angled Triangle Problems

Trigonometry is used to ﬁ nd an unknown side or angle of a triangle.

Finding a side

We can use trigonometry to ﬁ nd a side of a right-angled triangle.

EXAMPLES

1. Find the value of x , correct to 1 decimal place.

Solution

° . ° . . ° . . . cos cos cos cos x x x x 23 49 11 8 23 49 11 8 11 8 23 49 10 8 1 11 8 11 8 hypotenuse adjacent cm to decimal point ` # # i = = = = = l l l ^ h CONTINUED

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2. Find the value of y , correct to 3 signiﬁ cant ﬁ gures.

Solution

c 15 c 15 15 15 c c c c 15 15 c . . . . . . sin sin sin sin sin sin sin sin y y y y y y y 41 15 9 7 41 9 7 41 9 7 41 _{9 7} 41 9 7 14 7 3 41 41 hypotenuse opposite m to significant figures # # i = = = = = = = l l l l l l l ^ h

6.3 Exercises

1. Find the values of all pronumerals, correct to 1 decimal place.

(a) (b) (c) (d)

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(e) (f) (g) (h) (i) (j) (k) x 5.4 cm 31

_c

12l (l) x 4.7 cm 37

c

22l (m) x 6.3 cm 72

c

18l (n) 23 mm 63

_c

14l x (o) 3.7 m 39

_c

47l y (p) 14.3 cm 46

c

5l k (q) 4.8 m74

c

29l h

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(r) 0.45 m 68

c

41l d (s) _{5.75 cm} 19

c

17l x (t) 17.3 m 6

c

3l b

2. A roof is pitched at 60c. A room built inside the roof space is to have a 2.7 m high ceiling. How far in from the side of the roof will the wall for the room go?

60

_c

2.7 m

x

3. A diagonal in a rectangle with breadth 6.2 cm makes an angle of 73c with the vertex as shown. Find the length of the rectangle correct to 1 decimal place.

73

_c

6.2 cm

4. Hamish is standing at an angle of

67c from a goalpost and 12.8 m away as shown. How far does he need to kick a football for it to reach the goal?

x

67

_c

12.8 m

5. Square ABCD with side 6 cm has line CD produced to E as shown so that +EAD=64 12c l . Evaluate the length, correct to 1 decimal place, of (a) CE (b) AE E 6 cm 64

_c

12l B A C D

6. A right-angled triangle with hypotenuse 14.5 cm long has one interior angle of 43 36c l. Find the lengths of the other two sides of the triangle.

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7. A right-angled triangle ABC with the right angle at A has

B 56 44c

+ = l and AB =26 mm. Find the length of the

hypotenuse.

8. A triangular fence is made for a garden inside a park. Three holes

A , B and C for fence posts are

made at the corners so that A and

B are 10.2 m apart, AB and CB are

perpendicular, and angle CAB is

59 54c l. How far apart are A and C ?

9. Triangle ABC has +BAC=46c

and +ABC=54c. An altitude is drawn from C to meet AB at point D . If the altitude is 5.3 cm long, ﬁ nd, correct to 1 decimal place, the length of sides

(a) AC (b) BC (c) AB

10. A rhombus has one diagonal 12 cm long and the diagonal makes an angle of 28 23c l with the side of the rhombus.

Find the length of the side of (a)

the rhombus.

Find the length of the other (b)

diagonal.

11. Kite ABCD has diagonal

15.8 BD = cm as shown. If +ABD=57 29 andc l 72 51 DBC c + = l, ﬁ nd the length of the other diagonal AC.

B A C D 72

c

51l 57

_c

29l 15.8 cm

Finding an angle

Trigonometry can also be used to ﬁ nd one of the angles in a right-angled triangle.

EXAMPLES

1. Find the value of i, in degrees and minutes.

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Solution

_.. 7.3 5.8 cos 7 3 5 8 hypotenuse adjacent 1 i = -cos 37 23 ` c i = = = l c m

2. Find the value of a, in degrees and minutes.

Solution

_.. . . tan tan 2 1 4 9 2 1 4 9 66 48 adjacent opposite 1 ` c a a = = = = -l c m

6.4 Exercises

1. Find the value of each pronumeral, in degrees and minutes.

(a)

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(c) (d) (e) (f) (g) (h) (i) (j) (k) 2.4 cm 3.8 cm a (l) 8.3 cm 5.7 cm i (m) 6.9 mm 11.3 mm i (n) 3 m 7 m i

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(o) 5.1 cm 11.6 cm b (p) 15 m 13 m a (q) 7.6 cm 4.4 cm i (r) 14.3 cm 8.4 cm a (s) 3 m 5 m i (t) 10.3 cm 18.9 cm c

2. A kite is ﬂ ying at an angle of i above the ground as shown. If the kite is 12.3 m above the ground and has 20 m of string, ﬁ nd angle i .

12.3 m 20 m

i

3. A ﬁ eld is 13.7 m wide and Andre is on one side. There is a gate on the opposite side and 5.6 m along from where Andre is. At what angle will he walk to get to the gate?

Gate Andre 5.6 m 13.7 m i

4. A 60 m long bridge has an opening in the middle and both sides open up to let boats pass underneath. The two parts of the bridge ﬂ oor rise up to a height of 18 m.

Through what angle do they move?

18 m i

60 m

5. An equilateral triangle ABC with side 7 cm has an altitude AD that is 4.5 cm long. Evaluate the angle the altitude makes with vertex A

DAB +

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6. Rectangle ABCD has dimensions 18 cm # 7 cm. A line AE is drawn so that E bisects DC .

How long is line

(a) AE ? (Answer

to 1 decimal place). Evaluate (b) +DEA .

7. A 52 m tall tower has wire stays on either side to minimise wind movement. One stay is 61.3 m long and the other is 74.5 m long as shown. Find the angles that the tower makes with each stay.

52 m

61.3 m 74.5 m

b a

8. (a) The angle from the ground up to the top of a pole is 41c when standing 15 m on one side of it. Find the height h of the pole, to the nearest metre.

If Seb stands 6 m away on the (b)

other side, ﬁ nd angle i .

41

_c

h

6 m 15 m

i

9. Rectangle ABCD has a line BE drawn so that +AEB=90c and

1

DE = cm. The width of the rectangle is 5 cm. 5 cm B A C E D 1 cm Find (a) +BEC .

Find the length of the (b)

rectangle.

10. A diagonal of a rhombus with side 9 cm makes an angle of 16c

with the side as shown. Find the lengths of the diagonals.

16

c

9 cm

11. (a) Kate is standing at the side of a road at point A , 15.9 m away from an intersection. She is at an angle of 39c from point B on the other side of the road. What is the width w of the road? (b) Kate walks 7.4 m to point

C . At what angle is she from

point B ? w B C A 7.4 m 15.9 m 39

c

i

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Applications

DID YOU KNOW?

The Leaning Tower of Pisa was built as a belfry for the cathedral nearby. Work started on the tower in 1174, but when it was only half completed the soil underneath one side of it subsided. This made the tower lean to one side. Work stopped, and it wasn’t until 100 years later that architects found a way of completing the tower. The third and fi fth storeys were built close to the vertical to compensate for the lean. Later a vertical top storey was added.

The tower is about 55 m tall and 16 m in diameter. It is tilted about 5 m from the vertical, and tilts by an extra 0.6 cm each year.

Class Investigation

Discuss some of the problems with the Leaning Tower of Pisa.

Find the angle at which it is tilted from the vertical. •

Work out how far it will be tilted in 10 years. •

Use research to ﬁ nd out if the tower will fall over, and if so, when. •

Angle of elevation

The angle of elevation is used to measure the height of tall objects that cannot be measured directly, for example a tree, cliff, tower or building.

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Class Exercise

Stand outside the school building and look up to the top of the building. Think about which angle your eyes pass through to look up to the top of the building.

The angle of elevation, i, is the angle measured when looking from the ground up to the top of the object. We assume that the ground is horizontal.

EXAMPLE

The angle of elevation of a tree from a point 50 m out from its base is

.

38 14c l Find the height of the tree, to the nearest metre.

Solution

We assume that the tree is vertical!

tan tan tan h h h h 38 14 50 38 14 50 50 38 14 39 50# 50# c c c Z = = = l l l

So the tree is 39 m tall, to the nearest metre.

A clinometer is used to measure the angle of elevation or depression.

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Angle of depression

The angle of depression is the angle formed when looking down from a high place to an object below.

Class Exercise

If your classroom is high enough, stand at the window and look down to something below the window. If the classroom is not high enough, ﬁ nd a hill or other high place. Through which angle do your eyes pass as you look down?

The angle of depression, i, is the angle measured when looking down from the horizontal to an object below.

EXAMPLES

1. The angle of depression from the top of a 20 m building to a boy below is 61 3c 9l. How far is the boy from the building, to 1 decimal place?

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39 39 c 39 39 39 39 39 c c c c c c c ( , ) 61 39 . 61 tan tan tan tan tan tan tan DAC ACB AD BC x x x x x x x 61 61 20 61 20 61 20 61 20 20 10 8 61 alternate angles # # + + < Z = = = = = = = l l l l l l l l

So the boy is 10.8 m from the building.

2. A bird sitting on top of an 8 m tall tree looks down at a possum 3.5 m out from the base of the tree. Find the angle of elevation to the nearest minute.

Solution

3.5 m 8 m A B C D i

The angle of depression is i

AB DC

BDC

Since horizontal lines

alternate angles + =<i ] ^ g h . . tan tan 3 5 8 3 5 8 66 1 ` c i i = = = -22l c m

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Bearings

Bearings can be described in different ways: For example, N70 Wc :

Start at north and measure 70o around towards the west.

True bearings measure angles clockwise from north

EXAMPLES

1. Sketch the diagram when M is on a bearing of 315c from P .

Solution

2. X is on a bearing of 030c from Y . Sketch this diagram.

Solution

3. A house is on a bearing of 305c from a school. What is the bearing of the school from the house?

Measure clockwise, starting at north.

All bearings have 3 digits so 30° becomes 030° for a bearing.

We could write 315o T for true bearings.

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Solution

The diagram below shows the bearing of the house from the school.

North

School House

305

c

To ﬁ nd the bearing of the school from the house, draw in North from the house and use geometry to ﬁ nd the bearing as follows:

S H N1 N2 N N 305

_c

The bearing of the school from the house is +N HS₂ .

360 305 180 55 ( ) N SH N HS N H N S 55 125 angle of revolution cointerior angles, 1 2 2 1 c c c c c c + + < = -= = -= ^ h

So the bearing of the school from the house is 125c .

4. A plane leaves Sydney and ﬂ ies 100 km due east, then

125 km due north. Find the bearing of the plane from Sydney, to the nearest degree.

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Solution

c . ( . ) 51 ( ) tan tan x x x 100 125 1 25 1 25 90 90 51 39

to the nearest degree

1 c c c c c i = = = = = -= -=

So the bearing of the plane from Sydney is 039°.

5. A ship sails on a bearing of 140° from Sydney for 250 km. How far east of Sydney is the ship now, to the nearest km?

Solution

cos cos cos x x x x 140 90 50 50 250 50 250 250 50 161 250# 250# c c c c c c Z i = -= = = =

So the ship is 161 km east of Sydney, to the nearest kilometre. A navigator on a ship

uses a sextant to

measure angles.

Could you use a different triangle for this question?

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6.5 Exercises

1. Draw a diagram to show the bearing in each question .

A boat is on a bearing of 100

(a) c

from a beach house.

Jamie is on a bearing of 320 (b) c from a campsite. A seagull is on a bearing of (c) 200c from a jetty. Alistair is on a bearing of (d)

050c from the bus stop. A plane is on a bearing of (e)

285c from Broken Hill .

A farmhouse is on a bearing (f)

of 012c from a dam.

Mohammed is on a bearing of (g)

160c from his house.

A mine shaft is on a bearing (h)

of 080c from a town.

Yvonne is on a bearing of (i)

349c from her school.

A boat ramp is on a bearing of (j)

280c from an island.

2. Find the bearing of X from Y in each question in 3 ﬁ gure (true) bearings . X Y North 112

_c

(a) X 35

_c

Y North South East West (b) X 10

c

Y North South East West (c) 23

c

X Y North South East West (d) X Y North South East West (e)

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3. Jack is on a bearing of 260c from Jill. What is Jill’s bearing from Jack?

4. A tower is on a bearing of 030c from a house. What is the bearing of the house from the tower?

5. Tamworth is on a bearing of 340c from Newcastle. What is the bearing of Newcastle from Tamworth?

6. The angle of elevation from a point 11.5 m away from the base of a tree up to the top of the tree is 42 12c l. Find the height of the tree to one decimal point.

7. Geoff stands 25.8 m away from the base of a tower and measures the angle of elevation as 39 20c l.

Find the height of the tower to the nearest metre.

8. A wire is suspended from the top of a 100 m tall bridge tower down to the bridge at an angle of elevation of 52c. How long is the wire, to 1 decimal place?

9. A cat crouches at the top of a 4.2 m high cliff and looks down at a mouse 1.3 m out from the foot (base) of the cliff. What is the angle of depression, to the nearest minute?

10. A plane leaves Melbourne and ﬂ ies on a bearing of 065c for 2500 km.

How far north of Melbourne (a)

is the plane?

How far east of Melbourne (b)

is it?

What is the bearing of (c)

Melbourne from the plane?

11. The angle of elevation of a tower is 39 44c l when measured at a point 100 m from its base. Find the height of the tower, to 1 decimal place.

12. Kim leaves his house and walks for 2 km on a bearing of 155c. How far south is Kim from his house now, to 1 decimal place?

13. The angle of depression from the top of an 8 m tree down to a rabbit is 43 52c l. If an eagle is perched in the top of the tree, how far does it need to ﬂ y to reach the rabbit, to the nearest metre?

14. A girl rides a motorbike through her property, starting at her house. If she rides south for 1.3 km, then rides west for 2.4 km, what is her bearing from the house, to the nearest degree?

15. A plane ﬂ ies north from Sydney for 560 km, then turns and ﬂ ies east for 390 km. What is its bearing from Sydney, to the nearest degree?

16. Find the height of a pole, correct to 1 decimal place, if a 10 m rope tied to it at the top and stretched out straight to reach the ground makes an angle of elevation of

. 67 13c l

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17. The angle of depression from the top of a cliff down to a boat 100 m out from the foot of the cliff is 59 42c l. How high is the cliff, to the nearest metre?

18. A group of students are

bushwalking. They walk north from their camp for 7.5 km, then walk west until their bearing from camp is 320c. How far are they from camp, to 1 decimal place?

19. A 20 m tall tower casts a shadow 15.8 m long at a certain time of day. What is the angle of elevation from the edge of the shadow up to the top of the tower at this time?

15.8 m 20 m

20. A ﬂ at verandah roof 1.8 m deep is 2.6 m up from the ground. At a certain time of day, the sun makes an angle of elevation of 72 25c l. How much shade is provided on the ground by the verandah roof at that time, to 1 decimal place?

21. Find the angle of elevation of a

.

15 9 m cliff from a point 100 m

out from its base.

22. A plane leaves Sydney and ﬂ ies for 2000 km on a bearing of 195c.

How far due south of Sydney is it?

23. The angle of depression from the top of a 15 m tree down to a pond is 25 41c l. If a bird is perched in the top of the tree, how far does it need to ﬂ y to reach the pond, to the nearest metre?

24. A girl starting at her house, walks south for 2.7 km then walks east for 1.6 km. What is her bearing from the house, to the nearest degree?

25. The angle of depression from the top of a tower down to a car 250 m out from the foot of the tower is 38 19c l. How high is the tower, to the nearest metre?

26. A hot air balloon ﬂ ies south for 3.6 km then turns and ﬂ ies east until it is on a bearing of 127c

from where it started. How far east does it ﬂ y?

27. A 24 m wire is attached to the top of a pole and runs down to the ground where the angle of elevation is 22 32c l. Find the height of the pole.

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28. A train depot has train tracks running north for 7.8 km where they meet another set of tracks going east for 5.8 km into a station. What is the bearing of the depot from the station, to the nearest degree?

29. Jessica leaves home and walks for 4.7 km on a bearing of 075c. She then turns and walks for 2.9 km on a bearing of 115c and she is then due east of her home.

How far north does Jessica (a)

walk?

How far is she from home? (b)

30. Builder Jo stands 4.5 m out from the foot of a building and looks up at to the top of the building where the angle of elevation is

71c. Builder Ben stands at the top of the building looking down at his wheelbarrow that is 10.8 m out from the foot of the building on the opposite side from where Jo is standing.

Find the height of the (a)

building.

Find the angle of depression (b)

from Ben down to his wheelbarrow.

Exact Ratios

A right-angled triangle with one angle of 45° is isosceles. The exact length of its hypotenuse can be found.

c a b AC AC 1 1 2 2 2 2 2 2 2 2 = + = + = =

This means that the trigonometric ratios of 45c can be written as exact ratios. Pythagoras’ theorem is used

to fi nd the length of the hypotenuse. sin cos 45 2 1 45 2 1 c c = =

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This angle is commonly used; for example, 45° is often used for the pitch of a roof. The triangle with angles of 60° and 30° can also be written with exact sides.

2 1 3 AD AD 3 2₌ 2_- 2 = =

Halve the equilateral triangle to get TABD.

60 60 60 ° ° ° sin cos tan 2 3 2 1 3 = = =

30 sin cos tan 2 1 30 2 3 30 3 1 c c c = = =

It may be easier to remember the triangle rather than all these ratios.

DID YOU KNOW?

The ratios of all multiples of these angles follow a pattern:

A 0c 30c 45c 60c 90c 120c 135c 150c sin A 2 0 2 1 2 2 2 3 2 4 2 3 2 2 2 1 cos A 2 4 2 3 2 2 2 1 2 0 2 1 2 2 2 3 The rules of the pattern are:

for sin

• A , when you reach 4, reverse the numbers for cos

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EXAMPLES

1. Find the exact value of sec 45°.

Solution

° ° sec cos 45 45 1 2 1 1 2 = = =

2. A boat ramp is to be made with an angle of 30c and base length 5 m. What is the exact length of the surface of the ramp?

Solution

cos cos cos x x x 30 5 30 5 30 5 2 3 5 5 3 2 3 10 3 10 3 # c c c = = = = = = =

So the exact length of the ramp is . 3 10 3

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6.6 Exercises

Find the exact value in all questions, with rational denominator where relevant.

1. Evaluate

(a) sin60c+cos60c (b) cos245_c₊sin245_c (c) cosec 45c (d) 2sec60c

(e) cot30c+cot60c

(f) tan60c-tan30c

(g) sin260_c₊sin245_c

(h) sin45ccos30c+cos45csin30c (i) 3tan30c (j) tan tan tan tan 1 45 60 45 60 c c c c - +

(k) cos30ccos60c-sin30csin60c (l) cos230_c₊sin230_c sec cosec 2 45c- 30c (m) (n) sin sin 45 2 60 cc (o) 1₊tan230_c (p) cos cos 1 45 1 45 c c + - (q) sec cot 60 30 cc (r) sin 452 _{c -}1 (s) 5cosec260_c (t) sec tan 45 2 60 2 _c c -

2. Find the exact value of all pronumerals

(a)

(b)

(c)

3. A 2.4 m ladder reaches 1.2 m up a wall. At what angle is it resting against the wall?

4. A 2-person tent is pitched at an angle of 45c. Each side of the tent is 2 m long. A pole of what height is needed for the centre of the tent?

5. If the tent in the previous question was pitched at an angle of 60c, how high would the pole need to be?

6. The angle of elevation from a point 10 m out from the base of a tower to the top of the tower is 30c. Find the exact height of the tower, with rational denominator.

cos 452 (cos 45)2

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7. The pitch of a roof is 45c and spans a length of 12m.

What is the length

(a) l of the

roof?

If a wall is placed inside the (b)

roof one third of the way along from the corner, what height will the wall be?

8. A 1.8 m ladder is placed so that it makes a 60c angle where it meets

the ﬂ oor. How far out from the wall is it?

9. Find the exact length of AC .

10. The angle of depression from the top of a 100 m cliff down to a boat at the foot of the cliff is 30 .c How far out from the cliff is the boat?

Angles of Any Magnitude

The angles in a right-angled triangle are always acute. However, angles greater than 90c are used in many situations, such as in bearings. Negative angles are also used in areas such as engineering and science.

We can use a circle to ﬁ nd trigonometric ratios of angles of any magnitude (size) up to and beyond 360c.

Investigation

(a) Copy and complete the table for these acute angles 1. (between 0c and 90c). x 0c 10c 20c 30c 40c 50c 60c 70c 80c 90c sin x cos x tan x

(b) Copy and complete the table for these obtuse angles (between 90c and 180c).

x 100c 110c 120c 130c 140c 150c 160c 170c 180c

sin x cos x

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(c) Copy and complete the table for these reﬂ ex angles (between 180c and 270c). x 190c 200c 210c 220c 230c 240c 250c 260c 270c sin x cos x tan x

(d) Copy and complete the table for these reﬂ ex angles (between 270c and 360c).

x 280c 290c 300c 310c 320c 330c 340c 350c 360c

sin x cos x tan x

What do you notice about their signs? Can you see any patterns? 2.

Could you write down any rules for the sign of sin, cos and tan for different angle sizes?

Draw the graphs of

3. y=sinx, y=cosx and y=tanx for

. x

0c# #360c For y=tanx , you may need to ﬁ nd the ratios of angle close to and either side of 90c and 270c.

Drawing the graphs of the trigonometric ratios can help us to see the change in signs as angles increase.

We divide the domain 0c to 360c into 4 quadrants:

1 st quadrant: 0c to 90c 2 nd quadrant: 90c to 180c 3 rd quadrant: 180c to 270c 4 th quadrant: 270c to 360c

EXAMPLES

1. Describe the sign of sin x in each section (quadrant) of the graph y=sinx.

Solution

We can sketch the graph using the table below or using the values from the tables in the investigation above for more accuracy.

x 0c 90c 180c 270c 360c

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y 90

_c

180

c

270

c

360

c

1 -1 y=sin x x

The graph is above the x -axis for the ﬁ rst 2 quadrants, then below for the 3 rd and 4 th quadrants.

This means that sin x is positive in the 1 st and 2 nd quadrants and negative in the 3 rd and 4 th quadrants.

2. Describe the sign of cos x in each section (quadrant) of the graph of

. cos

y= x

Solution

x 0c 90c 180c 270c 360c y 1 0 _-1 ₀ 1 y 90

c

180

c

270

c

360

c

1 -1 y=cos x x

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This means that cos x is positive in the 1 st and 4 th quadrants and negative in the 2 nd and 3 rd quadrants.

3. Describe the sign of tan x in each section (quadrant) of the graph y=tanx .

Solution

x 0c 90c 180c 270c 360c

y 0 No result 0 No result 0

Neither tan 90c nor tan 270c exists (we say that they are undeﬁ ned). Find the tan of angles close to these angles, for example tan 89c59l and tan 90c01l, tan 279c59l and tan 270c l01.

There are asymptotes at 90c and 270c. On the left of 90c and 270c, tan x is positive and on the right, the ratio is negative.

y

x

90

c

180

c

270

c

360

c

y = tanx

The graph is above the x -axis in the 1 st quadrant, below for the 2 nd , above for the 3 rd and below for the 4 th quadrant.

This means that tan x is positive in the 1 st and 3 rd quadrants and negative in the 2 nd and 4 th quadrants.

You will see why these ratios are undefi ned later on in this chapter.

To show why these ratios have different signs in different quadrants, we look at angles around a unit circle (a circle with radius 1 unit).

We use congruent triangles when ﬁ nding angles of any magnitude. Page 310 shows an example of congruent triangles all with angles of 20c inside a circle with radius 1 unit.

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y x 1 unit 1 unit 20

c

20

c

20

_c

20

c

1 unit 1 unit

If we divide the circle into 4 quadrants, we notice that the x - and y -values have different signs in different quadrants. This is crucial to notice when looking at angles of any magnitude and explains the different signs you get when ﬁ nding sin, cos and tan for angles greater than 90c.

Quadrant 1

Looking at the ﬁ rst quadrant (see diagram below), notice that x and y are both positive and that angle i is turning anticlockwise from the x -axis.

(x, y) 1 unit First quadrant y x y x i

Point ( x , y ) forms a triangle with sides 1, x and y , so we can ﬁ nd the trigonometric ratios for angle i .

The angle at the x -axis is 0 and the angle at the y -axis is 90c, with all other angles in this quadrant between these two angles .

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Investigation

Since cos i =x and sin i =y , we can write the point ( x , y ) as (cos i , sin i ) . The polar coordinates (cos i , sin i ) give a circle.

The polar coordinates 6Asin]ai+c Bg, sin]big@ form a shape called a Lissajous ﬁ gure. These are sometimes called a Bowditch curve and they are often used as logos, for example the ABC logo.

Use the Internet to research these and other similar shapes.

Use a graphics calculator or a computer program such as Autograph to draw other graphs with polar coordinates using variations of sin i and cos i .

These are called polar coordinates.

Quadrant 2

In the second quadrant, angles are between 90c and 180c.

If we take the 1 st quadrant coordinates ( x , y ), where x20 and y 20 and put them in the 2 nd quadrant, we notice that all x values are negative in the second quadrant and y values are positive.

So the point in the 2 nd quadrant will be (- x , y )

x y 0

c

90

c

180

c

(-x, y) 1 unit Second quadrant y x 180

_c

- i i sin y y1 i = = cos x x1 i = = tani = _xy

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Since cos i =x , cos i will negative in the 2 nd quadrant. Since sin i =y , sin i will be positive in the 2 nd quadrant.

tan i =_xy so it will be negative (a positive number divided by a negative number).

To have an angle of i in the triangle, the angle around the circle is 180c- i .

Quadrant 3

In the third quadrant, angles are between 180c and 270c.

90

c

270

c

x y 0

c

180

c

(-x, -y) 1 unit Third quadrant i y x 180

_c

+i

Notice that x and y are both negative in the third quadrant, so cos i and sin i will be both negative.

tani = _xy so will be positive (a negative divided by a negative number). To have an angle of i in the triangle, the angle around the circle is 180c+ i .

Quadrant 4

In the fourth quadrant, angles are between 270c and 360c.

90

_c

x y 0

c

180

_c

1 unit y x 360

c

- ii 360

c

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While y remains negative in the fourth quadrant, x is positive again, so sin i is negative and cos i is positive.

tani = _xy so will be negative (a negative divided by a positive number) For an angle i in the triangle, the angle around the circle is 360c- i .

ASTC rule

Putting all of these results together gives a rule for all four quadrants that we usually call the ASTC rule.

4th quadrant 1st quadrant 3rd quadrant 2nd quadrant S A T C 180

c

+ i 360

c

- i 360

c

180

c

- i i 90

_c

270

c

0

c

180

_c

y x

You could remember this rule as A ll S tations T o C entral or A S illy T rigonometry C oncept, or

you could make up your own!

This rule also works for the reciprocal trigonometric ratios. For example, where cos is positive, sec is also positive, where sin is positive, so is cosec and where tan is positive, so is cot.

We can summarise the ASTC rules for all 4 quadrants: A: ALL ratios are positive in the 1 st quadrant

S: Sin is positive in the 2 nd quadrant (cos and tan are negative) T: Tan is positive in the 3 rd quadrant (sin and cos are negative) C: Cos is positive in the 4 th quadrant (sin and tan are negative)

First quadrant: Angle i : sin i is positive cos i is positive tan i is positive

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Second quadrant: Angle 180c i- : sin]180c-ig=sini cos]180c-ig= -cosi tan]180c-ig= -tani Third quadrant: Angle 180c i+ : sin]180c+ig= -sini cos]180c+ig= -cosi tan]180c+ig=tani Fourth quadrant: Angle 360c i- : sin]360c-ig= -sini cos]360c-ig=cosi tan]360c-ig= -tani

EXAMPLES

1. Find all quadrants where (a) sini20

(b) cosi10

(c) tani10andcosi20

Solution

(a) sini20 means sin i is positive.

Using the ASTC rule, sin i is positive in the 1 st and 2 nd quadrants. cos

(b) i is positive in the 1 st and 4 th quadrants, so cos i is negative in the 2 nd and 3 rd quadrants.

tan

(c) i is positive in the 1 st and 3 rd quadrants so tan i is negative in the 2 nd and 4 th quadrants. Also cos i is positive in the 1 st and 4 th quadrants.

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2. Find the exact ratio of tan 330c .

Solution

First we ﬁ nd the quadrant that 330c is in. It is in the 4 th quadrant.

The angle inside the triangle in the 4 th quadrant is 30c and tan is negative in the 4 th quadrant.

tan330 tan30 3 1 c= - c = -

3. Find the exact value of sin 225c .

Solution

The angle in the triangle in the 3 rd quadrant is 45c and sin is negative in the 3 rd quadrant. CONTINUED Notice that 360c-30c=3 0 .3 c Notice that 1 08c+45c=225c. 330

_c

₃₀

_c

y x 60

c

30

_c

2 1 :3

(43)

225

c

45

c

y x sin225 sin45 2 1 c= - c = -

4. Find the exact value of cos 510c .

Solution

To ﬁ nd cos 510c, we move around the circle more than once.

510

c

150

_c

30

c

y x 510 360 150 510 360 150 So c c c c c c - = = + 45

_c

45

c

1 1 :2

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The angle is in the 2 nd quadrant where cos is negative. The triangle has 30c in it.

cos510 cos30 2 3 c= - c = - 5. Simplify cos (180c+ x ) .

Solution

180c+ x is an angle in the 3 rd quadrant where cos is negative. So cos]180c +xg= -cosx

6. If sin

5 3

x = - and cos x 2 0, ﬁ nd the value of tan x and sec x .

Solution

sin x 1 0 in the 3 rd and 4 th quadrants and cos x 2 0 in the 1 st and 4 th quadrants.

So sin x 1 0 and cos x 2 0 in the 4 th quadrant. This means that tan x 1 0 and sec x 2 0.

sinx

hypotenuse opposite

=

So the opposite side is 3 and the hypotenuse is 5.

3 5

y

x x

By Pythagoras’ theorem, the adjacent side is 4.

sec x is the reciprocal of cos x so is positive in the 4 th quadrant . This is a 3-4-5 triangle . Notice that 180c-30c=150c. CONTINUED 60

c

30

c

2 1 :3

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So tanx 4 3 = - secx _cos1_x 4 5 = =

The ASTC rule also works for negative angles. These are measured in the opposite way (clockwise) from positive angles as shown.

4th quadrant 1st quadrant 3rd quadrant 2nd quadrant S A T C -i 0 -(180

_c

+ i ) -(180

c

- i ) -(360

c

- i ) -180

c

y x -90

_c

-270

c

-360

_c

The only difference with this rule is that the angles are labelled differently.

EXAMPLE

Find the exact value of tan (-120c)

.

Solution

Moving around the circle the opposite way, the angle is in the 3 rd quadrant, with 60c in the triangle.

y x 120

c

60

c

Notice that 180 0 1 0 . ( c 6c)= 2 c - -

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Tan is positive in the 3 rd quadrant. tan 120 tan60 3 c c - = = ] g

6.7 Exercises

1. Find all quadrants where (a) cosi20 (b) tani20 (c) sini20 (d) tani10 (e) sini10 (f) cosi10

(g) sini10 and tani20

(h) cosi10 and tani20

(i) sini20 and tani10

(j) sini10 and tani10

2. (a) Which quadrant is the angle

240c in?

Find the exact value of cos

(b) 240c .

315c in?

Find the exact value of sin

(b) 315c .

120c in?

Find the exact value of (b)

tan 120c .

5. (a) Which quadrant is the angle -225c in?

sin (-225c) .

6. (a) Which quadrant is the angle -330c in?

cos (-330c) .

7. Find the exact value of each ratio. tan 225 (a) c cos 315 (b) c tan 300 (c) c sin 150 (d) c cos 120 (e) c sin 210 (f) c cos 330 (g) c tan 150 (h) c sin 300 (i) c cos 135 (j) c

8. Find the exact value of each ratio. cos ( (a) -225c) cos ( (b) -210c) tan ( (c) -300c) cos ( (d) -150c) sin ( (e) -60c) tan ( (f) -240c) cos ( (g) -300c) tan ( (h) -30c) cos ( (i) -45c) sin ( (j) -135c) 60

c

30

_c

2 1 :3

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Trigonometric Equations

Whenever you ﬁ nd an unknown angle in a triangle, you solve a trigonometric equation e.g. cosx=0 34. . You can ﬁ nd this on your calculator.

Now that we know how to ﬁ nd the trigonometric ratios of angles of any magnitude, there can be more than one solution to a trigonometric equation if we look at a larger domain.

9. Find the exact value of cos 570 (a) c tan 420 (b) c sin 480 (c) c cos 660 (d) c sin 690 (e) c tan 600 (f) c sin 495 (g) c cos 405 (h) c tan 675 (i) c sin 390 (j) c 10. If tan 4 3 i = and cos i10 , ﬁ nd sin i and cos i as fractions.

11. Given sin 7 4

i = and tan i10 , ﬁ nd the exact value of cos i and tan i .

12. If sin x 1 0 and tanx 8 5 = - , ﬁ nd the exact value of cos x and cosec x.

13. Given cos

5 2

x = and tanx10, ﬁ nd the exact value of cosec x , cot x and tan x .

14. If cos x 1 0 and sin x 1 0, ﬁ nd cos x and sin x in surd form with rational denominator if tan x

7 5 = . 15. If sin 9 4 i = - and 270c1 1i 360c , ﬁ nd the exact value of tan i and sec i .

16. If cos 8 3 i = - and ° ° 180 1 1i 270 , ﬁ nd the exact value of tan x , sec x and cosec x .

17. Given sin x =0.3 and tan x 1 0, express sin

(a) x as a fraction

ﬁ nd the exact value of cos

(b) x and tan x . 18. If tan a = -1.2 and ° ° 270 1 1i 360 , ﬁ nd the exact values of cot a , sec a and cosec a .

19. Given that cos 0.7i = - and

90c1 1i 180c , ﬁ nd the exact value of sin i and cot i .

20. Simplify (a) sin 180] c-ig (b) cos]360c -xg (c) tan 180^ c+bh (d) sin 180] c+ag (e) tan 360] c-ig (f) sin]-ig (g) cos]-ag (h) tan]-xg Use Pythagoras’ theorem to

fi nd the third side .

This is called the principle solution.

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EXAMPLES

1. Solve cosx 2 3 = in the domain 0°# #x 360° .

Solution

2 3

is a positive ratio and cos is positive in the 1 st and 4 th quadrants . So there are two possible answers.

In the 1 st quadrant, angles are in the form of i and in the 4 th quadrant angles are in the form of 360c- i .

cos 30 2

3

c =

But there is also a solution in the 4 th quadrant where the angle is 360c- i . cosx 2 3 For = , , x 30 360 30 30cc 330cc c = -=

2. Solve 2sin2x_-1₌0_for₀_c_{# #}_x ₃₆₀_c_.

Solution

sin sin sin sin x x x x 2 1 0 2 1 2 1 2 1 2 1 2 2 2 ! ! - = = = = =

Since the ratio could be positive or negative, there are solutions in all 4 quadrants. 1 st quadrant: angle i 2 nd quadrant: angle 180c- i 3 rd quadrant: angle 180c+ i 4 th quadrant: angle 360c- i 60

_c

30

c

2 1 :3

This is called the principle solution.

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, , , , , , sin x 45 2 1 45 180 45 180 45 360 45 45 135 225 315 c c c c c c c c c c c c = = - + -=

3. Solve tanx= 3 for -180c# #x 180c .

Solution

3 is a positive ratio and tan is positive in the 1 st and 3 rd quadrants . So there are two possible answers.

In the domain -180c# #x 180c , we use positive angles for

x

0c# #180c and negative angles for -180c# #x 0c.

In the 1 st quadrant, angles are in the form of i and in the 3 rd quadrant angles are in the form of -^180c-ih .

tan 60c = 3

But there is also a solution in the 3 rd quadrant where the angle is

180c i -^ - h . 120c , , tanx x 3 60 180 60 30 For c c c c = = - -= -] g 45

c

45

c

1 1 :2 60

c

30

_c

2 1 :3 4th quadrant 1st quadrant 3rd quadrant 2nd quadrant S A T C -(180

c

- i) -i 180

c

- i 90

c

-90

c

0

c

0

_c

180

c

-180

_c

y x i

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4. Solve 2sin2x- =1 0 for 0c# #x 360c .

Solution

Notice that the angle is 2 x but the domain is for x .

If 0c# #x 360c then we multiply each part by 2 to get the domain for 2 x . 0c#2x#720c

This means that we can ﬁ nd the solutions by going around the circle twice!

sin sin sin sin x x x 2 2 1 0 2 2 1 2 2 1 30 2 1 c - = = = =

Sin is positive in the 1 st and 2 nd quadrants.

First time around the circle, 1 st quadrant is i and the 2 nd quadrant is 180c i- . Second time around the circle, we add 360c to the angles.

So 1 st quadrant answer is 360c i+ and the 2 nd quadrant answer is

360c+]180c-ig or 540c-i . , , , , , , , , , x x 2 30 180 30 360 30 540 30 30 150 390 510 15 75 195 255 So c c c c c c c c c c c c c c c = - + -= = `

The trigonometric graphs can also help solve some trigonometric equations.

EXAMPLE

Solve cosx=0 for 0c# #x 360c . cos 90c =0

However, looking at the graph of y=cosx shows that there is another solution in the domain 0c# #x 360c

.

0 90 , 270 cosx x For c c = =

y 90

c

180

c

270

c

360

c

1 -1 x

Notice that these solutions lie inside the original domain of

. 0c# #x 360c 60

c

30

c

2 1 :3

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Investigation

Here are the 3 trigonometric graphs that you explored earlier in the chapter. sin y= x

cos y= x

tan y= x

Use the values in the sin, cos and tan graphs to ﬁ nd values for the inverse trigonometric functions in the tables below and then sketch the inverse trigonometric functions.

For example sin 270°= -1

cosec 270 1 1 1 So c = -= -

Some values will be undeﬁ ned, so you will need to ﬁ nd values near them in order to see where the graph goes.

y=cosecx

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Here are the graphs of the inverse trigonometric functions. cosec y= x

sec y= x

cot y= x

sec y= x x 0c 90c 180c 270c 360c cos x sec x cot y= x x 0c 90c 180c 270c 360c tan x cot x y y = cotx x 90

c

180

c

270

c

360

c

360

c

-1 1 0

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1. Solve for 0c# #i 360c. (a) sini =0 35. (b) cos 2 1 i = - (c) tani = -1 (d) sin 2 3 i = (e) tan 3 1 i = - (f) 2cosi = 3 (g) tan 2i = 3 (h) 2sin3i = -1 (i) 2cos2i - =1 0 (j) tan 32 i =1 2. Solve for -180c# #i 180c. (a) cosi =0 187. (b) sin 2 1 i = (c) tani =1 (d) sin 2 3 i = - (e) tan 3 1 i = - (f) 3tan2_{i =}1 (g) tan 2i =1 (h) 2sin23_{i =}1 (i) tani + =1 0 (j) tan 22 _{i =}3

3. Sketch y=cosx for

0c#x#360 .c

4. Evaluate sin 270c.

5. Sketch y=tanx for

0c# #x 360 .c

6. Solve tanx =0 for

0c# #x 360 .c

7. Evaluate cos 180c.

8. Find the value of sin 90c.

9. Solve cosx=1 for

. x

0c# #360c

10. Sketch y=sinx for

. x 180c# #180c - 11. Evaluate cos 270c.

12. Solve sinx 1+ =0 for

. x

0c# #360c

13. Solve cos2x₌1_for . x

0c# #360c

14. Solve sinx=0 for

. x

0c# #360c

15. Solve sinx=1 for

. x

360c# #360c

-

16. Sketch y=secx for

. x

0c# #360c

17. Sketch y=cotx for

. x

0c# #360c

6.8 Exercises

Trigonometric Identities

Trigonometric identities are statements about the relationships of

trigonometric ratios. You have already met some of these—the reciprocal ratios, complementary angles and the rules for the angle of any magnitude.

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cosec sin sec cos cot tan 1 1 1 i _i i _i i i = = =

Reciprocal ratios

Complementary angles

sin cos cosec sec tan cot 90 90 90 c c c i i i i i i = -= -= -] ] ] g g g

Angles of any magnitude

sin sin cos cos tan tan 180 180 180 c c c i i i i i i - = - = -- = -] ] ] g g g ( ) ( ) ( ) sin sin cos cos tan tan 180 180 180 c c c i i i i i i + = -+ = -+ = ( ) ( ) ( ) sin sin cos cos tan tan 360 360 360 c c c i i i i i i - = -- = - = ) ) cos tan i i i i = = -) sin i = - i ( ( ( sin cos tan

In this section you will learn some other identities, based on the unit circle. In the work on angles of any magnitude, we deﬁ ned

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tan cos sin xy i i i = = tan cos sin i= _ii cot tan sin cos 1 i i i i = = cot sin cos i= _ii

Pythagorean identities

The circle has equation x2₊y2₌1.

Substituting x=cosi and y=sini into x2₊y2₌1_gives

cos2_i₊sin2_i₌1

This is an equation so can be rearranged to give sin cos cos sin 1 1 2 2 2i_i =₌ -_- 2i_i

There are two other identities that can be derived from this identity.

tan sec

1₊ 2_i₌ 2_i

Remeber that cos2_i means (cos )_{i .}2

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Proof

cos sin cos cos cos sin cos tan sec 1 1 1 2 2 2 2 2 2 2 2 2 i i i i i i i i i + = + = + =

This identity can be rearranged to give tan sec sec tan 1 1 2 2 2 2 i i i i = = -- cot2_i_{+ =}1 cosec2_i

Proof

cos sin sin cos sin sin sin cot cosec 1 1 1 2 2 2 2 2 2 2 2 2 i i i i i i i i i + = + = + =

This identity can be rearranged to give cot cosec cosec cot 1 1 2 2 2 2 i i i i = -= -

These are called Pythagorean identities since the equation of the circle comes from Pythagoras’ rule (see Chapter 5).

EXAMPLES

1. Simplify sinicoti.

Solution

sin cot sin

sin cos cos # i i i _ii i = =

2. Simplify sin^90c-bhsecb where b is an acute angle .

Solution

1

sin sec cos

cos 90c b b b# 1 b - = = ^ h CONTINUED

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3. Simplify sin4_i₊sin2_icos2_i.

Solution

sin sin cos sin sin cos

sin sin sin 1 4 2 2 2 2 2 2 2 i i i i i i i i i + = + = = = ^ ] h g

4. Prove cotx+tanx=cosecxsecx.

Solution

cot tan sin cos cos sin sin cos cos sin sin cos sin cos cosec sec x x xx xx x x x x x x x x x x 1 1 1 LHS RHS 2 2 # = + = + = + = = = =

cotx+tanx=cosecxsecx

` 5. Prove that . sin cos cos x x x 1 1 1 2 - ₌ +

Solution

sin cos cos cos cos cos cos cos x x x x x x x x 1 1 1 1 1 1 1 1 LHS RHS 2 2 = -= -= + -= ₊ = ] g] g 1 1 sin cos cos x x x 1 2 ` - = ₊