Trigonometric Graphs

y

x

60

60

120

120

180

180

240

240

300

300

– 60

– 60

0.5

0.5

1

1

1.5

1.5

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5 siny x

Period

Amplitude

The horizontal extent of the basic pattern is called the period.

Half of the vertical extent is called the amplitude.

Period = 3600

Amplitude = 1

y

x

60

60

120

120

180

180

240

240

300

300

– 60

– 60

0.5

0.5

1

1

1.5

1.5

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5 cosy x

Period

Amplitude

Period = 3600

Amplitude = 1

y

x

60

60

120

120

180

180

240

240

300

300

– 60

– 60

5

5

10

10

15

15

20

20

– 5

– 5

– 10

– 10

– 15

– 15

– 20

– 20

tany x

Period

Period = 1800

Amplitude = undefined

The dotted vertical lines are known as asymptotes. The graph approaches but never touches them.

1. Find the period and amplitude of the function 3sin 4 2y x

1 – 1 – 2 – 3 – 4 – 5

y

x

90

90

180

180

270

270

360

360

1

1

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

The graph repeats 4 times over 3600. 360

Period = 4

090

Vertical extent of graph (height) = 6 units. Amplitude = 3.

0 0For sin and cosy a bx y a bx

Amplitude = a0360

Period = b

0For tany a bx

Amplitude = can not be measured0180

Period = b

Sketching Trigonometric Graphs

The strategy for sketching trig graphs of the form y = a sin bx c or y = a cos bx c is :

1. Start with a simple sin or cosy bx y bx

2. Put in the scale to match the amplitude a.

3. Slide the graph vertically to match the constant c.

01. Sketch the graph of 2sin3 1 for 0 360y x x

Step 1. y

x

90

90

180

180

270

270

360

360

sin3y xStep 2.

01. Sketch the graph of 2sin3 1 for 0 360y x x

Step 1.

1 2 3 – 1 – 2 – 3

y

x

90

90

180

180

270

270

360

360

1

1

2

2

3

3

– 1

– 1

– 2

– 2

– 3

– 3

2sin3y xStep 2.

Step 3.

01. Sketch the graph of 2sin3 1 for 0 360y x x

Step 1.

1 2 3 – 1 – 2 – 3

y

x

90

90

180

180

270

270

360

360

1

1

2

2

3

3

– 1

– 1

– 2

– 2

– 3

– 3

2sin3 1y x Step 2.

Step 3.

Radians

Degrees are not the only units used to measure angles.

It is often, and usually, useful to measure angles in radians.

The angle subtended at the centre of a circle by an arc equal in length to the radius is 1 radian.

r

r

r1 radian

r

r

r1 radian

Circumference = 2 r

Hence the radius will fit the circumference 2 times.

So there are 2 radians in a complete turn.

0360 2 radians0180 radians

1. Convert 1200 into radians.

0180

180 120x

120

180x

2radians.

3

0120 x(cross multiply starting with the ‘x’ term)

2

3

This is usually written as 2

radians.3

52. Convert radians to degrees.

9

0180

180 5

9x

180 5

9x

20 5

1

0 5

9x

(cross multiply starting with the ‘x’ term)

20

1

0100

Special angles and triangles

It is useful and necessary to know exact values of common, and some not so common, trigonometric ratios.

If you remember only 2 of these you can work the others out easily.

This saves trying to remember them all. Although by the end of the course you will probably have remembered them all as we use them quite a lot in higher maths.

The two ratios that you MUST remember are:

0 1sin30

2 0tan 45 1

(remember SOH)

030

12

3

060

(remember TOA)

045

1045

1

2

This gives us exact trig ratios from all 4 quadrants.

The two ratios that you MUST remember are:

1sin

6 2

tan 1

4

(remember SOH)

6

1

2

3

(remember TOA)

1

1

2

This gives us exact trig ratios from all 4 quadrants.

3

4

4

1. What is the exact value of

(a) sin 3000 (b) cos(-135)0 (C) tan 11

4

3000

600

0 0sin300 sin 60

3

2

-1350

450

0 0cos( 135) cos 45

1

2

3

4

4

11tan tan

4 4

1

2. Calculate the exact length of the side marked x in the triangle.

x

600

10 m

0sin 6010

x

010sin 60x

310

2

10 3

2

1

5

5 3 m

Solving Problems using exact Values

An oil tanker is sailing north. At 0115 hours a lighthouse is due east of the tanker. By 0230 hours the lighthouse is 10km from the tanker on a bearing of 1500. Calculate the speed of the tanker.

N

T1Lighthouse

T2

10 km

1500

300

x

DistanceSpeed =

Time0cos3010

x

010cos30x

10 3

2

5 3 km

Time = 1 hour and 15 minutes5

hours4

10 32

54

10 3 4

2 5

4 3 km/h

Algebraic solution of equations

01. Solve for , 2sin 3 0 for 0 360x x x

2sin 3 0x 2sin 3x

3sin

2x

AS

CT

is in quadrant 3 and 4x

Acute value of x.

1 3sin

2x

060

180 60 and 360 60x 0 0240 ,300

2 02. Solve tan 3 for 0 360x x

2tan 3x

tan 3x

AS

CT

is in quadrants 1, 2, 3 and 4x

1tan 3x 060

60, 180 60, 180 60 and 360 60x

0 0 0 060 , 120 , 240 ,300

03. Solve 2sin 4 3 0 for 0 180x x

2sin 4 3 0x

2sin 4 3x

3sin 4

2x

AS

CT

is in quadrant 3 and 4x

Acute value of 4x.

1 34 sin

2x

060

Remember there are 4 repeats of the sine graph giving 8 solutions over 3600.

4 240, 300, 600, 660, 960, 1020x

0 0 0 060 , 75 , 150 , 165 .x

2 04. Solve 3sin 4sin 1 0 for 0 360x x x

23sin 4sin 1 0x x

(3sin 1)(sin 1) 0x x

3sin 1 0x sin 1 0x

3sin 1x 1

sin3

x

AS

CT

1 1sin

3x

0 019.5 , 160.5x

sin 1x 090x

0 0 019.5 , 90 , 160.5x

5. Solve 5cos 2 0.72 for 0 2x x

5cos 2 0.72x

5cos 1.28x

cos 0.256x

AS

CT

1cos 0.256x 1.312 radians

1.312 rads, 2 1.312 radsx

1.312rads, 4.971 radsx

Algebraic Solution of Compound Angle Equations

Algebraic solutions of trigonometric equations can be extended to problems involving compound angles such as (3x + 45)0 etc.

0 01. Solve for , 4 tan(3 45) 2.5, 0 180x x x

04 tan(3 45) 2.5 x5

tan(3 45)8

x AS

CT

Acute value of (3x+45).

0(3 45) 32x

0 0 0Since 0 180 , 45 3 45 585x x

3 45 180 32 and 360 32x 0 0148 , 328 0, 508 .0 0 03 103 , 283 , 463 .x

0 0 034.33 , 94.33 , 154.33 .x

2. Solve for , 2sin 2 1, 0 23

x x x

1sin 2

3 2

xAS

CT

2 ,3 6

x 5,

6

11Since 0 2 , 2

3 3 3x x

13,

6 17

.6

7 15 192 , , , .

2 6 6 6x

7 5 19, , , .

4 12 4 12x