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Introduction

In geometry we learn about how the sides of polygons relate to the angles in the polygons, but we have not learned how to calculate an angle if we only know the lengths of the sides. Trigonometry (pronounced: trig-oh-nom-eh-tree) deals with the relationship between the angles and the sides of a right-angled triangle. We will learn about trigonometric functions, which form the basis of trigonometry.

Investigation : History of Trigonometry

Work in pairs or groups and investigate the history of the foundation of trigonometry. Describe the various stages of development and how the following cultures used trigonometry to improve their lives.

The works of the following people or cultures can be investigated:

  1. Cultures
    1. Ancient Egyptians
    2. Mesopotamians
    3. Ancient Indians of the Indus Valley
  2. People
    1. Lagadha (circa 1350-1200 BC)
    2. Hipparchus (circa 150 BC)
    3. Ptolemy (circa 100)
    4. Aryabhata (circa 499)
    5. Omar Khayyam (1048-1131)
    6. Bhaskara (circa 1150)
    7. Nasir al-Din (13th century)
    8. al-Kashi and Ulugh Beg (14th century)
    9. Bartholemaeus Pitiscus (1595)

Note: Interesting Fact :

You should be familiar with the idea of measuring angles from geometry but have you ever stopped to think why there are 360 degrees in a circle? The reason is purely historical. There are 360 degrees in a circle because the ancient Babylonians had a number system with base 60. A base is the number at which you add another digit when you count. The number system that we use everyday is called the decimal system (the base is 10), but computers use the binary system (the base is 2). 360=6×60360=6×60 so for them it made sense to have 360 degrees in a circle.

Where Trigonometry is Used

There are many applications of trigonometry. Of particular value is the technique of triangulation, which is used in astronomy to measure the distance to nearby stars, in geography to measure distances between landmarks, and in satellite navigation systems. GPSs (global positioning systems) would not be possible without trigonometry. Other fields which make use of trigonometry include astronomy (and hence navigation, on the oceans, in aircraft, and in space), music theory, acoustics, optics, analysis of financial markets, electronics, probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy, chemistry, number theory (and hence cryptology), seismology, meteorology, oceanography, many physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical engineering, mechanical engineering, civil engineering, computer graphics, cartography, crystallography and game development.

Discussion : Uses of Trigonometry

Select one of the uses of trigonometry from the list given and write a 1-page report describing how trigonometry is used in your chosen field.

Similarity of Triangles

If ABCABC is similar to DEFDEF, then this is written as:

A B C D E F A B C D E F
(1)

Figure 1
Figure 1 (MG10C15_001.png)

Then, it is possible to deduce ratios between corresponding sides of the two triangles, such as the following:

A B B C = D E E F A B A C = D E D F A C B C = D F E F A B D E = B C E F = A C D F A B B C = D E E F A B A C = D E D F A C B C = D F E F A B D E = B C E F = A C D F
(2)

The most important fact about similar triangles ABCABC and DEFDEF is that the angle at vertex A is equal to the angle at vertex D, the angle at B is equal to the angle at E, and the angle at C is equal to the angle at F.

A = D B = E C = F A = D B = E C = F
(3)

Investigation : Ratios of Similar Triangles

In your exercise book, draw three similar triangles of different sizes, but each with A^=30A^=30; B^=90B^=90 and C^=60C^=60. Measure angles and lengths very accurately in order to fill in the table below (round answers to one decimal place).

Figure 2
Figure 2 (MG10C15_002.png)

Table 1
Dividing lengths of sides (Ratios)
A B B C = A B B C = A B A C = A B A C = C B A C = C B A C =
A ' B ' B ' C ' = A ' B ' B ' C ' = A ' B ' A ' C ' = A ' B ' A ' C ' = C ' B ' A ' C ' = C ' B ' A ' C ' =
A ' ' B ' ' B ' ' C ' ' = A ' ' B ' ' B ' ' C ' ' = A ' ' B ' ' A ' ' C ' ' = A ' ' B ' ' A ' ' C ' ' = C ' ' B ' ' A ' ' C ' ' = C ' ' B ' ' A ' ' C ' ' =

What observations can you make about the ratios of the sides?

These equal ratios are used to define the trigonometric functions.

Note: In algebra, we often use the letter xx for our unknown variable (although we can use any other letter too, such as aa, bb, kk, etc). In trigonometry, we often use the Greek symbol θθ for an unknown angle (we also use αα , ββ , γγ etc).

Definition of the Trigonometric Functions

We are familiar with a function of the form f(x)f(x) where ff is the function and xx is the argument. Examples are:

f ( x ) = 2 x (exponential function) g ( x ) = x + 2 (linear function) h ( x ) = 2 x 2 (parabolic function) f ( x ) = 2 x (exponential function) g ( x ) = x + 2 (linear function) h ( x ) = 2 x 2 (parabolic function)
(4)

The basis of trigonometry are the trigonometric functions. There are three basic trigonometric functions:

  1. sine
  2. cosine
  3. tangent

These are abbreviated to:

  1. sin
  2. cos
  3. tan

These functions are defined from a right-angled triangle, a triangle where one internal angle is 90 .

Consider a right-angled triangle.

Figure 3
Figure 3 (MG10C15_003.png)

In the right-angled triangle, we refer to the lengths of the three sides according to how they are placed in relation to the angle θθ. The side opposite to the right angle is labelled the hypotenus, the side opposite θθ is labelled opposite, the side next to θθ is labelled adjacent. Note that the choice of non-90 degree internal angle is arbitrary. You can choose either internal angle and then define the adjacent and opposite sides accordingly. However, the hypotenuse remains the same regardless of which internal angle you are referring to.

We define the trigonometric functions, also known as trigonometric identities, as:

sin θ = o p p o s i t e h y p o t e n u s e cos θ = a d j a c e n t h y p o t e n u s e tan θ = o p p o s i t e a d j a c e n t sin θ = o p p o s i t e h y p o t e n u s e cos θ = a d j a c e n t h y p o t e n u s e tan θ = o p p o s i t e a d j a c e n t
(5)

These functions relate the lengths of the sides of a right-angled triangle to its interior angles.

One way of remembering the definitions is to use the following mnemonic that is perhaps easier to remember:

Table 2
Silly Old Hens S in = O pposite H ypotenuse S in = O pposite H ypotenuse
Cackle And Howl C os = A djacent H ypotenuse C os = A djacent H ypotenuse
Till Old Age T an = O pposite A djacent T an = O pposite A djacent

You may also hear people saying Soh Cah Toa. This is just another way to remember the trig functions.

Tip:

The definitions of opposite, adjacent and hypotenuse are only applicable when you are working with right-angled triangles! Always check to make sure your triangle has a right-angle before you use them, otherwise you will get the wrong answer. We will find ways of using our knowledge of right-angled triangles to deal with the trigonometry of non right-angled triangles in Grade 11.

Investigation : Definitions of Trigonometric Functions

  1. In each of the following triangles, state whether aa, bb and cc are the hypotenuse, opposite or adjacent sides of the triangle with respect to the marked angle.
    Figure 4
    Figure 4 (MG10C15_004.png)
  2. Complete each of the following, the first has been done for you
    Figure 5
    Figure 5 (MG10C15_005.png)
    a)sinA^= opposite hypotenuse =CBACb)cosA^=c)tanA^=a)sinA^= opposite hypotenuse =CBACb)cosA^=c)tanA^=
    (6)
    d)sinC^=e)cosC^=f)tanC^=d)sinC^=e)cosC^=f)tanC^=
    (7)
  3. Complete each of the following without a calculator:
    Figure 6
    Figure 6 (MG10C15_006.png)
    sin60=cos30=tan60=sin60=cos30=tan60=
    (8)
    Figure 7
    Figure 7 (MG10C15_007.png)
    sin45=cos45=tan45=sin45=cos45=tan45=
    (9)

For most angles θθ, it is very difficult to calculate the values of sinθsinθ, cosθcosθ and tanθtanθ. One usually needs to use a calculator to do so. However, we saw in the above Activity that we could work these values out for some special angles. Some of these angles are listed in the table below, along with the values of the trigonometric functions at these angles. Remember that the lengths of the sides of a right angled triangle must obey Pythagoras' theorum. The square of the hypothenuse (side opposite the 90 degree angle) equals the sum of the squares of the two other sides.

Table 3
  0 0 30 30 45 45 60 60 90 90 180 180
cos θ cos θ 1 3 2 3 2 1 2 1 2 1 2 1 2 0 - 1 - 1
sin θ sin θ 0 1 2 1 2 1 2 1 2 3 2 3 2 1 0
tan θ tan θ 0 1 3 1 3 1 3 3 - - 0

These values are useful when asked to solve a problem involving trig functions without using a calculator.

Exercise 1: Finding Lengths

Find the length of x in the following triangle.

Figure 8
Figure 8 (MG10C15_008.png)

Solution

  1. Step 1. Identify the trig identity that you need :

    In this case you have an angle (5050), the opposite side and the hypotenuse.

    So you should use sinsin

    sin 50 = x 100 sin 50 = x 100
    (10)
  2. Step 2. Rearrange the question to solve for xx :
    x = 100 × sin 50 x = 100 × sin 50
    (11)
  3. Step 3. Use your calculator to find the answer :

    Use the sin button on your calculator

    x = 76 . 6 m x = 76 . 6 m
    (12)

Exercise 2: Finding Angles

Find the value of θθ in the following triangle.

Figure 9
Figure 9 (MG10C15_009.png)

Solution

  1. Step 1. Identify the trig identity that you need :

    In this case you have the opposite side and the hypotenuse to the angle θθ.

    So you should use tantan

    tan θ = 50 100 tan θ = 50 100
    (13)
  2. Step 2. Calculate the fraction as a decimal number :
    tan θ = 0 . 5 tan θ = 0 . 5
    (14)
  3. Step 3. Use your calculator to find the angle :

    Since you are finding the angle,

    use tan-1tan-1 on your calculator

    Don't forget to set your calculator to `deg' mode!

    θ = 26 . 6 θ = 26 . 6
    (15)

The following videos provide a summary of what you have learnt so far.

Figure 10
Trigonometry - 1

Figure 11
Khan academy video on trigonometry - 2

Finding Lengths

Find the length of the sides marked with letters. Give answers correct to 2 decimal places.

Figure 12
Figure 12 (MG10C15_010.png)
Figure 13
Figure 13 (MG10C15_011.png)

Click here for the solution.

Simple Applications of Trigonometric Functions

Trigonometry was probably invented in ancient civilisations to solve practical problems such as building construction and navigating by the stars. In this section we will show how trigonometry can be used to solve some other practical problems.

Height and Depth

Figure 14: Determining the height of a building using trigonometry.
Figure 14 (MG10C15_012.png)

One simple task is to find the height of a building by using trigonometry. We could just use a tape measure lowered from the roof, but this is impractical (and dangerous) for tall buildings. It is much more sensible to measure a distance along the ground and use trigonometry to find the height of the building.

Figure 14 shows a building whose height we do not know. We have walked 100 m away from the building and measured the angle from the ground up to the top of the building. This angle is found to be 38,738,7. We call this angle the angle of elevation. As you can see from Figure 14, we now have a right-angled triangle. As we know the length of one side and an angle, we can calculate the height of the triangle, which is the height of the building we are trying to find.

If we examine the figure, we see that we have the opposite and the adjacent of the angle of elevation and we can write:

tan 38 , 7 = opposite adjacent = height 100 m height = 100 m × tan 38 , 7 = 80 m tan 38 , 7 = opposite adjacent = height 100 m height = 100 m × tan 38 , 7 = 80 m
(16)

Exercise 3: Height of a tower

A block of flats is 100m away from a cellphone tower. Someone stands at BB. They measure the angle from BB up to the top of the tower EE to be 62 . This is the angle of elevation. They then measure the angle from BB down to the bottom of the tower at CC to be 34 . This is the angle of depression.What is the height of the cellph one tower correct to 1 decimal place?

Figure 15
Figure 15 (MG10C15_013.png)

Solution
  1. Step 1. Identify a strategy :

    To find the height of the tower, all we have to do is find the length of CDCD and DEDE. We see that BDEBDE and BDCBDC are both right-angled triangles. For each of the triangles, we have an angle and we have the length ADAD. Thus we can calculate the sides of the triangles.

  2. Step 2. Calculate CDCD :

    We are given that the length ACAC is 100m. CABDCABD is a rectangle so BD=AC=100mBD=AC=100m.

    tan ( C B ^ D ) = C D B D C D = B D × tan ( C B ^ D ) = 100 × tan 34 tan ( C B ^ D ) = C D B D C D = B D × tan ( C B ^ D ) = 100 × tan 34
    (17)

    Use your calculator to find that tan34=0,6745tan34=0,6745. Using this, we find that CD=67,45CD=67,45m

  3. Step 3. Calculate DEDE :
    tan ( D B ^ E ) = D E B D D E = B D × tan ( D B ^ E ) = 100 × tan 62 = 188 , 07 m tan ( D B ^ E ) = D E B D D E = B D × tan ( D B ^ E ) = 100 × tan 62 = 188 , 07 m
    (18)
  4. Step 4. Combine the previous answers :

    We have that the height of the tower CE=CD+DE=67,45m+188,07m=255.5mCE=CD+DE=67,45m+188,07m=255.5m.

Maps and Plans

Maps and plans are usually scale drawings. This means that they are an exact copy of the real thing, but are usually smaller. So, only lengths are changed, but all angles are the same. We can use this idea to make use of maps and plans by adding information from the real world.

Exercise 4: Scale Drawing

A ship approaching Cape Town Harbour reaches point A on the map, due south of Pretoria and due east of Cape Town. If the distance from Cape Town to Pretoria is 1000km, use trigonometry to find out how far east the ship is from Cape Town, and hence find the scale of the map.

Figure 16
Figure 16 (MG10C15_014.png)

Solution
  1. Step 1. Identify what happens in the question :

    We already know the distance between Cape Town and AA in blocks from the given map (it is 5 blocks). Thus if we work out how many kilometers this same distance is, we can calculate how many kilometers each block represents, and thus we have the scale of the map.

  2. Step 2. Identify given information :

    Let us denote Cape Town with CC and Pretoria with PP. We can see that triangle APCAPC is a right-angled triangle. Furthermore, we see that the distance ACAC and distance APAP are both 5 blocks. Thus it is an isoceles triangle, and so AC^P=AP^C=45AC^P=AP^C=45.

  3. Step 3. Carry out the calculation :
    C A = C P × cos ( A C ^ P ) = 1000 × cos ( 45 ) = 1000 2 km C A = C P × cos ( A C ^ P ) = 1000 × cos ( 45 ) = 1000 2 km
    (19)

    To work out the scale, we see that

    5 blocks = 1000 2 km 1 block = 200 2 km 5 blocks = 1000 2 km 1 block = 200 2 km
    (20)

Exercise 5: Building plan

Mr Nkosi has a garage at his house, and he decides that he wants to add a corrugated iron roof to the side of the garage. The garage is 4m high, and his sheet for the roof is 5m long. If he wants the roof to be at an angle of 55, how high must he build the wall BDBD, which is holding up the roof? Give the answer to 2 decimal places.

Figure 17
Figure 17 (MG10C15_015.png)

Solution
  1. Step 1. Set out strategy :

    We see that the triangle ABCABC is a right-angled triangle. As we have one side and an angle of this triangle, we can calculate ACAC. The height of the wall is then the height of the garage minus ACAC.

  2. Step 2. Execute strategy :

    If BCBC=5m, and angle AB^C=5AB^C=5, then

    A C = B C × sin ( A B ^ C ) = 5 × sin 5 = 5 × 0 , 0871 = 0 . 4358 m A C = B C × sin ( A B ^ C ) = 5 × sin 5 = 5 × 0 , 0871 = 0 . 4358 m
    (21)

    Thus we have that the height of the wall BD=4m-0.4358m=3.56mBD=4m-0.4358m=3.56m.

Applications of Trigonometric Functions

  1. A boy flying a kite is standing 30 m from a point directly under the kite. If the string to the kite is 50 m long, find the angle of elevation of the kite.
    Click here for the solution.
  2. What is the angle of elevation of the sun when a tree 7,15 m tall casts a shadow 10,1 m long?
    Click here for the solution.

Graphs of Trigonometric Functions

This section describes the graphs of trigonometric functions.

Graph of sinθsinθ

Graph of sinθsinθ

Complete the following table, using your calculator to calculate the values. Then plot the values with sinθsinθ on the yy-axis and θθ on the xx-axis. Round answers to 1 decimal place.

Table 4
θ θ 0 30 60 90 120 150  
sin θ sin θ              
θ θ 180 210 240 270 300 330 360
sin θ sin θ              
 
Figure 18
Figure 18 (MG10C15_016.png)

Let us look back at our values for sinθsinθ

Table 5
θ θ 0 0 30 30 45 45 60 60 90 90 180 180
sin θ sin θ 0 1 2 1 2 1 2 1 2 3 2 3 2 1 0

As you can see, the function sinθsinθ has a value of 0 at θ=0θ=0. Its value then smoothly increases until θ=90θ=90 when its value is 1. We also know that it later decreases to 0 when θ=180θ=180. Putting all this together we can start to picture the full extent of the sine graph. The sine graph is shown in Figure 19. Notice the wave shape, with each wave having a length of 360360. We say the graph has a period of 360360. The height of the wave above (or below) the xx-axis is called the wave's amplitude. Thus the maximum amplitude of the sine-wave is 1, and its minimum amplitude is -1.

Figure 19: The graph of sinθsinθ.
Figure 19 (MG10C15_017.png)

Functions of the form y=asin(x)+qy=asin(x)+q

In the equation, y=asin(x)+qy=asin(x)+q, aa and qq are constants and have different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 20 for the function f(θ)=2sinθ+3f(θ)=2sinθ+3.

Figure 20: Graph of f(θ)=2sinθ+3f(θ)=2sinθ+3
Figure 20 (trigrep.png)

Functions of the Form y=asin(θ)+qy=asin(θ)+q :

  1. On the same set of axes, plot the following graphs:
    1. a(θ)=sinθ-2a(θ)=sinθ-2
    2. b(θ)=sinθ-1b(θ)=sinθ-1
    3. c(θ)=sinθc(θ)=sinθ
    4. d(θ)=sinθ+1d(θ)=sinθ+1
    5. e(θ)=sinθ+2e(θ)=sinθ+2
    Use your results to deduce the effect of qq.
  2. On the same set of axes, plot the following graphs:
    1. f(θ)=-2·sinθf(θ)=-2·sinθ
    2. g(θ)=-1·sinθg(θ)=-1·sinθ
    3. h(θ)=0·sinθh(θ)=0·sinθ
    4. j(θ)=1·sinθj(θ)=1·sinθ
    5. k(θ)=2·sinθk(θ)=2·sinθ
    Use your results to deduce the effect of aa.

You should have found that the value of aa affects the height of the peaks of the graph. As the magnitude of aa increases, the peaks get higher. As it decreases, the peaks get lower.

qq is called the vertical shift. If q=2q=2, then the whole sine graph shifts up 2 units. If q=-1q=-1, the whole sine graph shifts down 1 unit.

These different properties are summarised in Table 6.

Table 6: Table summarising general shapes and positions of graphs of functions of the form y=asin(x)+qy=asin(x)+q.
  a > 0 a > 0 a < 0 a < 0
q > 0 q > 0
Figure 21
Figure 21 (MG10C15_019.png)
Figure 22
Figure 22 (MG10C15_020.png)
q < 0 q < 0
Figure 23
Figure 23 (MG10C15_021.png)
Figure 24
Figure 24 (MG10C15_022.png)

Domain and Range

For f(θ)=asin(θ)+qf(θ)=asin(θ)+q, the domain is {θ:θR}{θ:θR} because there is no value of θRθR for which f(θ)f(θ) is undefined.

The range of f(θ)=asinθ+qf(θ)=asinθ+q depends on whether the value for aa is positive or negative. We will consider these two cases separately.

If a>0a>0 we have:

- 1 sin θ 1 - a a sin θ a ( Multiplication by a positive number maintains the nature of the inequality ) - a + q a sin θ + q a + q - a + q f ( θ ) a + q - 1 sin θ 1 - a a sin θ a ( Multiplication by a positive number maintains the nature of the inequality ) - a + q a sin θ + q a + q - a + q f ( θ ) a + q
(22)

This tells us that for all values of θθ, f(θ)f(θ) is always between -a+q-a+q and a+qa+q. Therefore if a>0a>0, the range of f(θ)=asinθ+qf(θ)=asinθ+q is {f(θ):f(θ)[-a+q,a+q]}{f(θ):f(θ)[-a+q,a+q]}.

Similarly, it can be shown that if a<0a<0, the range of f(θ)=asinθ+qf(θ)=asinθ+q is {f(θ):f(θ)[a+q,-a+q]}{f(θ):f(θ)[a+q,-a+q]}. This is left as an exercise.

Tip:
The easiest way to find the range is simply to look for the "bottom" and the "top" of the graph.

Intercepts

The yy-intercept, yintyint, of f(θ)=asin(x)+qf(θ)=asin(x)+q is simply the value of f(θ)f(θ) at θ=0θ=0.

y i n t = f ( 0 ) = a sin ( 0 ) + q = a ( 0 ) + q = q y i n t = f ( 0 ) = a sin ( 0 ) + q = a ( 0 ) + q = q
(23)

Graph of cosθcosθ

Graph of cosθcosθ :

Complete the following table, using your calculator to calculate the values correct to 1 decimal place. Then plot the values with cosθcosθ on the yy-axis and θθ on the xx-axis.

Table 7
θ θ 0 30 60 90 120 150  
cos θ cos θ              
θ θ 180 210 240 270 300 330 360
cos θ cos θ              
 
Figure 25
Figure 25 (MG10C15_023.png)

Let us look back at our values for cosθcosθ

Table 8
θ θ 0 0 30 30 45 45 60 60 90 90 180 180
cos θ cos θ 1 3 2 3 2 1 2 1 2 1 2 1 2 0 - 1 - 1

If you look carefully, you will notice that the cosine of an angle θθ is the same as the sine of the angle 90-θ90-θ. Take for example,

cos 60 = 1 2 = sin 30 = sin ( 90 - 60 ) cos 60 = 1 2 = sin 30 = sin ( 90 - 60 )
(24)

This tells us that in order to create the cosine graph, all we need to do is to shift the sine graph 9090 to the left. The graph of cosθcosθ is shown in Figure 26. As the cosine graph is simply a shifted sine graph, it will have the same period and amplitude as the sine graph.

Figure 26: The graph of cosθcosθ.
Figure 26 (MG10C15_024.png)

Functions of the form y=acos(x)+qy=acos(x)+q

In the equation, y=acos(x)+qy=acos(x)+q, aa and qq are constants and have different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 27 for the function f(θ)=2cosθ+3f(θ)=2cosθ+3.

Figure 27: Graph of f(θ)=2cosθ+3f(θ)=2cosθ+3
Figure 27 (trigrep1.png)

Functions of the Form y=acos(θ)+qy=acos(θ)+q :

  1. On the same set of axes, plot the following graphs:
    1. a(θ)=cosθ-2a(θ)=cosθ-2
    2. b(θ)=cosθ-1b(θ)=cosθ-1
    3. c(θ)=cosθc(θ)=cosθ
    4. d(θ)=cosθ+1d(θ)=cosθ+1
    5. e(θ)=cosθ+2e(θ)=cosθ+2
    Use your results to deduce the effect of qq.
  2. On the same set of axes, plot the following graphs:
    1. f(θ)=-2·cosθf(θ)=-2·cosθ
    2. g(θ)=-1·cosθg(θ)=-1·cosθ
    3. h(θ)=0·cosθh(θ)=0·cosθ
    4. j(θ)=1·cosθj(θ)=1·cosθ
    5. k(θ)=2·cosθk(θ)=2·cosθ
    Use your results to deduce the effect of aa.

You should have found that the value of aa affects the amplitude of the cosine graph in the same way it did for the sine graph.

You should have also found that the value of qq shifts the cosine graph in the same way as it did the sine graph.

These different properties are summarised in Table 9.

Table 9: Table summarising general shapes and positions of graphs of functions of the form y=acos(x)+qy=acos(x)+q.
  a > 0 a > 0 a < 0 a < 0
q > 0 q > 0
Figure 28
Figure 28 (MG10C15_026.png)
Figure 29
Figure 29 (MG10C15_027.png)
q < 0 q < 0
Figure 30
Figure 30 (MG10C15_028.png)
Figure 31
Figure 31 (MG10C15_029.png)

Domain and Range

For f(θ)=acos(θ)+qf(θ)=acos(θ)+q, the domain is {θ:θR}{θ:θR} because there is no value of θRθR for which f(θ)f(θ) is undefined.

It is easy to see that the range of f(θ)f(θ) will be the same as the range of asin(θ)+qasin(θ)+q. This is because the maximum and minimum values of acos(θ)+qacos(θ)+q will be the same as the maximum and minimum values of asin(θ)+qasin(θ)+q.

Intercepts

The yy-intercept of f(θ)=acos(x)+qf(θ)=acos(x)+q is calculated in the same way as for sine.

y i n t = f ( 0 ) = a cos ( 0 ) + q = a ( 1 ) + q = a + q y i n t = f ( 0 ) = a cos ( 0 ) + q = a ( 1 ) + q = a + q
(25)

Comparison of Graphs of sinθsinθ and cosθcosθ

Figure 32: The graph of cosθcosθ (solid-line) and the graph of sinθsinθ (dashed-line).
Figure 32 (MG10C15_030.png)

Notice that the two graphs look very similar. Both oscillate up and down around the xx-axis as you move along the axis. The distances between the peaks of the two graphs is the same and is constant along each graph. The height of the peaks and the depths of the troughs are the same.

The only difference is that the sinsin graph is shifted a little to the right of the coscos graph by 90. That means that if you shift the whole coscos graph to the right by 90 it will overlap perfectly with the sinsin graph. You could also move the sinsin graph by 90 to the left and it would overlap perfectly with the coscos graph. This means that:

sin θ = cos ( θ - 90 ) ( shift the cos graph to the right ) a nd cos θ = sin ( θ + 90 ) ( shift the sin graph to the left ) sin θ = cos ( θ - 90 ) ( shift the cos graph to the right ) a nd cos θ = sin ( θ + 90 ) ( shift the sin graph to the left )
(26)

Graph of tanθtanθ

Graph of tanθtanθ

Complete the following table, using your calculator to calculate the values correct to 1 decimal place. Then plot the values with tanθtanθ on the yy-axis and θθ on the xx-axis.

Table 10
θ θ 0 30 60 90 120 150  
tan θ tan θ              
θ θ 180 210 240 270 300 330 360
tan θ tan θ              
 
Figure 33
Figure 33 (MG10C15_031.png)

Let us look back at our values for tanθtanθ

Table 11
θ θ 0 0 30 30 45 45 60 60 90 90 180 180
tan θ tan θ 0 1 3 1 3 1 3 3 0

Now that we have graphs for sinθsinθ and cosθcosθ, there is an easy way to visualise the tangent graph. Let us look back at our definitions of sinθsinθ and cosθcosθ for a right-angled triangle.

sin θ cos θ = opposite hypotenuse adjacent hypotenuse = opposite adjacent = tan θ sin θ cos θ = opposite hypotenuse adjacent hypotenuse = opposite adjacent = tan θ
(27)

This is the first of an important set of equations called trigonometric identities. An identity is an equation, which holds true for any value which is put into it. In this case we have shown that

tan θ = sin θ cos θ tan θ = sin θ cos θ
(28)

for any value of θθ.

So we know that for values of θθ for which sinθ=0sinθ=0, we must also have tanθ=0tanθ=0. Also, if cosθ=0cosθ=0 our value of tanθtanθ is undefined as we cannot divide by 0. The graph is shown in Figure 34. The dashed vertical lines are at the values of θθ where tanθtanθ is not defined.

Figure 34: The graph of tanθtanθ.
Figure 34 (trgirep2.png)

Functions of the form y=atan(x)+qy=atan(x)+q

In the figure below is an example of a function of the form y=atan(x)+qy=atan(x)+q.

Figure 35: The graph of 2tanθ+12tanθ+1.
Figure 35 (trigrep3.png)

Functions of the Form y=atan(θ)+qy=atan(θ)+q :

  1. On the same set of axes, plot the following graphs:
    1. a(θ)=tanθ-2a(θ)=tanθ-2
    2. b(θ)=tanθ-1b(θ)=tanθ-1
    3. c(θ)=tanθc(θ)=tanθ
    4. d(θ)=tanθ+1d(θ)=tanθ+1
    5. e(θ)=tanθ+2e(θ)=tanθ+2
    Use your results to deduce the effect of qq.
  2. On the same set of axes, plot the following graphs:
    1. f(θ)=-2·tanθf(θ)=-2·tanθ
    2. g(θ)=-1·tanθg(θ)=-1·tanθ
    3. h(θ)=0·tanθh(θ)=0·tanθ
    4. j(θ)=1·tanθj(θ)=1·tanθ
    5. k(θ)=2·tanθk(θ)=2·tanθ
    Use your results to deduce the effect of aa.

You should have found that the value of aa affects the steepness of each of the branches. The larger the absolute magnitude of a, the quicker the branches approach their asymptotes, the values where they are not defined. Negative aa values switch the direction of the branches. You should have also found that the value of qq affects the vertical shift as for sinθsinθ and cosθcosθ. These different properties are summarised in Table 12.

Table 12: Table summarising general shapes and positions of graphs of functions of the form y=atan(x)+qy=atan(x)+q.
  a > 0 a > 0 a < 0 a < 0
q > 0 q > 0
Figure 36
Figure 36 (MG10C15_034.png)
Figure 37
Figure 37 (MG10C15_035.png)
q < 0 q < 0
Figure 38
Figure 38 (MG10C15_036.png)
Figure 39
Figure 39 (MG10C15_037.png)

Domain and Range

The domain of f(θ)=atan(θ)+qf(θ)=atan(θ)+q is all the values of θθ such that cosθcosθ is not equal to 0. We have already seen that when cosθ=0cosθ=0, tanθ=sinθcosθtanθ=sinθcosθ is undefined, as we have division by zero. We know that cosθ=0cosθ=0 for all θ=90+180nθ=90+180n, where nn is an integer. So the domain of f(θ)=atan(θ)+qf(θ)=atan(θ)+q is all values of θθ, except the values θ=90+180nθ=90+180n.

The range of f(θ)=atanθ+qf(θ)=atanθ+q is {f(θ):f(θ)(-,)}{f(θ):f(θ)(-,)}.

Intercepts

The yy-intercept, yintyint, of f(θ)=atan(x)+qf(θ)=atan(x)+q is again simply the value of f(θ)f(θ) at θ=0θ=0.

y i n t = f ( 0 ) = a tan ( 0 ) + q = a ( 0 ) + q = q y i n t = f ( 0 ) = a tan ( 0 ) + q = a ( 0 ) + q = q
(29)

Asymptotes

As θθ approaches 9090, tanθtanθ approaches infinity. But as θθ is undefined at 9090, θθ can only approach 9090, but never equal it. Thus the tanθtanθ curve gets closer and closer to the line θ=90θ=90, without ever touching it. Thus the line θ=90θ=90 is an asymptote of tanθtanθ. tanθtanθ also has asymptotes at θ=90+180nθ=90+180n, where nn is an integer.

Graphs of Trigonometric Functions
  1. Using your knowldge of the effects of aa and qq, sketch each of the following graphs, without using a table of values, for θ[0;360]θ[0;360]
    1. y=2sinθy=2sinθ
    2. y=-4cosθy=-4cosθ
    3. y=-2cosθ+1y=-2cosθ+1
    4. y=sinθ-3y=sinθ-3
    5. y=tanθ-2y=tanθ-2
    6. y=2cosθ-1y=2cosθ-1
    Click here for the solution.
  2. Give the equations of each of the following graphs:
    Figure 40
    Figure 40 (trigrep4.png)
    Figure 41
    Figure 41 (trigrep5.png)
    Figure 42
    Figure 42 (trigrep6.png)
    Click here for the solution.

The following presentation summarises what you have learnt in this chapter. Ignore the last slide.

Figure 43

End of Chapter Exercises

  1. Calculate the unknown lengths
    Figure 44
    Figure 44 (MG10C15_041.png)
    Click here for the solution.
  2. In the triangle PQRPQR, PR=20PR=20 cm, QR=22QR=22 cm and PR^Q=30PR^Q=30. The perpendicular line from PP to QRQR intersects QRQR at XX. Calculate
    1. the length XRXR,
    2. the length PXPX, and
    3. the angle QP^XQP^X
    Click here for the solution.
  3. A ladder of length 15 m is resting against a wall, the base of the ladder is 5 m from the wall. Find the angle between the wall and the ladder?
    Click here for the solution.
  4. A ladder of length 25 m is resting against a wall, the ladder makes an angle 3737 to the wall. Find the distance between the wall and the base of the ladder?
    Click here for the solution.
  5. In the following triangle find the angle AB^CAB^C
    Figure 45
    Figure 45 (MG10C15_042.png)
    Click here for the solution.
  6. In the following triangle find the length of side CDCD
    Figure 46
    Figure 46 (MG10C15_043.png)
    Click here for the solution.
  7. A(5;0)A(5;0) and B(11;4)B(11;4). Find the angle between the line through A and B and the x-axis.
    Click here for the solution.
  8. C(0;-13)C(0;-13) and D(-12;14)D(-12;14). Find the angle between the line through C and D and the y-axis.
    Click here for the solution.
  9. A 5m5m ladder is placed 2m2m from the wall. What is the angle the ladder makes with the wall?
    Click here for the solution.
  10. Given the points: E(5;0), F(6;2) and G(8;-2), find angle FE^GFE^G.
    Click here for the solution.
  11. An isosceles triangle has sides 9 cm ,9 cm 9 cm ,9 cm and 2 cm 2 cm . Find the size of the smallest angle of the triangle.
    Click here for the solution.
  12. A right-angled triangle has hypotenuse 13 mm 13 mm . Find the length of the other two sides if one of the angles of the triangle is 5050.
    Click here for the solution.
  13. One of the angles of a rhombus (rhombus - A four-sided polygon, each of whose sides is of equal length) with perimeter 20 cm 20 cm is 3030.
    1. Find the sides of the rhombus.
    2. Find the length of both diagonals.
    Click here for the solution.
  14. Captain Hook was sailing towards a lighthouse with a height of 10m10m.
    1. If the top of the lighthouse is 30m30m away, what is the angle of elevation of the boat to the nearest integer?
    2. If the boat moves another 7m7m towards the lighthouse, what is the new angle of elevation of the boat to the nearest integer?
    Click here for the solution.
  15. (Tricky) A triangle with angles 40,4040,40 and 100100 has a perimeter of 20 cm 20 cm . Find the length of each side of the triangle.
    Click here for the solution.

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