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# Trigonometry

## 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
2. Hipparchus (circa 150 BC)
3. Ptolemy (circa 100)
4. Aryabhata (circa 499)
5. Omar Khayyam (1048-1131)
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 fields that uses trigonometry from the list given above and write a 1-page report describing how trigonometry is used in the field that you chose.

## 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)

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).

 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 can be defined from a right-angled triangle, a triangle where one internal angle is 90 .

Consider a right-angled triangle.

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 labeled the hypotenuse, the side opposite θθ is labeled opposite, the side next to θθ is labeled 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 (because it is ALWAYS opposite the right angle and ALWAYS the longest side).

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

sin θ = opposite hypotenuse cos θ = adjacent hypotenuse tan θ = opposite adjacent sin θ = opposite hypotenuse cos θ = adjacent hypotenuse tan θ = opposite adjacent
(5)

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

### Note:

The trig ratios are independent of the lengths of the sides of a triangle and depend only on the angles, this is why we can consider them to be functions of the angles.

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

 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.
2. Complete each of the following, the first has been done for you
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:
sin60=cos30=tan60=sin60=cos30=tan60=
(8)
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' theorem. The square of the hypotenuse (side opposite the 90 degree angle) equals the sum of the squares of the two other sides.

 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.

Each of the trigonometric functions has a reciprocal that has a special name. The three reciprocals are cosecant (or cosec), secant (or sec) and cotangent (or cot). These reciprocals are given below:

cosecθ= 1 sinθ secθ= 1 cosθ cotθ= 1 tanθ cosecθ= 1 sinθ secθ= 1 cosθ cotθ= 1 tanθ
(10)

We can also define these reciprocals for any right angled triangle:

cosec θ = hypotenuse opposite sec θ = hypotenuse adjacent cot θ = adjacent opposite cosec θ = hypotenuse opposite sec θ = hypotenuse adjacent cot θ = adjacent opposite
(11)

### Exercise 1: Finding Lengths

Find the length of x in the following triangle.

#### 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
(12)
2. Step 2. Rearrange the question to solve for xx :
x = 100 × sin 50 x = 100 × sin 50
(13)

Use the sin button on your calculator

x = 76 . 6 m x = 76 . 6 m
(14)

### Exercise 2: Finding Angles

Find the value of θθ in the following triangle.

#### 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
(15)
2. Step 2. Calculate the fraction as a decimal number :
tan θ = 0 . 5 tan θ = 0 . 5
(16)
3. Step 3. Use your calculator to find the angle :

Since you are finding the angle,

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

θ = 26 . 6 θ = 26 . 6
(17)

In the previous example we used tan-1tan-1. This is simply the inverse of the tan function. Sin and cos also have inverses. All this means is that we want to find the angle that makes the expression true and so we must move the tan (or sin or cos) to the other side of the equals sign and leave the angle where it is. Sometimes the reciprocal trigonometric functions are also referred to as the 'inverse trigonometric functions'. You should note, however that tan-1tan-1 and cotcot are definitely NOT the same thing.

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.

## Two-dimensional problems

We can use the trig functions to solve problems in two dimensions that involve right angled triangles. For example if you are given a quadrilateral and asked to find the one of the angles, you can construct a right angled triangle and use the trig functions to solve for the angle. This will become clearer after working through the following example.

### Exercise 3

Let ABCD be a trapezium with AB=4cmAB=4cm, CD=6cmCD=6cm, BC=5cmBC=5cm and AD=5cmAD=5cm. Point E on diagonal AC divides the diagonal such that AE=3cmAE=3cm. Find A B ^ CA B ^ C.

#### Solution

1. Step 1. Draw a diagram: We draw a diagram and construct right angled triangles to help us visualize the problem.
2. Step 2. Determine how to solve the problem: We will use triangle ABE and triangle BEC to get the two angles, and then we will add these two angles together to find the angle we want.
3. Step 3. Determine which trig function to use: We use sin for both triangles since we have the hypotenuse and the opposite side.
4. Step 4. Solve the problem: In triangle ABE we find:
sin ( A B ^ E ) = opp hyp sin ( A B ^ E ) = 34 A B ^ E = sin-1 (34) A B ^ E = 48,59 sin ( A B ^ E ) = opp hyp sin ( A B ^ E ) = 34 A B ^ E = sin-1 (34) A B ^ E = 48,59
(18)
We use the theorem of Pythagoras to find EC=4,4cmEC=4,4cm. In triangle BEC we find:
sin ( C B ^ E ) = opp hyp sin ( C B ^ E ) = 4,45 A B ^ E = sin-1 (4,45 ) C B ^ E = 61,64 sin ( C B ^ E ) = opp hyp sin ( C B ^ E ) = 4,45 A B ^ E = sin-1 (4,45 ) C B ^ E = 61,64
(19)
5. Step 5. Write down the final answer: We add the two angles together to get: 48,59+61,64=110,2348,59+61,64=110,23

## The trig functions for any angle

So far we have defined the trig functions using right angled triangles. We can now extend these definitions to any angle. We do this by noting that the definitions do not rely on the lengths of the sides of the triangle, but only on the angle. So if we plot any point on the Cartesian plane and then draw a line from the origin to that point, we can work out the angle of that line. In Figure 15 points P and Q have been plotted. A line from the origin to each point is drawn. The dotted lines show how we can construct right angle triangles for each point. Now we can find the angles A and B.

You should find the angle A is 63,4363,43. For angle B, you first work out x (33,6933,69) and then B is 180-33,69=146,31180-33,69=146,31. But what if we wanted to do this without working out these angles and figuring out whether to add or subtract 180 or 90? Can we use the trig functions to do this? Consider point P in Figure 15. To find the angle you would have used one of the trig functions, e.g. tanθtanθ. You should also have noted that the side adjacent to the angle was just the x-co-ordinate and that the side opposite the angle was just the y-co-ordinate. But what about the hypotenuse? Well, you can find that using Pythagoras since you have two sides of a right angled triangle. If we were to draw a circle centered on the origin, then the length from the origin to point P is the radius of the circle, which we denote r. Now we can rewrite all our trig functions in terms of x, y and r. But how does this help us to find angle B? Well, we know that from point Q to the origin is r, and we have the co-ordinates of Q. So we simply use the newly defined trig functions to find angle B! (Try it for yourself and confirm that you get the same answer as before.) One final point to note is that when we go anti-clockwise around the Cartesian plane the angles are positive and when we go clockwise around the Cartesian plane, the angles are negative.

So we get the following definitions for the trig functions:

sin θ = x r cos θ = y r tan θ = y x sin θ = x r cos θ = y r tan θ = y x
(20)

But what if the x- or y-co-ordinate is negative? Do we ignore that, or is there some way to take that into account? The answer is that we do not ignore it. The sign in front of the x- or y-co-ordinate tells us whether or not sin, cos and tan are positive or negative. We divide the Cartesian plane into quadrants and then we can use Figure 16 to tell us whether the trig function is positive or negative. This diagram is known as the CAST diagram.

We can also extend the definitions of the reciprocals in the same way:

cosec θ = r x sec θ = r y cot θ = x y cosec θ = r x sec θ = r y cot θ = x y
(21)

### Exercise 4: Finding the angle

Points R(-1;-3) and point S(3;-3) are plotted in the diagram below. Find the angles αα and ββ.

#### Solution

1. Step 1. Write down what is given and what is required :

We have the co-ordinates of the points R and S. We are required to find two angles. Angle ββ is positive and angle αα is negative.

2. Step 2. Calculate ββ:

We use tan to find ββ, since we are only given x and y. We note that we are in the third quadrant, where tan is positive.

tan ( β ) = y x tan ( β ) = -3-1 β = tan-1 (3) β = 71,57 tan ( β ) = y x tan ( β ) = -3-1 β = tan-1 (3) β = 71,57
(22)
3. Step 3. Calculate αα:

We use tan to calculate αα, since we are only given x and y. We also note that we are in the fourth quadrant, where tan is positive.

tan ( α ) = y x tan ( α ) = -33 α = tan-1 (-1) α = -45 tan ( α ) = y x tan ( α ) = -33 α = tan-1 (-1) α = -45
(23)
4. Step 4. Write the final answer:

Angle αα is -45-45 and angle ββ is 71,5771,57

### Note:

You should observe that in the worked example above, the angle αα is simply the angle that line OS makes with the x-axis. So if you are asked to find out what angle a line makes with the x- or y-axes, then you can use trigonometry!

### Note:

The CAST diagram can be generalized to a very powerful tool for solving trigonometric functions by hand (that is, without a calculator) called the unit circle. You may or may not touch on this in Grade 11 or 12.

## Solving simple trigonometric equations

Using what we have learnt about trig functions we can now solve some simple trig equations. We use the principles from Equations and Inequalities to help us solve trig equations.

### Note:

It is important to note that in general 2sinθsin(2θ)2sinθsin(2θ). In other words doubling (or multiplying by 2) has a different effect from doubling the angle.

### Exercise 5

Solve the following trig equation: 3cos(2x+38)+3=23cos(2x+38)+3=2

#### Solution

1. Step 1. Rearrange the equation:
3cos(2x+38)=2-3 cos(2x+38)=-13 (2x+38)=107,46 2x=107,46-38 2x=69,46 x=34,73 3cos(2x+38)=2-3 cos(2x+38)=-13 (2x+38)=107,46 2x=107,46-38 2x=69,46 x=34,73
(24)
2. Step 2. Write the final answer: x=34,73x=34,73

## 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

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 18 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 18, 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
(25)

#### Exercise 6: 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?

##### 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
(26)

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
(27)
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 7: 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.

##### 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
(28)

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
(29)

#### Exercise 8: 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.

##### 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
(30)

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.
2. What is the angle of elevation of the sun when a tree 7,15 m tall casts a shadow 10,1 m long?

## 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.

 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

Let us look back at our values for sinθsinθ

 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 23. 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.

### 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 24 for the function f(θ)=2sinθ+3f(θ)=2sinθ+3.

#### 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.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### 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 - a + q a sin θ + q a + q - a + q f ( θ ) a + q - 1 sin θ 1 - a a sin θ a - a + q a sin θ + q a + q - a + q f ( θ ) a + q
(31)

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
(32)

### 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.

 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

Let us look back at our values for cosθcosθ

 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 )
(33)

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 30. As the cosine graph is simply a shifted sine graph, it will have the same period and amplitude as the sine graph.

### 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 31 for the function f(θ)=2cosθ+3f(θ)=2cosθ+3.

#### 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.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### 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
(34)

### Comparison of Graphs of sinθsinθ and cosθcosθ

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 )
(35)

### 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.

 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

Let us look back at our values for tanθtanθ

 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
(36)

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
(37)

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 38. The dashed vertical lines are at the values of θθ where tanθtanθ is not defined.

### 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.

#### 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.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

#### 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
(38)

#### 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
2. Give the equations of each of the following graphs: Click here for the solution.

The following presentation summarises what you have learnt in this chapter.

## Summary

• We can define three trigonometric functions for right angled triangles: sine (sin), cosine (cos) and tangent (tan).
• Each of these functions have a reciprocal: cosecant (cosec), secant (sec) and cotangent (cot).
• We can use the principles of solving equations and the trigonometric functions to help us solve simple trigonometric equations.
• We can solve problems in two dimensions that involve right angled triangles.
• For some special angles, we can easily find the values of sin, cos and tan.
• We can extend the definitions of the trigonometric functions to any angle.
• Trigonometry is used to help us solve problems in 2-dimensions, such as finding the height of a building.
• We can draw graphs for sin, cos and tan

## End of Chapter Exercises

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
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?
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?
5. In the following triangle find the angle AB^CAB^CClick here for the solution.
6. In the following triangle find the length of side CDCDClick 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.
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.
9. A 5m5m ladder is placed 2m2m from the wall. What is the angle the ladder makes with the wall?
10. Given the points: E(5;0), F(6;2) and G(8;-2), find angle FE^GFE^G.
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.
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.
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.
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?
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.

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