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.

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.

If ▵ABC▵ABC is similar to ▵DEF▵DEF, then this is written as:

Figure 1

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

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.

Investigation : Ratios of Similar Triangles

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

Figure 2

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).

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:

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

These are abbreviated to:

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

Consider a right-angled triangle.

Figure 3

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:

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:

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

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.

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:

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

Exercise 1: Finding Lengths

Find the length of x in the following triangle.

Figure 8

Solution

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

   In this case you have an angle (50∘50∘), 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)
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

   (14)

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Exercise 2: Finding Angles

Find the value of θθ in the following triangle.

Figure 9

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,

   use tan-1tan-1 on your calculator

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

   θ = 26 . 6 ∘ θ = 26 . 6 ∘

   (17)

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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.

Finding Lengths

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

Figure 12

Figure 13

Click here for the solution.

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.

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.

Figure 15

You should find the angle A is 63,43∘63,43∘. For angle B, you first work out x (33,69∘33,69∘) and then B is 180∘-33,69∘=146,31∘180∘-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:

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.

Figure 16

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

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.

Figure 17

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 71,57∘71,57∘

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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.

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 18: Determining the height of a building using trigonometry.

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 38,7∘38,7∘. 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?

Figure 19

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 ▵BDE▵BDE and ▵BDC▵BDC 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.

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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.

Figure 20

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=45∘AC^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)

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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 5∘5∘, how high must he build the wall BDBD, which is holding up the roof? Give the answer to 2 decimal places.

Figure 21

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=5∘AB^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.

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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.

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 22

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=0∘θ=0∘. Its value then smoothly increases until θ=90∘θ=90∘ when its value is 1. We also know that it later decreases to 0 when θ=180∘θ=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 360∘360∘. We say the graph has a period of 360∘360∘. 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 23: The graph of sinθsinθ.

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.

Figure 24: Graph of 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.

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 25	Figure 26
q < 0 q < 0	Figure 27	Figure 28

Domain and Range

For f(θ)=asin(θ)+qf(θ)=asin(θ)+q, the domain is {θ:θ∈R}{θ:θ∈R} because there is no value of θ∈Rθ∈R 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=0∘θ=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.

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 29

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 ∘ )

(33)

This tells us that in order to create the cosine graph, all we need to do is to shift the sine graph 90∘90∘ 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.

Figure 30: The graph of cosθcosθ.