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The trig functions and 2-D problems

Module by: Free High School Science Texts Project. E-mail the author

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)
(1)

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.

Figure 1
Figure 1 (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 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
(2)

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:

Table 1
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 2
    Figure 2 (MG10C15_004.png)
  2. Complete each of the following, the first has been done for you
    Figure 3
    Figure 3 (MG10C15_005.png)
    a)sinA^= opposite hypotenuse =CBACb)cosA^=c)tanA^=a)sinA^= opposite hypotenuse =CBACb)cosA^=c)tanA^=
    (3)
    d)sinC^=e)cosC^=f)tanC^=d)sinC^=e)cosC^=f)tanC^=
    (4)
  3. Complete each of the following without a calculator:
    Figure 4
    Figure 4 (MG10C15_006.png)
    sin60=cos30=tan60=sin60=cos30=tan60=
    (5)
    Figure 5
    Figure 5 (MG10C15_007.png)
    sin45=cos45=tan45=sin45=cos45=tan45=
    (6)

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.

Table 2
  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
(7)

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
(8)

Exercise 1: Finding Lengths

Find the length of x in the following triangle.

Figure 6
Figure 6 (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
    (9)
  2. Step 2. Rearrange the question to solve for xx :
    x = 100 × sin 50 x = 100 × sin 50
    (10)
  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
    (11)

Exercise 2: Finding Angles

Find the value of θθ in the following triangle.

Figure 7
Figure 7 (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
    (12)
  2. Step 2. Calculate the fraction as a decimal number :
    tan θ = 0 . 5 tan θ = 0 . 5
    (13)
  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
    (14)

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 8
Trigonometry - 1

Figure 9
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 10
Figure 10 (MG10C15_010.png)
Figure 11
Figure 11 (MG10C15_011.png)
Click here for the solution.

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.
    Figure 12
    Figure 12 (q1trig.png)
  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
    (15)
    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
    (16)
  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

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