Sectieoverzicht

  • We use the fact that it is the size of the angles in a triangle that determine the ratio of the sides as the basis of trigonometry. We can define three basic trigonometric ratios for any triangle.

    Have a look at the right-angled triangle below. We can name the sides of the triangle with reference to the angle called \( \theta \) (theta) at   as follows. The hypotenuse ( ) is always the side opposite the right-angle. The side next to  ( ) is called the adjacent side (adjacent means 'next to') and the side opposite \( \theta \) ( ) is called the opposite side.

    Figure 2

    If we defined the sides with respect to \( \beta \) (beta) at  , then the adjacent side would be   and the opposite side would be  . The hypotenuse would remain unchanged.


    Note
    The definitions of opposite, adjacent and hypotenuse are only applicable when working with right-angled triangles. Always check to make sure your triangle has a right-angle before you use them.

    We can define the three ratios of the lengths of the sides within any right-angled triangle. We can give each of these three ratios a special name – sine, cosine and tangent – all with respect to the angle \( \theta \) :

    \( \sin \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the hypotenuse}} \)

    \(\cos \theta=\displaystyle \frac{\text{length of the adjacent side}}{\text{length of the hypotenuse}} \)

    \(\tan \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the adjacent side}} \)

    With respect to angle \( \theta \) in \( \Delta ABC \) above, these three ratios would be:

    \( \sin \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the hypotenuse}}=\frac{BC}{AB}\)

    \(\cos \theta=\displaystyle \frac{\text{length of the adjacent side}}{\text{length of the hypotenuse}}=\frac{AC}{AB} \)

    \(\tan \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the adjacent side}}=\frac{BC}{AC} \)

    Can you work out what these three ratios would be when defined with respect to angle \( \beta )\. Write them down.

    With respect to angle \( \beta \) in \( \Delta ABC \) above, these three ratios would be:

    \( \sin \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the hypotenuse}}=\displaystyle \frac{AC}{AB}\)

    \(\cos \theta=\displaystyle \frac{\text{length of the adjacent side}}{\text{length of the hypotenuse}}=\displaystyle \frac{BC}{AB} \)

    \(\tan \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the adjacent side}}=\displaystyle \frac{AC}{BC} \)

    The three ratios of sine, cosine and tangent form the basis of all of trigonometry.
    Note
    The three basic trigonometric ratios are:

    \( \sin \theta=\displaystyle \frac{\text{opposite}}{\text{hypotenuse}} \)

    \(\cos \theta=\displaystyle \frac{\text{adjacent}}{\text{hypotenuse}} \)

    \(\tan \theta=\displaystyle \frac{\text{opposite}}{\text{adjacent}} \)

    We never define the trigonometric ratios with respect to the right-angle.
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      Activity 1: Example 1

      State the three basic trigonometric ratios for the following triangles with respect to the angle \( \theta \).

      1. Example 1a
      2. Example 1b
      3. Example 1c
      Solution:
      1. \( \sin \theta=\displaystyle \frac{\text{opp}}{\text{hyp}} =\displaystyle\frac{EF}{DE} \)
      2. \( \cos \theta=\displaystyle \frac{\text{adj}}{\text{hyp}} =\displaystyle\frac{DF}{DE} \)
      3. \( \tan \theta=\displaystyle \frac{\text{opp}}{\text{adj}} =\displaystyle\frac{EF}{DF} \)
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      Activity 2: Example 2

      State each of the following ratios for the given triangle:

      Example 2

      1. \( \sin\hat{A} \)
      2. \( \cos\hat{A} \)
      3. \( \tan\hat{A} \)
      4. \( \sin\hat{C} \)
      5. \( \cos\hat{C} \)
      6. \( \tan\hat{C} \)

      Solution:

      1. \( \sin\hat{A}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{BC}{AC} \)
      2. \( \cos\hat{A}=\displaystyle\frac{\text{adj}}{\text{hyp}}=\displaystyle\frac{AB}{AC} \)
      3. \( \tan\hat{A}=\displaystyle\frac{\text{opp}}{\text{adj}}=\displaystyle\frac{BC}{AB} \)
      4. \( \sin\hat{C}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{AB}{AC} \)
      5. \( \cos\hat{C}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{BC}{AC} \)
      6. \( \tan\hat{C}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{AB}{BC} \)

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      SOH CAH TOA

      You can use Soh Cah Toa to help you remember how each of the trig ratios are defined.

      sohcahtoa

    • The basic trigonometric ratios Test
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      Activity 3: Example 3

      Have a look at this next example.

      Given \( \Delta ABC \) with \( \hat{C}=90^\circ \),\( \hat{A}=\theta \) and\( \hat{B}=\alpha \).

      Example 3 

      Determine the value of the following, showing all calculations.

      1. \( \cos\alpha \)
      2. \( \tan\theta \)
      3. \( \displaystyle\frac{\sin\alpha}{\cos\alpha} \)
      4. \( 1+\tan^2\alpha \)

      Solution

      1. \( \cos\alpha=\displaystyle\frac{\text{adj}}{\text{hyp}}=\displaystyle\frac{BC}{AB}=\displaystyle\frac{4}{5} \)
      2. \( \tan\theta=\displaystyle\frac{\text{opp}}{\text{adj}}=\displaystyle\frac{BC}{AC} \).
        But we do not know the length of AC. However, we can use the theorem of Pythagoras to find AC.
        \( \begin{align*}AB^2&=AC^2+BC^2 \\\therefore AC^2 &= AB^2-BC^2 \\\therefore AC&=\sqrt{AB^2-BC^2} \\&=\sqrt{5^2-4^2} \\&=\sqrt{25-16} \\&=\sqrt{9} \\&=3 \end{align*} \)
        Therefore \( \tan\theta=\displaystyle\frac{\text{opp}}{\text{adj}}=\displaystyle\frac{BC}{AC}=\displaystyle\frac{4}{3} \)
      3. \( \displaystyle\frac{\sin\alpha}{\cos\alpha}=\displaystyle\frac{\frac{3}{5}}{\frac{4}{5}}=\displaystyle\frac{3}{5} \times\displaystyle\frac{5}{4}=\displaystyle\frac{3}{4} \)
      4. Remember that \( \tan\theta \) is actually just a number (a fraction). Therefore, we can apply all the normal operations to the trigonometric ratios. In this case we have to square the ratio and add 1 to it.
        \( 1+\tan^2\theta=1+(\tan\theta)^2=1+ \left( \displaystyle\frac{4}{3}\right)^2=1+\displaystyle\frac{16}{9} = \displaystyle\frac{9}{9}+\displaystyle\frac{16}{9}=\displaystyle\frac{25}{9} \)
      Note
      If we mean to raise a trigonometric ratio to a power, we always write the power directly after the ratio like \( \sin^2\theta \) or \( \cos^3\alpha \). If you see \( \sin\theta^2 \), for example, this means that the angle is being raised to the power not the whole ratio.
    • Calculating with the basic trigonometric ratios Test
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