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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.4.3 Trigonometric Ratios of Special Angles

Subsubsection 2.4.3.1 Special Angles: 0, 45, and 90 Degrees

In this section you will use isosceles right-angle triangle to explore on how to find Tangent, Cosine and Sine of \(45^\circ\)
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An isosceles triangle is a triangle whose two sides and base angles are equal.
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Learner Experience 2.4.5.

\(\textbf{Work in groups}\)
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What you require: A ruler, a piece of paper and a protractor.
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  1. Draw the isosceles right-angle triangle like the one below,(ensure that two sides are equal)
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  2. Measure the angle \(\theta\text{.}\)
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  3. Measure the length \(PQ \, \text{and}\, QR\)
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    What do you notice about \(PQ \, \text{and}\, QR\)
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  4. Find the following trigonometric ratios:
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    1. \(\displaystyle \tan \, \theta\)
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    2. \(\displaystyle \cos \, \theta\)
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    3. \(\displaystyle \sin \, \theta\)
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  5. Compare the value of \(\cos \, \theta \, \text{and} \,\sin \, \theta \text{,}\) How do they relate?
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Exploration 2.4.6. Special Angles (0°, 45°, 90°).

Use the slider to investigate how the trigonometric ratios behave at special angles. Before moving the slider, make a prediction. Then explore the interactive and follow the guiding questions below to help you think carefully about what you observe.
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Instructions.
Move the slider slowly and pause at each marked angle. Ask yourself:
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  • When the angle is \(0^\circ\text{,}\) what shape does the triangle form? What do you notice about the opposite side? What value does the calculator show for \(\sin 0^\circ\text{,}\) \(\cos 0^\circ\text{,}\) and \(\tan 0^\circ\text{?}\) Why might these values make sense from the diagram?
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  • At \(45^\circ\text{,}\) compare the lengths of the opposite and adjacent sides. What relationship do you observe between them? How does this relate to the values of \(\sin 45^\circ\) and \(\cos 45^\circ\text{?}\) Why do you think this triangle is often called a special triangle?
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  • As the angle approaches \(90^\circ\text{,}\) how does the triangle change? What happens to the adjacent side? What value is displayed for \(\tan 90^\circ\text{?}\) Why might this value not exist?
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  • Compare the three angles \(0^\circ\text{,}\) \(45^\circ\text{,}\) and \(90^\circ\text{.}\) What patterns do you notice in the sine values? What patterns do you notice in the cosine values? How are these patterns reflected in the geometry of the triangle?
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Discuss your observations with a partner and explain how the diagram supports the calculator values.
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Figure 2.4.25. Special Angles Calculator
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Key Takeaway 2.4.26.

Look at the figure below;
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To find the tangent, cosine and sine of \(45^\circ\text{,}\) find first the length \(PR\)
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\begin{align*} PR^2=\amp 1^2+1^2 \\ = \amp 1+1 \\ =\amp 2\\ PR=\amp \sqrt{2} \end{align*}
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Therefore,
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\begin{align*} \sin\,45^\circ=\amp \frac{1}{\sqrt{2}} \\ \cos\,45^\circ=\amp \frac{1}{\sqrt{2}}\\ \tan\,45^\circ=\amp 1 \end{align*}
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Example 2.4.27.

Simplify the following without using tables or calculators:
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  1. \(\displaystyle 8\; \cos \;45^\circ \;sin\; 45^\circ\)
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  2. \(\displaystyle \sin \;90^\circ\; \cos \;45^\circ + \cos \;0^\circ \tan \;45^\circ\)
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  3. \(\displaystyle \tan \;45^\circ\; \cos \;90^\circ - \cos \;0^\circ \sin \;90^\circ\)
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Solution.
  1. \begin{align*} 8 \;cos \;45^\circ \;sin 45^\circ \amp = 8 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \\ \amp = \frac{8}{\sqrt{2} \times \sqrt{2}} \\ \amp = \frac{8}{2} \\ \amp = 4 \end{align*}
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  2. \(\cos \;45^\circ = \frac{1}{\sqrt{2}}\)
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    \(\tan \;45^\circ = 1\)
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    \(\sin \;90^\circ = 1\) and \(\cos \;0^\circ = 1\)
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    \(\sin \;90^\circ\; \cos \;45^\circ + \cos \;0^\circ \tan \;45^\circ = \cos \;45^\circ + \tan \;45^\circ\)
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    \(\sin \;90^\circ\; \cos \;45^\circ + \cos \;0^\circ \tan \;45^\circ = 1 + \frac{1}{\sqrt{2}}\)
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  3. \(\tan \;45^\circ = 1\)
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    \(\cos \;90^\circ = 0\)
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    \(\cos \;0^\circ = 1\)
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    \(\sin \;90^\circ = 1\)
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    \(\tan \;45^\circ\; \cos \;90^\circ - \cos \;0^\circ \sin \;90^\circ = 0 \times 1 - 1 \times 1 = -1\)
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Example 2.4.28.

The angle made by the arms of an upright pair of dividers and the horizontal is 45°. The vertical distance from the horizontal to the vertex is 15 cm. Find without using tables:
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  1. The horizontal distance between the tips of the arms.
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  2. The length of the arms.
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Solution.
  1. Since the angle is \(45^\circ\text{,}\) \(\tan \;45^\circ = 1\)
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    Let horizontal distance \(= x\)
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    \(\tan \;45^\circ = \frac{\text{vertical distance}}{\text{horizontal distance} \div 2}\)
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    \(1 = \frac{15}{\frac{x}{2}}\)
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    \(\frac{x}{2} = 15\)
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    \(x = 30\; cm\)
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  2. Using Pythagoras theorem:
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    \(\text{Length}^2 = 15^2 + 15^2 = 225 + 225 = 450\)
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    Length \(= \sqrt{450} = 15\sqrt{2} \;cm\)
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    The length of each arm is \(15\sqrt{2} \;cm\text{.}\)
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Checkpoint 2.4.29. Finding Trigonometric Ratios in a Right Angled Triangle (45°).

Load the question by clicking the button below.
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Checkpoint 2.4.30. Determining the Length of a Divider Arm from Vertical Height.

Load the question by clicking the button below.
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Exercises Exercises

1.
Find the exact values:
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  1. \(\displaystyle \sin 0^\circ\)
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  2. \(\displaystyle \cos 0^\circ\)
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  3. \(\displaystyle \tan 0^\circ\)
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Answer.
  1. \(\displaystyle \sin 0^\circ = 0\)
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  2. \(\displaystyle \cos 0^\circ = 1\)
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  3. \(\displaystyle \tan 0^\circ = 0\)
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2.
Find the exact values:
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  1. \(\displaystyle \sin 45^\circ\)
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  2. \(\displaystyle \cos 45^\circ\)
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  3. \(\displaystyle \tan 45^\circ\)
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Answer.
  1. \(\displaystyle \sin 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
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  2. \(\displaystyle \cos 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
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  3. \(\displaystyle \tan 45^\circ = 1\)
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3.
Find the exact values (and state any undefined):
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  1. \(\displaystyle \sin 90^\circ\)
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  2. \(\displaystyle \cos 90^\circ\)
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  3. \(\displaystyle \tan 90^\circ\)
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Answer.
  1. \(\displaystyle \sin 90^\circ = 1\)
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  2. \(\displaystyle \cos 90^\circ = 0\)
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  3. \(\tan 90^\circ\) is undefined (division by zero: \(\dfrac{\sin 90^\circ}{\cos 90^\circ}\)).
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4.
Given the addition formula
\begin{equation*} \sin(a+b)=\sin a\cos b + \cos a\sin b\text{,} \end{equation*}
evaluate exactly:
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  1. \(\displaystyle \sin 135^\circ\)
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  2. \(\displaystyle \sin 180^\circ\)
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Answer.
  1. Using the formula with \(a=90^\circ\) and \(b=45^\circ\text{:}\)
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    \begin{align*} \sin 135^\circ \amp = \sin(90^\circ+45^\circ)\\ \amp = \sin90^\circ\cos45^\circ + \cos90^\circ\sin45^\circ\\ \amp = 1\cdot \dfrac{\sqrt{2}}{2} + 0\cdot \dfrac{\sqrt{2}}{2}\\ \amp = \dfrac{\sqrt{2}}{2} \end{align*}
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  2. Using the formula with \(a=90^\circ\) and \(b=90^\circ\text{:}\)
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    \begin{align*} \sin 180^\circ \amp = \sin(90^\circ+90^\circ)\\ \amp = \sin90^\circ\cos90^\circ + \cos90^\circ\sin90^\circ\\ \amp = 1\cdot 0 + 0\cdot 1\\ \amp = 0 \end{align*}
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Subsubsection 2.4.3.2 Special Angles: 30 and 60 Degrees

In this section, you will be using equilateral triangle to find the Tangent, Cosine and Sine of \(30^\circ \, \text{and}\, 60^\circ\) .
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Learner Experience 2.4.7.

\(\textbf{Work in groups}\)
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What you require: Ruler, pencil, protractor, graph paper.
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  1. Draw \(30-60-90\text{:}\)
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    1. Draw an equilateral triangle.
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    2. Draw a line from a corner to the middle of the opposite side.
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    3. Label angles (\(30^\circ, \,60^\circ,\, 90^\circ\)).
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  2. Measure and Calculate:
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    • Measure sides of each triangle.
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    • Calculate sin, cos, tan for \(30^\circ, 60^\circ\) angles.
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    • Record your results in the table below.
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      Angle Triangle type Opposite side length Adjacent side length Hypotenuse length \(\sin(\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}}\) \(\cos(\theta)=\frac{\text{Adjacent}}{\text{Hypotenuse}}\) \(\tan(\theta)=\frac{\text{Opposite}}{\text{Adjacen}}\)
      \({\color{blue} 30^\circ}\) \(30-60-90\)
      \({\color{blue} 60^\circ}\) \(30-60-90\)
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  3. Discuss your result with other . What do you notice for the \(\textbf{Special Angles}\) (\(30^\circ, 60^\circ\))?
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Exploration 2.4.8. Special Angles (30° and 60°).

Use the slider to explore how the trigonometric ratios behave at the special angles \(30^\circ\) and \(60^\circ\text{.}\) Carefully explore the interactive by working through the guiding questions in the instructions section. Use them to help you investigate the relationships between the angles and their trigonometric ratios.
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Instructions.
Move the slider between \(30^\circ\) and \(60^\circ\) and observe carefully.
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  • When the angle is \(30^\circ\text{,}\) compare the lengths of the three sides. Which side is the shortest? Which is the longest? How do these side lengths relate to the displayed values of \(\sin 30^\circ\text{,}\) \(\cos 30^\circ\text{,}\) and \(\tan 30^\circ\text{?}\)
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  • Now move the slider to \(60^\circ\text{.}\) What changes in the triangle? Which sides appear to have exchanged roles compared to the \(30^\circ\) case?
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  • Compare \(\sin 30^\circ\) with \(\cos 60^\circ\text{.}\) What do you notice? Why might these values be the same?
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  • Compare \(\cos 30^\circ\) with \(\sin 60^\circ\text{.}\) What pattern do you observe?
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  • How does the tangent value change between \(30^\circ\) and \(60^\circ\text{?}\) What does this tell you about the ratio of opposite to adjacent sides?
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  • Based on your observations, what relationship can you describe between complementary angles (angles that add up to \(90^\circ\))?
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Discuss your findings with a partner and explain how the geometry of the triangle supports the calculator values.
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Figure 2.4.31. Exploring Trigonometric Ratios of 30° and 60°
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Key Takeaway 2.4.32.

The figure below shows an equilateral triangle \(ABC\text{.}\) \(AD\) Is the perpendicular bisector of \(BC\text{.}\)
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Notice that;
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length \(AD\) is given by,
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\begin{align*} AD^2\amp = 2^2-1^2\\ \amp = 4-1\\ AD\amp = \sqrt{3} \end{align*}
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Therefore, \(\sin\,30^\circ=\frac{1}{2},\quad \cos\,30^\circ=\frac{\sqrt{3}}{2},\quad \text{and} \quad \tan\,30^\circ=\frac{1}{\sqrt{3}}\)
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Similarly,
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\(\sin\,60^\circ=\frac{\sqrt{3}}{2},\quad \cos\,60^\circ=\frac{1}{2},\quad \text{and} \quad \tan\,60^\circ=\sqrt{3}\)
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Conclusion
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  • Sin and Cos are just swapped between \(30^\circ \) and \(60^\circ \text{.}\)
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  • \(\tan 30^\circ \) is small while tan \(60^\circ \) is large.
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  • \(\sin\,\theta \times \sin \, \theta \text{,}\) can be written as \(\sin^2\,\theta\)
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Example 2.4.33.

Simplify the following without using tables or calculator.
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  1. \(\displaystyle \sin \,30^\circ \, \cos\,45^\circ\)
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  2. \(\displaystyle 8\,\cos\,45^\circ\, \sin\,45^\circ \)
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  3. \(\displaystyle \sin\,60^\circ\, \cos\,45^\circ +\sin\,30^\circ\,\tan\,45^\circ\)
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Solution.
  1. \begin{align*} sin \,30^\circ \, cos\,45^\circ= \amp \frac{1}{2} \times \frac{1}{\sqrt{2}} \\ \amp = \frac{1}{2\sqrt{2}} \end{align*}
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  2. \begin{align*} 8\,cos\,45^\circ\, sin\,45^\circ\amp = \left( 8\times \frac{1}{\sqrt{2}} \right) \times \frac{1}{\sqrt{2}}\\ \amp = \frac{8}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \\ \amp = \frac{8}{\sqrt{2} \times \sqrt{2}} \\ \amp = \frac{8}{2} \\ \amp = 4 \end{align*}
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  3. \begin{align*} sin\,60^\circ\, cos\,45^\circ +sin\,30^\circ\,tan\,45^\circ \amp = \left(\frac{\sqrt{3}}{3} \times \frac{1}{\sqrt{2}}\right) +\left(\frac{1}{2} \times 1 \right)\\ \amp = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2} \\ \amp = \frac{\sqrt{3}+ \sqrt{2}}{2\sqrt{2}} \end{align*}
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Example 2.4.34.

The angle at the vertex of a cone is \(90^\circ\text{.}\) If the slant height is \(3\sqrt{2\,cm}\text{.}\) Find without using tables:
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  1. The diameter of the cone
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  2. The height of the cone.
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Solution.
Since the vertex angle is \(90^\circ\text{,}\) the cone can be thought of as half of a right circular cone, meaning that the base of the cone forms the hypotenuse of a right-angled triangle.
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Let:
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  • \(r\) be the radius of the base,
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  • \(h\) be the height of the cone,
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  • \(l=3\sqrt{2\,cm}\) be the slant height (hypotenuse of the right-angled triangle).
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Since the triangle formed is a right-angled isosceles triangle (because of the \(90^\circ\) vertex angle), we can say:
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\(r=h\)
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Since the right-angled triangle has radius \(r\) and height \(h\text{,}\) we use:
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\(r^2=h^2=l^2\)
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Since \(r=h\text{,}\) we substitute:
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\begin{align*} r^2+r^2\amp = \left(3\sqrt{2}\right) \\ 2r^2\amp = 9 \times 2\\ r^2\amp = 18 \\ ^2\amp = 9 \\ r\amp = 3\,cm \end{align*}
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  1. The diameter of the cone is:
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    Diameter \(=2r=2(3)=6\,cm\)
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  2. The height of the cone;
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    Since \(h=r\text{,}\) we conclude:
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    \(h=3\,cm\)
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Checkpoint 2.4.35. Finding the Trigonometry Ratios in a 30°, 60° and 90° Triangle.

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Exercises Exercises

1.
Using the triangle below, calculate the value of:
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  1. \(\displaystyle \sin \,30^\circ\)
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  2. \(\displaystyle \cos \,60^\circ\)
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  3. \(\displaystyle \tan \,30^\circ\)
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Answer.
  1. \(\displaystyle \sin \,30^\circ = \frac{1}{2}\)
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  2. \(\displaystyle \cos \,60^\circ = \frac{1}{2}\)
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  3. \(\displaystyle \tan \,30^\circ = \frac{1}{\sqrt{3}}\)
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2.
Without using a calculator, find
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\(\sin\,60^\circ+\cos\,30^\circ\)
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Answer.
\(\sin\,60^\circ+\cos\,30^\circ =\sqrt{3}\)
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3.
Given that \(\theta=45^\circ\text{,}\) calculate:
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\(\tan^2\,\theta-\sin^2\,\theta\)
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Answer.
\(\tan^2\,\theta-\sin^2\,\theta = \frac{1}{2}\)
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4.
Given: \(\sin\,\theta = \frac{1}{2}\)
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Find the angle \(\theta\text{,}\) where \(\theta^\circ \leq \theta \leq 90^\circ\)
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Answer.
Angle \(\theta = 30^\circ\text{.}\)
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