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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 \(45^\circ\)
An isosceles triangle is a triangle whose two sides and base angles are equal.

Learner Experience 2.4.29.

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

Exploration 2.4.30. 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.

Instructions.
Move the slider slowly and pause at each marked angle. Ask yourself:
  • 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?
  • 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?
  • 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?
  • 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?
Discuss your observations with a partner and explain how the diagram supports the calculator values.
Figure 2.4.31. Special Angles Calculator

Key Takeaway 2.4.32.

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

Example 2.4.33.

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

Example 2.4.34.

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:
  1. The horizontal distance between the tips of the arms.
  2. The length of the arms.
Solution.
  1. Since the angle is \(45^\circ\text{,}\) \(\tan \;45^\circ = 1\)
    Let horizontal distance \(= x\)
    \(\tan \;45^\circ = \frac{\text{vertical distance}}{\text{horizontal distance} \div 2}\)
    \(1 = \frac{15}{\frac{x}{2}}\)
    \(\frac{x}{2} = 15\)
    \(x = 30\; cm\)
  2. Using Pythagoras theorem:
    \(\text{Length}^2 = 15^2 + 15^2 = 225 + 225 = 450\)
    Length \(= \sqrt{450} = 15\sqrt{2} \;cm\)
    The length of each arm is \(15\sqrt{2} \;cm\text{.}\)

Checkpoint 2.4.35. Finding Trigonometric Ratios in a Right Angled Triangle (45°).

Load the question by clicking the button below.

Checkpoint 2.4.36. Determining the Length of a Divider Arm from Vertical Height.

Load the question by clicking the button below.

Exercises Exercises

1.

Find the exact values:
  1. \(\displaystyle \sin 0^\circ\)
  2. \(\displaystyle \cos 0^\circ\)
  3. \(\displaystyle \tan 0^\circ\)
Answer.
  1. \(\displaystyle \sin 0^\circ = 0\)
  2. \(\displaystyle \cos 0^\circ = 1\)
  3. \(\displaystyle \tan 0^\circ = 0\)

2.

Find the exact values:
  1. \(\displaystyle \sin 45^\circ\)
  2. \(\displaystyle \cos 45^\circ\)
  3. \(\displaystyle \tan 45^\circ\)
Answer.
  1. \(\displaystyle \sin 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
  2. \(\displaystyle \cos 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
  3. \(\displaystyle \tan 45^\circ = 1\)

3.

Find the exact values (and state any undefined):
  1. \(\displaystyle \sin 90^\circ\)
  2. \(\displaystyle \cos 90^\circ\)
  3. \(\displaystyle \tan 90^\circ\)
Answer.
  1. \(\displaystyle \sin 90^\circ = 1\)
  2. \(\displaystyle \cos 90^\circ = 0\)
  3. \(\tan 90^\circ\) is undefined (division by zero: \(\dfrac{\sin 90^\circ}{\cos 90^\circ}\)).

4.

Given the addition formula
\begin{equation*} \sin(a+b)=\sin a\cos b + \cos a\sin b\text{,} \end{equation*}
evaluate exactly:
  1. \(\displaystyle \sin 135^\circ\)
  2. \(\displaystyle \sin 180^\circ\)
Answer.
  1. Using the formula with \(a=90^\circ\) and \(b=45^\circ\text{:}\)
    \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*}
  2. Using the formula with \(a=90^\circ\) and \(b=90^\circ\text{:}\)
    \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*}

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 \(30^\circ \, \text{and}\, 60^\circ\) .

Learner Experience 2.4.37.

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

Exploration 2.4.38. 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 below. Use them to help you investigate the relationships between the angles and their trigonometric ratios.

Instructions.
Move the slider between \(30^\circ\) and \(60^\circ\) and observe carefully.
  • 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 \(\tan 30^\circ\text{?}\)
  • 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?
  • Compare \(\sin 30^\circ\) with \(\cos 60^\circ\text{.}\) What do you notice? Why might these values be the same?
  • Compare \(\cos 30^\circ\) with \(\sin 60^\circ\text{.}\) What pattern do you observe?
  • 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?
  • Based on your observations, what relationship can you describe between complementary angles (angles that add up to \(90^\circ\))?
Discuss your findings with a partner and explain how the geometry of the triangle supports the calculator values.
Figure 2.4.39. Exploring Trigonometric Ratios of 30° and 60°

Key Takeaway 2.4.40.

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

Example 2.4.41.

Simplify the following without using tables or calculator.
  1. \(\displaystyle \sin \,30^\circ \, \cos\,45^\circ\)
  2. \(\displaystyle 8\,\cos\,45^\circ\, \sin\,45^\circ \)
  3. \(\displaystyle \sin\,60^\circ\, \cos\,45^\circ +\sin\,30^\circ\,\tan\,45^\circ\)
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*}
  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*}
  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*}

Example 2.4.42.

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:
  1. The diameter of the cone
  2. The height of the cone.
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.
Let:
  • \(r\) be the radius of the base,
  • \(h\) be the height of the cone,
  • \(l=3\sqrt{2\,cm}\) be the slant height (hypotenuse of the right-angled triangle).
Since the triangle formed is a right-angled isosceles triangle (because of the \(90^\circ\) vertex angle), we can say:
\(r=h\)
Since the right-angled triangle has radius \(r\) and height \(h\text{,}\) we use:
\(r^2=h^2=l^2\)
Since \(r=h\text{,}\) we substitute:
\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*}
  1. The diameter of the cone is:
    Diameter \(=2r=2(3)=6\,cm\)
  2. The height of the cone;
    Since \(h=r\text{,}\) we conclude:
    \(h=3\,cm\)

Checkpoint 2.4.43. Finding the Trigonometry Ratios in a 30°, 60° and 90° Triangle.

Exercises Exercises

1.

Using the triangle below, calculate the value of:
  1. \(\displaystyle \sin \,30^\circ\)
  2. \(\displaystyle \cos \,60^\circ\)
  3. \(\displaystyle \tan \,30^\circ\)
Answer.
  1. \(\displaystyle \sin \,30^\circ = \frac{1}{2}\)
  2. \(\displaystyle \cos \,60^\circ = \frac{1}{2}\)
  3. \(\displaystyle \tan \,30^\circ = \frac{1}{\sqrt{3}}\)

2.

Without using a calculator, find
\(\sin\,60^\circ+\cos\,30^\circ\)
Answer.
\(\sin\,60^\circ+\cos\,30^\circ =\sqrt{3}\)

3.

Given that \(\theta=45^\circ\text{,}\) calculate:
\(\tan^2\,\theta-\sin^2\,\theta\)
Answer.
\(\tan^2\,\theta-\sin^2\,\theta = \frac{1}{2}\)

4.

Given: \(\sin\,\theta = \frac{1}{2}\)
Find the angle \(\theta\text{,}\) where \(\theta^\circ \leq \theta \leq 90^\circ\)
Answer.
Angle \(\theta = 30^\circ\text{.}\)
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