Curriculum Alignment
- Strand
- 2.0 Measurements and Geometry
- Sub-Strand
- 2.4 Trigonometry 1
- Specific Learning Outcomes
- Relate sines and cosines of complementary angles
Subsection 2.4.5 Sines and Cosines of Complementary Angles
Teacher Resource 2.4.48.
To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.
Learner Experience 2.4.12.
\(\textbf{Work in pairs}\)
What you require: A pencil, a ruler, a scientific calculator (\(\textbf{for verification}\)), working material and a printed sine and cosine table.
You can use Table 2.4.40 if needed.
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Read from the tables the values of the following pairs of angles.
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\(\displaystyle \sin\,40^\circ \quad \cos\,50^\circ\)
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\(\displaystyle \cos\,30^\circ \quad \sin\,60^\circ\)
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\(\displaystyle \sin\,70^\circ \quad \cos\,20^\circ\)
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\(\displaystyle \sin\,80^\circ \quad \cos\,90^\circ\)
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What do you notice on the results obtained in "1"?
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Discuss your work with other learners in your class.
Exploration 2.4.13. Relationship of Complementary Ratios.
In a right triangle, the two acute angles are always complementary. But how are the trigonometric ratios of one angle related to those of its complement?
Explore the interactive below and investigate your ideas by following the guiding questions.
Instructions.
In a right-angled triangle, the two acute angles always sum to \(90^\circ\text{.}\) Use the slider to change the angle \(\theta\) and observe the results:
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Observe the Angles: As you increase \(\theta\) (the blue angle), what happens to the size of its complement, \(\phi\) (the green angle)? What is their sum at any given point?
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Track the Sides: Look at the side \(BC\text{.}\) Relative to the blue angle (\(\theta\)), is this the "Opposite" or "Adjacent" side? Now, look at it relative to the green angle (\(\phi\)). What role does it play there?
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Compare the Ratios: Compare the value of \(\sin(\theta)\) in the left column to \(\cos(\phi)\) in the right column. Why do you think these two different trigonometric functions result in the exact same decimal value?
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Find the Crossing Point: Move the slider until \(\theta\) is exactly \(45^\circ\text{.}\) What do you notice about all four ratios (Sine and Cosine for both angles) at this specific position?
Key Takeaway 2.4.50.
\(\textbf{Complementary angles}\) are two angles whose sum is \(90^\circ\) (or \(\frac{\pi}{2}\)radians). The sine and cosine functions of complementary angles have a special relationship:
\begin{align*}
\sin(90^\circ-\theta)=\amp cos \, \theta \\
\cos(90^\circ-\theta)=\amp \sin\, \theta
\end{align*}
\begin{align*}
y+x =\amp 90^\circ
\end{align*}
Generally,For any two complementary angles \(x \) and \(y\text{,}\) the following relationships hold:
This means that the sine of one angle is equal to the cosine of the other and vice versa.
Example 2.4.51.
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\(\cos\,45^\circ=\sin\,\alpha\text{.}\)
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\(\cos\,\beta=\sin\,5\beta\text{.}\)
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\(\sin\,2\alpha=\cos\,30^\circ\text{.}\)
Solution.
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\(\cos\,45^\circ=\sin\,\alpha\text{.}\)Implies that,\begin{align*} 45^\circ+\alpha=\amp 90^\circ \\ =\amp 90^\circ -45^\circ\\ =\amp 45^\circ \end{align*}
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\(\cos\,\beta=\sin\,5\beta\text{.}\)Implies that,\begin{align*} \beta+5\beta=\amp 90^\circ \\ 6\beta=\amp 90^\circ \\ \beta=\amp\frac{90^\circ}{6} \\ = \amp 15^\circ \end{align*}
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\(\sin\,2\alpha=\cos\,30^\circ\text{.}\)Implies that,\begin{align*} 2\alpha+30^\circ=\amp 90^\circ \\ 2\alpha=\amp 90^\circ-30^\circ \\ \alpha=\amp \frac{60^\circ}{2} \\ = \amp 30^\circ \end{align*}
Example 2.4.52.
Solution.
Also we have, \(A=\frac{1}{2}B\)
\begin{align*}
A= \amp \frac{1}{2}B\\
B= \amp 2A
\end{align*}
\begin{align*}
A+2A=\amp 90^\circ\\
3A=\amp 90^\circ\\
A= \amp \frac{90^\circ}{3}\\
=\amp 30^\circ
\end{align*}
Therefore,
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\begin{align*} sin\,A=\amp sin\,30^\circ \end{align*}By exact valuyes, \(\sin\,30^\circ = \frac{1}{2}\)
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\begin{align*} cos\,A=\amp cos\,30^\circ \end{align*}By exact values, \(\cos\,30^\circ = \frac{\sqrt{3}}{2}\)
Checkpoint 2.4.53. Finding Complementary Angles and Trigonometric Ratios.
Checkpoint 2.4.54. Finding Sines and Cosines of Complementary Angles.
Checkpoint 2.4.55. Finding Trigonometric Ratios in a Right Angled Triangle (1).
Load the question by clicking the button below.
Exercises Exercises
1.
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If \(\theta\) is an acute angle such that \(\sin\,(\theta)=\frac{3}{5}\text{,}\) find \(\cos(90^\circ-\theta)\text{.}\)
2.
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In the right-angled triangle below, find \(\sin\, \theta \) and \(\cos \,\theta\text{,}\) then verify that \(\sin \,\theta= \cos\, \theta\text{.}\)
Answer.
In the right-angled triangle, \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}\) and \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}\text{.}\) Since \(\sin \theta = \frac{3}{5}\) and \(\cos \theta = \frac{4}{5}\text{,}\) we observe that \(\sin \theta \neq \cos \theta\text{.}\)
3.
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Given that \(\cos(32^\circ) = 0.848\text{,}\) find the value of \(\sin(58^\circ)\) without using a calculator.
4.
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A ladder leans against a wall, making a \(65^\circ\) angle with the ground. Find the height at which the ladder touches the wall if the ladder is \(10\,m\) long.
