4.2 Trigonometrische Funktionen
Aus Online Mathematik Brückenkurs 1
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{{Fristående formel||<math>\tan 40^\circ = \frac{x}{5 \mbox{ m }}</math>}} | {{Fristående formel||<math>\tan 40^\circ = \frac{x}{5 \mbox{ m }}</math>}} | ||
| - | and since <math>\tan 40^\circ \approx 0{ | + | and since <math>\tan 40^\circ \approx 0\textrm{.}84</math> we get |
{{Fristående formel||<math> | {{Fristående formel||<math> | ||
| - | x = 5\,\mbox{m} \cdot \tan 40^\circ \approx 5\,\mbox{m} \cdot 0{ | + | x = 5\,\mbox{m} \cdot \tan 40^\circ \approx 5\,\mbox{m} \cdot 0\textrm{.}84 |
| - | = 4{ | + | = 4\textrm{.}2\,\mbox{m}\,\mbox{.}</math>}} |
</div> | </div> | ||
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| - | Equality of the two expressions for <math>\tan u</math> | + | Equality of the two expressions for <math>\tan u</math> gives |
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| - | gives | + | |
{{Fristående formel||<math>\frac{22}{40} = \frac{x}{60}</math>}} | {{Fristående formel||<math>\frac{22}{40} = \frac{x}{60}</math>}} | ||
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The definition of sine gives that | The definition of sine gives that | ||
{{Fristående formel||<math>\sin 38^\circ = \frac{x}{5}</math>}} | {{Fristående formel||<math>\sin 38^\circ = \frac{x}{5}</math>}} | ||
| - | and if we know that <math>\sin 38^\circ \approx 0{ | + | and if we know that <math>\sin 38^\circ \approx 0\textrm{.}616</math> then we get |
| - | {{Fristående formel||<math>x = 5 \cdot \sin 38^\circ \approx 5 \cdot 0{ | + | {{Fristående formel||<math>x = 5 \cdot \sin 38^\circ \approx 5 \cdot 0\textrm{.}616 \approx 3\textrm{.}1\,\mbox{.}</math>}} |
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<center>{{:4.2 - Figur - Rätvinklig triangel med vinkel u och sidor ½ och 1}}</center> | <center>{{:4.2 - Figur - Rätvinklig triangel med vinkel u och sidor ½ och 1}}</center> | ||
| - | With the help of | + | With the help of the Pythagorean theorem the side on the right can be determined |
<center> | <center> | ||
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| - | Using | + | Using the Pythagorean theorem, we can determine the length <math>x</math> of the diagonal, |
{{Fristående formel||<math> | {{Fristående formel||<math> | ||
x^2 = 1^2 + 1^2 | x^2 = 1^2 + 1^2 | ||
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| - | + | The Pythagorean theorem shows that the vertical side of either half-triangle is <math>x=\sqrt{3}/2</math>. From one of these half-triangles we get that | |
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= \frac{\pi}{2} + \frac{\pi}{6}</math>}} | = \frac{\pi}{2} + \frac{\pi}{6}</math>}} | ||
| - | shows that the angle <math>2\pi/3</math> lands in the the second quadrant and makes the angle <math>\pi/6</math> with the positive ''y''-axis. If we draw an extra triangle as in the figure below on the right, we see that | + | shows that the angle <math>2\pi/3</math> lands in the the second quadrant and makes the angle <math>\pi/6</math> with the positive ''y''-axis. If we draw an extra triangle as in the figure below on the right, we see that the <math>2\pi/3</math>- point on the unit circle has a ''y''-coordinate, which is equal to the adjacent side <math>\cos \frac{\pi}{6} = \sqrt{3}/2</math>. So we have that |
{{Fristående formel||<math> | {{Fristående formel||<math> | ||
\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}\,\mbox{.}</math>}} | \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}\,\mbox{.}</math>}} | ||
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<br> | <br> | ||
<br> | <br> | ||
| - | By drawing the graphs <math>y=\cos x</math> and <math>y=x^2</math> we see that the curves intersect in two points. So there are two ''x''- values for which the corresponding ''y''-values are equal. In other words, the equation has two solutions. | + | By drawing the graphs <math>y=\cos x</math> and <math>y=x^2</math> we see that the curves intersect in two points. So there are two ''x''-values for which the corresponding ''y''-values are equal. In other words, the equation has two solutions. |
<center>{{:4.2 - Figur - Kurvorna y = cos x och y = x²}}</center> | <center>{{:4.2 - Figur - Kurvorna y = cos x och y = x²}}</center> | ||
Version vom 14:09, 19. Aug. 2008
Contents:
- The trigonometric functions cosine, sine and tangent.
Learning outcomes:
After this section, you will have learned :
- The concepts of acute, obtuse and right angles.
- The definition of cosine, sine and tangent in the unit circle.
- The values of cosine, sine and tangent for the standard angles \displaystyle 0, \displaystyle \pi/6 , \displaystyle \pi/4 , \displaystyle \pi/3 and \displaystyle \pi/2 by heart.
- To determine the values of cosine, sine and tangent of arguments that can be reduced to a standard angle in a quadrant of the unit circle.
- To sketch graphs of cosine, sine and tangent.
- To solve trigonometric problems involving right-angled triangles.
Trigonometry of right-angled triangles
In the right-angled triangle below the ratio between the opposite side \displaystyle a and the adjacent side \displaystyle b is called the tangent of the angle \displaystyle u and is written as \displaystyle \tan u.
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4.2 - Figur - Rätvinklig triangel med vinkeln u och kateterna a och b |
\displaystyle \tan u = \displaystyle \frac{a}{b} |
The value of the ratio \displaystyle \frac{a}{b} is not dependent on the size of the triangle, but only on the angle \displaystyle u. For different values of the angle, you can get the equivalent value of the tangent either from a trigonometric table or by using a calculator (the relevent button is usually named tan).
Example 1
How high is the flagpole?
The flagpole and its shadow form a rectangular triangle where the vertical side is unknown (marked with \displaystyle x below).
From the definition of tangent, we have that Vorlage:Fristående formel
and since \displaystyle \tan 40^\circ \approx 0\textrm{.}84 we get Vorlage:Fristående formel
Example 2
Determine the length of the side designated with the \displaystyle x in the figure.
If we call the angle at the far left \displaystyle u there are two ways to construct an expression for \displaystyle \tan u.
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4.2 - Figur - Dubbeltriangel med den lilla triangeln framhävd |
\displaystyle \tan u = \displaystyle \frac{22}{40} |
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4.2 - Figur - Dubbeltriangel med den stora triangeln framhävd |
\displaystyle \tan u = \dfrac{x}{60} |
Equality of the two expressions for \displaystyle \tan u gives Vorlage:Fristående formel
which leads to \displaystyle x=60 \cdot \displaystyle \frac{22}{40} = 33.
There are two other ratios in right-angled triangles that have special names, and one is \displaystyle \cos u = b/c ("cosine of \displaystyle u") and the other \displaystyle \sin u = a/c (" sine of \displaystyle u").
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4.2 - Figur - Rätvinklig triangel med vinkeln u och sidorna a, b och c |
\displaystyle \begin{align*} \cos u &= \frac{b}{c}\\[8pt] \sin u &= \frac{a}{c} \end{align*} |
Like the tangent the ratios that define the cosine and sine do not depend on the size of the triangle, but only on the angle \displaystyle u.
Example 3
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4.2 - Figur - Rätvinklig triangel med vinkeln u och sidor 3, 4 och 5 |
In the triangle on the left Vorlage:Fristående formel | ||
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4.2 - Figur - Rätvinklig triangel med vinkeln 38° och sidor x och 5 |
The definition of sine gives that Vorlage:Fristående formel and if we know that \displaystyle \sin 38^\circ \approx 0\textrm{.}616 then we get Vorlage:Fristående formel | ||
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4.2 - Figur - Rätvinklig triangel med vinkeln 34° och sidor 3 och x |
Cosine is the ratio between the adjacent side and the hypotenuse Vorlage:Fristående formel Thus Vorlage:Fristående formel |
Example 4
Determine \displaystyle \sin u in the triangle
With the help of the Pythagorean theorem the side on the right can be determined
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4.2 - Figur - Rätvinklig triangel med vinkel u och sidor ½, x och 1 |
\displaystyle 1^2= \bigl( \tfrac{1}{2} \bigr)^2 + x^2 \quad\Leftrightarrow\quad x = \frac{\sqrt{3}}{2} |
and thus \displaystyle \sin u = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}.
Some standard angles
For some angles namely 30°, 45° and 60° it is relatively easy to calculate the exact values of the trigonometric functions.
Example 5
We start with a square having sides of length 1. A diagonal of the square divides the right angles in opposite corners into two equal parts of 45°.
Using the Pythagorean theorem, we can determine the length \displaystyle x of the diagonal,
Vorlage:Fristående formel
Each triangle has the diagonal as the hypotenuse, thus we can obtain the value of the trigonometric functions for the angle \displaystyle 45^\circ.
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4.2 - Figur - Enhetskvadrat vars halva är en rätvinklig triangel |
\displaystyle \begin{align*} \cos 45^\circ &= \frac{1}{\sqrt{2}}\\[8pt] \sin 45^\circ &= \frac{1}{\sqrt{2}}\\[8pt] \tan 45^\circ &= \frac{1}{1}= 1\\ \end{align*} |
Example 6
Imagine an equilateral triangle where all sides have length 1. The angles of the triangle are all 60°. The triangle can be divided into two halves by a line that divides the angle at the top in equal parts.
The Pythagorean theorem shows that the vertical side of either half-triangle is \displaystyle x=\sqrt{3}/2. From one of these half-triangles we get that
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\displaystyle \begin{align*} \cos 30^\circ &= \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,;\\[8pt] \sin 30^\circ &= \frac{1/2}{1} = \frac{1}{2}\,;\\[8pt] \tan 30^\circ &= \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}\,;\\ \end{align*} \qquad\quad \begin{align*} \cos 60^\circ &= \frac{1/2}{1} = \frac{1}{2}\\[8pt] \sin 60^\circ &= \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\\[8pt] \tan 60^\circ &= \frac{\sqrt{3}/2}{1/2}=\sqrt{3}\\ \end{align*} |
Trigonometric functions for general angles
For angles of less than 0° or greater than 90° the trigonometric functions are defined using the unit circle (that is the circle that has its centre at the origin and has a radius 1).
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The trigonometric functions \displaystyle \cos u and \displaystyle \sin u are x- and y- coordinates of the intersection between the unit circle and the radial line that forms the angle \displaystyle u with the positive x-axis. |
4.2 - Figur - Enhetscirkeln med vinkeln u och punkten (cos u, sin u) |
Tangent function is defined as
and the value of the tangent can be interpreted as the slope for the radial line.
Example 7
From the figures below, we obtain the values of cosine and sine.
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4.2 - Figur - Enhetscirkeln med vinkeln 104° och punkten (-0,24; 0,97) |
\displaystyle \begin{align*} \cos 104^\circ &\approx -0{,}24\\[8pt] \sin 104^\circ &\approx 0{,}97\\[8pt] \tan 104^\circ &\approx \dfrac{0{,}97}{-0{,}24} \approx -4{,}0\\ \end{align*} | |
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4.2 - Figur - Enhetscirkeln med vinkeln 201° och punkten (-0,93; -0,36) |
\displaystyle \begin{align*} \cos 201^\circ &\approx -0{,}93\\[8pt] \sin 201^\circ &\approx -0{,}36\\[8pt] \tan 201^\circ &\approx \dfrac{-0{,}36}{-0{,}93} \approx 0{,}4\\ \end{align*} |
Example 8
Which sign do the following have?
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4.2 - Figur - Enhetscirkeln med vinkeln 209° och linjen x = cos 209° | |
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4.2 - Figur - Enhetscirkeln med vinkeln 133° och linjen y = sin 133° | |
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4.2 - Figur - Enhetscirkeln med vinkeln -40° och linjen med riktningskoefficient tan -40° |
Example 9
Calculate \displaystyle \,\sin\frac{2\pi}{3}.
Rewriting
Vorlage:Fristående formel
shows that the angle \displaystyle 2\pi/3 lands in the the second quadrant and makes the angle \displaystyle \pi/6 with the positive y-axis. If we draw an extra triangle as in the figure below on the right, we see that the \displaystyle 2\pi/3- point on the unit circle has a y-coordinate, which is equal to the adjacent side \displaystyle \cos \frac{\pi}{6} = \sqrt{3}/2. So we have that Vorlage:Fristående formel
The trigonometric functions graphs
In the last section, we used a unit circle to define cosine and sine of arbitrary angles and we often will use the unit circle in the future, for example, to derive trigonometric relationships and solve trigonometric equations. However, there are certain characteristics of the trigonometric functions that are better illustrated by drawing their graphs.
In these graphs, we might observe several things more clearly than in the unit circle. Some examples are:
- The curves for cosine and sine repeat themselves after a change in angle of \displaystyle 2\pi, that is the \displaystyle \cos (x+2\pi) = \cos x and \displaystyle \sin (x+2\pi) = \sin x. For the unit circle \displaystyle 2\pi corresponds to a revolution, and after a complete revolution angles return to the same location on the unit circle and therefore have the same coordinates.
- The curve for the tangent repeats itself after a change in angle of \displaystyle \pi, that is \displaystyle \tan (x+\pi) = \tan x. Two angles which differ by \displaystyle \pi share the same line through the origin of the unit circle and thus their radial lines have the same slope.
- Except for a phase shift of \displaystyle \pi/2 the curves for cosine and sine are identical, that is \displaystyle \cos x = \sin (x+ \pi/2); more about this in the next section.
The curves can also be important when examining trigonometric equations. With a simple sketch, you can often get an idea of how many solutions an equation has, and where the solutions lie.
Example 10
How many solutions has the equation \displaystyle \cos x = x^2 ( where \displaystyle x is measured in radians)?
By drawing the graphs \displaystyle y=\cos x and \displaystyle y=x^2 we see that the curves intersect in two points. So there are two x-values for which the corresponding y-values are equal. In other words, the equation has two solutions.
Study advice
Basic and final tests
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
Keep in mind that:
If you have studied trigonometry, then you should not be afraid to use it in geometric problems. It often produces a simpler solution.
You may need to spend a lot of time on understanding how to use a unit circle to define the trigonometric functions.
Get into the habit of calculating with precise trigonometric values. It provides a good training in calculating fractions and eventually algebraic rational expressions.
Reviews
For those of you who want to deepen your studies or need more detailed explanations consider the following references
Learn more about trigonometry in Per Edström "Interactive Mathematics"
Learn more about trigonometry in the English Wikipedia
Learn more about the unit circle in the English Wikipedia
Useful web sites
