4.2 Trigonometric functions

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'''Learning outcomes:'''
'''Learning outcomes:'''
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After this section, you will have learned :
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After this section, you will have learned:
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*The concepts of acute, obtuse and right angles.
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*The definition of acute, obtuse and right angles.
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*The definition of cosine, sine and tangent in the unit circle.
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*The definition of cosine, sine and tangent.
*The values of cosine, sine and tangent for the standard angles <math>0</math>, <math>\pi/6</math> , <math>\pi/4</math> , <math>\pi/3</math> and <math>\pi/2</math> by heart.
*The values of cosine, sine and tangent for the standard angles <math>0</math>, <math>\pi/6</math> , <math>\pi/4</math> , <math>\pi/3</math> and <math>\pi/2</math> by heart.
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*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.
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*To determine the values of cosine, sine and tangent of arguments that can be reduced to a standard angle.
* To sketch graphs of cosine, sine and tangent.
* To sketch graphs of cosine, sine and tangent.
*To solve trigonometric problems involving right-angled triangles.
*To solve trigonometric problems involving right-angled triangles.
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== Trigonometry of right-angled triangles ==
== Trigonometry of right-angled triangles ==
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In the right-angled triangle below the ratio between the opposite side <math>a</math> and the adjacent side <math>b</math> is called the tangent of the angle <math>u</math> and is written as <math>\tan u</math>.
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In the right-angled triangle below, the ratio between the length <math>a</math> of the side opposite the angle and the length <math>b</math> of the adjacent side is called the tangent of the angle <math>u</math>, and is written as <math>\tan u</math>.
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{{:4.2 - Figur - Rätvinklig triangel med vinkeln u och kateterna a och b}}
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{{:4.2 - Figure - A right-angled triangle with angle u and legs a and b}}
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The value of the ratio <math>\frac{a}{b}</math> is not dependent on the size of the triangle, but only on the angle <math>u</math>. 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).
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The value of the ratio <math>\frac{a}{b}</math> is not dependent on the size of the triangle, but only on the angle <math>u</math>. For different values of the angle, you can get the value of the tangent either from a trigonometric table or by using a calculator (the relevent button is usually named tan).
<div class="exempel">
<div class="exempel">
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How high is the flagpole?
How high is the flagpole?
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<center>{{:4.2 - Figur - Flaggstång}}</center>
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<center>{{:4.2 - Figure - Flagpole}}</center>
The flagpole and its shadow form a rectangular triangle where the vertical side is unknown (marked with <math>x</math> below).
The flagpole and its shadow form a rectangular triangle where the vertical side is unknown (marked with <math>x</math> below).
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<center>{{:4.2 - Figur - Flaggstångstriangel}}</center>
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<center>{{:4.2 - Figure - Flagpole triangle}}</center>
From the definition of tangent, we have that
From the definition of tangent, we have that
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Determine the length of the side designated with the <math>x</math> in the figure.
Determine the length of the side designated with the <math>x</math> in the figure.
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<center>{{:4.2 - Figur - Dubbeltriangel}}</center>
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<center>{{:4.2 - Figure - A double triangle}}</center>
If we call the angle at the far left <math>u</math> there are two ways to construct an expression for <math>\tan u</math>.
If we call the angle at the far left <math>u</math> there are two ways to construct an expression for <math>\tan u</math>.
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{{:4.2 - Figur - Dubbeltriangel med den lilla triangeln framhävd}}
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{{:4.2 - Figure - A double triangle with the small triangle highlighted}}
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{{:4.2 - Figur - Dubbeltriangel med den stora triangeln framhävd}}
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{{:4.2 - Figure - A double triangle with the larger triangle highlighted}}
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</div>
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There are two other ratios in right-angled triangles that have special names, and one is <math>\cos u = b/c</math> ("cosine of <math>u</math>") and the other <math>\sin u = a/c</math> (" sine of <math>u</math>").
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There are two other ratios in right-angled triangles that have special names. The first is <math>\cos u = b/c</math> ("cosine of <math>u</math>") and the second is <math>\sin u = a/c</math> (" sine of <math>u</math>").
<center>
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{{:4.2 - Figur - Rätvinklig triangel med vinkeln u och sidorna a, b och c}}
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{{:4.2 - Figure - A right-angled triangle with angle u and sides a, b and c}}
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{{:4.2 - Figur - Rätvinklig triangel med vinkeln u och sidor 3, 4 och 5}}
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{{:4.2 - Figure - A right-angled triangle with angle u and sides 3, 4 and 5}}
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{{:4.2 - Figur - Rätvinklig triangel med vinkeln 38° och sidor x och 5}}
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{{:4.2 - Figure - A right-angled triangle with angle 38° and sides x and 5}}
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The definition of sine gives that
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From the definition of sine we have
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{{Displayed math||<math>\sin 38^\circ = \frac{x}{5}</math>}}
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{{Displayed math||<math>\sin 38^\circ = \frac{x}{5}</math>,}}
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and if we know that <math>\sin 38^\circ \approx 0\textrm{.}616</math> then we get
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and if we know that <math>\sin 38^\circ \approx 0\textrm{.}616</math> then
{{Displayed math||<math>x = 5 \cdot \sin 38^\circ \approx 5 \cdot 0\textrm{.}616 \approx 3\textrm{.}1\,\mbox{.}</math>}}
{{Displayed math||<math>x = 5 \cdot \sin 38^\circ \approx 5 \cdot 0\textrm{.}616 \approx 3\textrm{.}1\,\mbox{.}</math>}}
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{{:4.2 - Figur - Rätvinklig triangel med vinkeln 34° och sidor 3 och x}}
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{{:4.2 - Figure - A right-angled triangle with angle 34° and sides 3 and x}}
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Cosine is the ratio between the adjacent side and the hypotenuse
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Cosine is the ratio between the adjacent side and the hypotenuse, so
{{Displayed math||<math>\cos 34^\circ = \frac{3}{x}\,\mbox{.}</math>}}
{{Displayed math||<math>\cos 34^\circ = \frac{3}{x}\,\mbox{.}</math>}}
Thus
Thus
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Determine <math>\sin u</math> in the triangle
Determine <math>\sin u</math> in the triangle
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<center>{{:4.2 - Figur - Rätvinklig triangel med vinkel u och sidor ½ och 1}}</center>
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<center>{{:4.2 - Figure - A right-angled triangle with angle u and sides ½ and 1}}</center>
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With the help of the Pythagorean theorem the side on the right can be determined
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With the help of the Pythagorean theorem the side on the right can be determined:
<center>
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{{:4.2 - Figur - Rätvinklig triangel med vinkel u och sidor ½, x och 1}}
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{{:4.2 - Figure - A right-angled triangle with angle u and sides ½, x and 1}}
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== Some standard angles ==
== Some standard angles ==
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For some angles namely 30°, 45° and 60° it is relatively easy to calculate the exact values of the trigonometric functions.
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For some angles, namely 30°, 45° and 60°, it is relatively easy to calculate the exact values of the trigonometric functions.
<div class="exempel">
<div class="exempel">
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<center>{{:4.2 - Figur - Två enhetskvadrater}}</center>
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<center>{{:4.2 - Figure - Two unit squares}}</center>
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Using the Pythagorean theorem, we can determine the length <math>x</math> of the diagonal,
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Using the Pythagorean theorem, we can determine the length <math>x</math> of the diagonal:
{{Displayed math||<math>
{{Displayed math||<math>
x^2 = 1^2 + 1^2
x^2 = 1^2 + 1^2
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x = \sqrt{1^2 + 1^2} = \sqrt{2}\,\mbox{.}</math>}}
x = \sqrt{1^2 + 1^2} = \sqrt{2}\,\mbox{.}</math>}}
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Each triangle has the diagonal as the hypotenuse, thus we can obtain the value of the trigonometric functions for the angle <math>45^\circ</math>.
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Each triangle has the diagonal as the hypotenuse. Thus we can obtain the value of the trigonometric functions for the angle <math>45^\circ</math>:
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{{:4.2 - Figur - Enhetskvadrat vars halva är en rätvinklig triangel}}
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{{:4.2 - Figure - The unit square and half of it as a right-angled triangle}}
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<center>{{:4.2 - Figur - Två liksidiga trianglar}}</center>
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<center>{{:4.2 - Figure - Two equilateral triangles}}</center>
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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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The Pythagorean theorem allows us to calculate that the length of the vertical side of half-triangle is <math>x=\sqrt{3}/2</math>. Using the definitions we then get that
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{{:4.2 - Figur - En halv liksidig triangel}}
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{{:4.2 - Figure - A half of an equilateral triangle}}
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<math>\begin{align*}
<math>\begin{align*}
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\cos 30^\circ &= \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,;\\[8pt]
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\cos 30^\circ &= \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,\\[8pt]
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\sin 30^\circ &= \frac{1/2}{1} = \frac{1}{2}\,;\\[8pt]
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\sin 30^\circ &= \frac{1/2}{1} = \frac{1}{2}\,\\[8pt]
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\tan 30^\circ &= \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}\,;\\
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\tan 30^\circ &= \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}\,\\
\end{align*}
\end{align*}
\qquad\quad
\qquad\quad
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== Trigonometric functions for general angles ==
== Trigonometric functions for general angles ==
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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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For angles 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 radius 1).
<div class="regel">
<div class="regel">
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The trigonometric functions <math>\cos u</math> and <math>\sin u</math> are ''x''- and ''y''- coordinates of the intersection between the unit circle and the radial line that forms the angle <math>u</math> with the positive ''x''-axis.
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The trigonometric functions <math>\cos u</math> and <math>\sin u</math> are the ''x''- and ''y''- coordinates of the point on the unit circle reached by turning through the angle <math>u</math>, as shown in the diagram on the right.
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{{:4.2 - Figur - Enhetscirkeln med vinkeln u och punkten (cos u, sin u)}}
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{{:4.2 - Figure - The unit circle with angle u and the point (cos u, sin u)}}
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</div>
</div>
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Tangent function is defined as
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The tangent function is then defined as
{{Displayed math||<math>\tan u = \displaystyle\frac{\sin u}{\cos u}</math>}}
{{Displayed math||<math>\tan u = \displaystyle\frac{\sin u}{\cos u}</math>}}
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</ol>
</ol>
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{{:4.2 - Figur - Enhetscirkeln med vinkeln 104° och punkten (-0,24; 0,97)}}
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{{:4.2 - Figure - The unit circle with the angle 104° and the point (-0.24,0.97)}}
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<math>\begin{align*}
<math>\begin{align*}
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\cos 104^\circ &\approx -0{,}24\\[8pt]
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\cos 104^\circ &\approx -0\mbox{.}24\\[8pt]
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\sin 104^\circ &\approx 0{,}97\\[8pt]
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\sin 104^\circ &\approx 0\mbox{.}97\\[8pt]
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\tan 104^\circ &\approx \dfrac{0{,}97}{-0{,}24} \approx -4{,}0\\
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\tan 104^\circ &\approx \dfrac{0\mbox{.}97}{-0\mbox{.}24} \approx -4\mbox{.}0\\
\end{align*}</math>
\end{align*}</math>
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</ol>
</ol>
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{{:4.2 - Figur - Enhetscirkeln med vinkeln 201° och punkten (-0,93; -0,36)}}
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{{:4.2 - Figure - The unit circle with angle 201° and the point (-0.93,-0.36)}}
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<math>\begin{align*}
<math>\begin{align*}
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\cos 201^\circ &\approx -0{,}93\\[8pt]
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\cos 201^\circ &\approx -0\mbox{.}93\\[8pt]
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\sin 201^\circ &\approx -0{,}36\\[8pt]
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\sin 201^\circ &\approx -0\mbox{.}36\\[8pt]
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\tan 201^\circ &\approx \dfrac{-0{,}36}{-0{,}93} \approx 0{,}4\\
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\tan 201^\circ &\approx \dfrac{-0\mbox{.}36}{-0\mbox{.}93} \approx 0\mbox{.}4\\
\end{align*}</math>
\end{align*}</math>
|}
|}
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{{:4.2 - Figur - Enhetscirkeln med vinkeln 209° och linjen x = cos 209°}}
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{{:4.2 - Figure - The unit circle with angle 209° and the line 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 - Figure - The unit circle with the angle 133° and the line y = sin 133°}}
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<br>
<br>
<br>
<br>
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By drawing angle<math>-40^\circ</math> in the unit circle one obtains a radial line which has a negative slope, i.e. <math>\tan (-40^\circ)</math> is negative. </li>
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By drawing angle <math>-40^\circ</math> in the unit circle one obtains a radial line which has a negative slope, so that <math>\tan (-40^\circ)</math> is negative. </li>
</ol>
</ol>
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{{:4.2 - Figur - Enhetscirkeln med vinkeln -40° och linjen med riktningskoefficient tan -40°}}
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{{:4.2 - Figure - The unit circle with the angle -40° and the line with slope tan -40°}}
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<br>
<br>
<br>
<br>
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Rewriting
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Note that
{{Displayed math||<math>
{{Displayed math||<math>
\frac{2\pi}{3} = \frac{4\pi}{6}
\frac{2\pi}{3} = \frac{4\pi}{6}
= \frac{3\pi+ \pi}{6}
= \frac{3\pi+ \pi}{6}
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= \frac{\pi}{2} + \frac{\pi}{6}</math>}}
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= \frac{\pi}{2} + \frac{\pi}{6}</math>.}}
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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
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This shows that the point on the unit circle corresponding to the angle <math>2\pi/3</math> is 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
{{Displayed math||<math>
{{Displayed math||<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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== The trigonometric functions graphs ==
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== Graphs of the trigonometric functions ==
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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.
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In the last section we used a unit circle to define the cosine and sine of arbitrary angles, and we will often 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.
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<center><small>The graph of the sine function </small></center>
<center><small>The graph of the sine function </small></center>
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<center>{{:4.2 - Figur - Cosinuskurva}}</center>
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<center>{{:4.2 - Figure - A cosine curve}}</center>
<center><small>The graph of the cosine function </small></center>
<center><small>The graph of the cosine function </small></center>
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In these graphs, we might observe several things more clearly than in the unit circle. Some examples are:
In these graphs, we might observe several things more clearly than in the unit circle. Some examples are:
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* The curves for cosine and sine repeat themselves after a change in angle of <math>2\pi</math>, that is the <math>\cos (x+2\pi) = \cos x</math> and <math>\sin (x+2\pi) = \sin x</math>. For the unit circle <math>2\pi</math> 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.
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* The curves for cosine and sine repeat themselves after a change in angle of <math>2\pi</math>, that is <math>\cos (x+2\pi) = \cos x</math> and <math>\sin (x+2\pi) = \sin x</math>. To see why this is true, note that on the unit circle <math>2\pi</math> corresponds to a complete revolution, and after a complete revolution we return to the same point on the circle.
*The curve for the tangent repeats itself after a change in angle of <math>\pi</math>, that is <math>\tan (x+\pi) = \tan x</math>. Two angles which differ by <math>\pi</math> share the same line through the origin of the unit circle and thus their radial lines have the same slope.
*The curve for the tangent repeats itself after a change in angle of <math>\pi</math>, that is <math>\tan (x+\pi) = \tan x</math>. Two angles which differ by <math>\pi</math> share the same line through the origin of the unit circle and thus their radial lines have the same slope.
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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.
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<center>{{:4.2 - Figur - Kurvorna y = cos x och y = x²}}</center>
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<center>{{:4.2 - Figure - The curves y = cos x and y = x²}}</center>
</div>
</div>
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'''Keep in mind that: '''
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'''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.
If you have studied trigonometry, then you should not be afraid to use it in geometric problems. It often produces a simpler solution.
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You may need to spend a lot of time on understanding how to use a unit circle to define the trigonometric functions.
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You may need to spend a lot of time understanding how to use a unit circle to define the trigonometric functions.
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Get into the habit of calculating with precise trigonometric values. It provides a good training in calculating fractions and eventually algebraic rational expressions.
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You should get into the habit of calculating with precise trigonometric values. It is good training in calculating fractions and will eventually help you handle algebraic rational expressions.
'''Reviews'''
'''Reviews'''
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For those of you who want to deepen your studies or need more detailed explanations consider the following references
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For those of you who want to deepen your understanding or need more detailed explanations consider the following references:
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[http://dooku.miun.se/per.edstrom/interaktiv_matematik/trigonometri/cos_even.html Learn more about trigonometry in Per Edström "Interactive Mathematics"]
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[http://en.wikipedia.org/wiki/Trigonometric_function Learn more about trigonometry from Wikipedia]
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[http://en.wikipedia.org/wiki/Trigonometric_function Learn more about trigonometry in the English Wikipedia]
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[http://en.wikipedia.org/wiki/Unit_circle Learn more about the unit circle from Wikipedia]
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[http://en.wikipedia.org/wiki/Unit_circle Learn more about the unit circle in the English Wikipedia]
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'''Useful web sites'''
'''Useful web sites'''
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[http://www.math.kth.se/online/images/sinus_och_cosinus_i_enhetscirkeln.swf Experiment with the sine and cosine in the unit circle]
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[http://www.math.kth.se/online/images/sinus_och_cosinus_i_enhetscirkeln.swf Experiment with the sine and cosine functions]
[http://www.math.psu.edu/dlittle/java/geometry/euclidean/toolbox.html Experiment with Euclidean geometry]
[http://www.math.psu.edu/dlittle/java/geometry/euclidean/toolbox.html Experiment with Euclidean geometry]
</div>
</div>

Current revision

       Theory          Exercises      

Contents:

  • The trigonometric functions cosine, sine and tangent.

Learning outcomes:

After this section, you will have learned:

  • The definition of acute, obtuse and right angles.
  • The definition of cosine, sine and tangent.
  • 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.
  • 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 length \displaystyle a of the side opposite the angle and the length \displaystyle b of the adjacent side is called the tangent of the angle \displaystyle u, and is written as \displaystyle \tan u.

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\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 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?

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The flagpole and its shadow form a rectangular triangle where the vertical side is unknown (marked with \displaystyle x below).

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From the definition of tangent, we have that

\displaystyle \tan 40^\circ = \frac{x}{5 \mbox{ m }}

and since \displaystyle \tan 40^\circ \approx 0\textrm{.}84 we get

\displaystyle
 x = 5\,\mbox{m} \cdot \tan 40^\circ \approx 5\,\mbox{m} \cdot 0\textrm{.}84
   = 4\textrm{.}2\,\mbox{m}\,\mbox{.}

Example 2

Determine the length of the side designated with the \displaystyle x in the figure.

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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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\displaystyle \tan u = \displaystyle \frac{22}{40}

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\displaystyle \tan u = \dfrac{x}{60}

Equality of the two expressions for \displaystyle \tan u gives

\displaystyle \frac{22}{40} = \frac{x}{60}

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. The first is \displaystyle \cos u = b/c ("cosine of \displaystyle u") and the second is \displaystyle \sin u = a/c (" sine of \displaystyle u").

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\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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In the triangle on the left

\displaystyle \begin{align*}

\cos u &= \tfrac{4}{5}\\[6pt] \sin u &= \tfrac{3}{5} \end{align*}

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From the definition of sine we have

\displaystyle \sin 38^\circ = \frac{x}{5},

and if we know that \displaystyle \sin 38^\circ \approx 0\textrm{.}616 then

\displaystyle x = 5 \cdot \sin 38^\circ \approx 5 \cdot 0\textrm{.}616 \approx 3\textrm{.}1\,\mbox{.}

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Cosine is the ratio between the adjacent side and the hypotenuse, so

\displaystyle \cos 34^\circ = \frac{3}{x}\,\mbox{.}

Thus

\displaystyle x=\frac{3}{\cos 34^\circ}\,\mbox{.}

Example 4

Determine \displaystyle \sin u in the triangle

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With the help of the Pythagorean theorem the side on the right can be determined:

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\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°.


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Using the Pythagorean theorem, we can determine the length \displaystyle x of the diagonal:

\displaystyle
 x^2 = 1^2 + 1^2
 \quad \Leftrightarrow \quad
 x = \sqrt{1^2 + 1^2} = \sqrt{2}\,\mbox{.}

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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\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.


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The Pythagorean theorem allows us to calculate that the length of the vertical side of half-triangle is \displaystyle x=\sqrt{3}/2. Using the definitions we then 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 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 radius 1).

The trigonometric functions \displaystyle \cos u and \displaystyle \sin u are the x- and y- coordinates of the point on the unit circle reached by turning through the angle \displaystyle u, as shown in the diagram on the right.

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The tangent function is then defined as

\displaystyle \tan u = \displaystyle\frac{\sin u}{\cos u}

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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\displaystyle \begin{align*} \cos 104^\circ &\approx -0\mbox{.}24\\[8pt] \sin 104^\circ &\approx 0\mbox{.}97\\[8pt] \tan 104^\circ &\approx \dfrac{0\mbox{.}97}{-0\mbox{.}24} \approx -4\mbox{.}0\\ \end{align*}

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\displaystyle \begin{align*} \cos 201^\circ &\approx -0\mbox{.}93\\[8pt] \sin 201^\circ &\approx -0\mbox{.}36\\[8pt] \tan 201^\circ &\approx \dfrac{-0\mbox{.}36}{-0\mbox{.}93} \approx 0\mbox{.}4\\ \end{align*}

Example 8

Which sign do the following have?

  1. \displaystyle \cos 209^\circ

    Since the angle \displaystyle 209^\circ can be written as \displaystyle 209^\circ = 180^\circ + 29^\circ the angle corresponds to a point on the unit circle which lies in the third quadrant. The point has a negative x-coordinate, which means that \displaystyle \cos 209^\circ is negative .

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  1. \displaystyle \sin 133^\circ

    The angle \displaystyle 133^\circ is equal to \displaystyle 90^\circ + 43^\circ and gives a point on the unit circle which lies in the second quadrant. The quadrant has points with positive y-coordinate, and therefore \displaystyle \sin 133^\circ is positive.

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  1. \displaystyle \tan (-40^\circ)

    By drawing angle \displaystyle -40^\circ in the unit circle one obtains a radial line which has a negative slope, so that \displaystyle \tan (-40^\circ) is negative.

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Example 9

Calculate \displaystyle \,\sin\frac{2\pi}{3}.

Note that

\displaystyle
 \frac{2\pi}{3} = \frac{4\pi}{6}
                = \frac{3\pi+ \pi}{6}
                = \frac{\pi}{2} + \frac{\pi}{6}.

This shows that the point on the unit circle corresponding to the angle \displaystyle 2\pi/3 is 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

\displaystyle
 \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}\,\mbox{.}

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Graphs of the trigonometric functions

In the last section we used a unit circle to define the cosine and sine of arbitrary angles, and we will often 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.


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The graph of the sine function

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The graph of the cosine function

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The graph of the tangent function


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 \displaystyle \cos (x+2\pi) = \cos x and \displaystyle \sin (x+2\pi) = \sin x. To see why this is true, note that on the unit circle \displaystyle 2\pi corresponds to a complete revolution, and after a complete revolution we return to the same point on the circle.
  • 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.

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Exercises

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 understanding how to use a unit circle to define the trigonometric functions.

You should get into the habit of calculating with precise trigonometric values. It is good training in calculating fractions and will eventually help you handle algebraic rational expressions.

Reviews

For those of you who want to deepen your understanding or need more detailed explanations consider the following references:

Learn more about trigonometry from Wikipedia

Learn more about the unit circle from Wikipedia


Useful web sites

Experiment with the sine and cosine functions

Experiment with Euclidean geometry