Lösung 4.4:7c

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If we want to solve the equation
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<center> [[Image:4_4_7c-1(3).gif]] </center>
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<math>\text{cos 3}x=\text{sin 4}x</math>, we need an additional result which tells us for which values of
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<math>u</math>
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and
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<center> [[Image:4_4_7c-2(3).gif]] </center>
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<math>v</math>
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the equality
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<math>\text{cos }u=\text{sin }v</math>
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<center> [[Image:4_4_7c-3(3).gif]] </center>
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holds, but to get that we have to start with the equality
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<math>\cos u=\cos v</math>.
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So, we start by looking at the equality
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<math>\cos u=\cos v</math>
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We know that for fixed
 +
<math>u</math>
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there are two angles
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<math>v=u\text{ }</math>
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and
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<math>v=-\text{u}</math>
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in the unit circle which have the cosine value
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<math>\cos u</math>, i.e. their
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<math>x</math>
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-coordinate is equal to
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<math>\cos u</math>.
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[[Image:4_4_7_c1.gif|center]]
[[Image:4_4_7_c1.gif|center]]
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 +
Imagine now that the whole unit circle is rotated anti-clockwise an angle
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<math>{\pi }/{2}\;</math>. The line
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<math>x=\cos u</math>
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will become the line
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<math>y=\cos u</math>
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and the angles
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<math>u</math>
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and
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<math>-u</math>
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are rotated to
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<math>u+{\pi }/{2}\;</math>
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and
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<math>-u+{\pi }/{2}\;</math>, respectively.
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[[Image:4_4_7_c2.gif|center]]
[[Image:4_4_7_c2.gif|center]]
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The angles
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<math>u+{\pi }/{2}\;</math>
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and
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<math>-u+{\pi }/{2}\;</math>
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therefore have their
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<math>y</math>
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-coordinate, and hence sine value, equal to
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<math>\cos u</math>. In other words, the equality
 +
 +
 +
<math>\text{cos }u=\text{sin }v</math>
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 +
 +
holds for fixed
 +
<math>u</math>
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in the unit circle when
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<math>v=\pm u+{\pi }/{2}\;</math>, and more generally when
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 +
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<math>v=\pm u+\frac{\pi }{2}+2n\pi </math>
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(
 +
<math>n</math>
 +
an arbitrary integer).
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 +
For our equation
 +
<math>\text{cos 3}x=\text{sin 4}x</math>, this result means that
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<math>x\text{ }</math>
 +
must satisfy
 +
 +
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<math>4x=\pm 3x+\frac{\pi }{2}+2n\pi </math>
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This means that the solutions to the equation are
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 +
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<math>\left\{ \begin{array}{*{35}l}
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x=\frac{\pi }{2}+2n\pi \\
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x=\frac{\pi }{14}+\frac{2}{7}\pi n \\
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\end{array} \right.</math>
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(
 +
<math>n</math>
 +
an arbitrary integer)

Version vom 13:20, 1. Okt. 2008

If we want to solve the equation \displaystyle \text{cos 3}x=\text{sin 4}x, we need an additional result which tells us for which values of \displaystyle u and \displaystyle v the equality \displaystyle \text{cos }u=\text{sin }v holds, but to get that we have to start with the equality \displaystyle \cos u=\cos v.

So, we start by looking at the equality


\displaystyle \cos u=\cos v


We know that for fixed \displaystyle u there are two angles \displaystyle v=u\text{ } and \displaystyle v=-\text{u} in the unit circle which have the cosine value \displaystyle \cos u, i.e. their \displaystyle x -coordinate is equal to \displaystyle \cos u.


Imagine now that the whole unit circle is rotated anti-clockwise an angle \displaystyle {\pi }/{2}\;. The line \displaystyle x=\cos u will become the line \displaystyle y=\cos u and the angles \displaystyle u and \displaystyle -u are rotated to \displaystyle u+{\pi }/{2}\; and \displaystyle -u+{\pi }/{2}\;, respectively.


The angles \displaystyle u+{\pi }/{2}\; and \displaystyle -u+{\pi }/{2}\; therefore have their \displaystyle y -coordinate, and hence sine value, equal to \displaystyle \cos u. In other words, the equality


\displaystyle \text{cos }u=\text{sin }v


holds for fixed \displaystyle u in the unit circle when \displaystyle v=\pm u+{\pi }/{2}\;, and more generally when


\displaystyle v=\pm u+\frac{\pi }{2}+2n\pi ( \displaystyle n an arbitrary integer).

For our equation \displaystyle \text{cos 3}x=\text{sin 4}x, this result means that \displaystyle x\text{ } must satisfy


\displaystyle 4x=\pm 3x+\frac{\pi }{2}+2n\pi


This means that the solutions to the equation are


\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{2}+2n\pi \\ x=\frac{\pi }{14}+\frac{2}{7}\pi n \\ \end{array} \right. ( \displaystyle n an arbitrary integer)

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:7c“