Lösung 4.4:5b

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Let's first investigate when the equality
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<center> [[Image:4_4_5b-1(2).gif]] </center>
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<math>\tan u=\tan v</math>
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<center> [[Image:4_4_5b-2(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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is satisfied. Because
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<math>u</math>
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can be interpreted as the slope (gradient) of the line which makes an angle
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<math>u</math>
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with the positive
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<math>x</math>
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-axis, we see that for a fixed value of tan u, there are two angles
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<math>v</math>
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in the unit circle with this slope:
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<math>v=u</math>
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and
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<math>v=u+\pi </math>
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[[Image:4_4_5_b.gif|center]]
[[Image:4_4_5_b.gif|center]]
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slope
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<math>=\text{ tan }u</math>
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slope
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<math>=\text{ tan }u</math>
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 +
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The angle
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<math>v</math>
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has the same slope after every half turn, so if we add multiples of
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<math>\pi \text{ }</math>
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to
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<math>u</math>, we will obtain all the angles
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<math>v</math>
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which satisfy the equality
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<math>v=u+n\pi </math>
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 +
 +
where
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<math>n</math>
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is an arbitrary integer.
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If we apply this result to the equation
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<math>\tan x=\tan 4x</math>
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we see that the solutions are given by
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<math>4x=x+n\pi </math>
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(
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<math>n</math>
 +
an arbitrary integer),
 +
 +
and solving for
 +
<math>x</math>
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gives
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<math>x=\frac{1}{3}n\pi </math>
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(
 +
<math>n</math>
 +
an arbitrary integer).

Version vom 11:06, 1. Okt. 2008

Let's first investigate when the equality


\displaystyle \tan u=\tan v


is satisfied. Because \displaystyle u can be interpreted as the slope (gradient) of the line which makes an angle \displaystyle u with the positive \displaystyle x -axis, we see that for a fixed value of tan u, there are two angles \displaystyle v in the unit circle with this slope:


\displaystyle v=u and \displaystyle v=u+\pi


slope \displaystyle =\text{ tan }u slope \displaystyle =\text{ tan }u


The angle \displaystyle v has the same slope after every half turn, so if we add multiples of \displaystyle \pi \text{ } to \displaystyle u, we will obtain all the angles \displaystyle v which satisfy the equality


\displaystyle v=u+n\pi


where \displaystyle n is an arbitrary integer.

If we apply this result to the equation


\displaystyle \tan x=\tan 4x


we see that the solutions are given by


\displaystyle 4x=x+n\pi ( \displaystyle n an arbitrary integer),

and solving for \displaystyle x gives


\displaystyle x=\frac{1}{3}n\pi ( \displaystyle n an arbitrary integer).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:5b“