Lösung 4.4:5a
Aus Online Mathematik Brückenkurs 1
K (Lösning 4.4:5a moved to Solution 4.4:5a: Robot: moved page) |
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- | + | If we consider for a moment the equality | |
- | < | + | |
- | + | ||
- | + | <math>\sin u=\sin v\quad \quad \quad (*)</math> | |
- | < | + | |
- | + | ||
+ | |||
+ | where | ||
+ | <math>u</math> | ||
+ | has a fixed value, there are usually two angles | ||
+ | <math>v</math> | ||
+ | in the unit circle which ensure that the equality holds: | ||
+ | |||
+ | |||
+ | <math>v=u</math> | ||
+ | and | ||
+ | <math>v=\pi -u</math> | ||
+ | |||
[[Image:4_4_5_a.gif]] | [[Image:4_4_5_a.gif]] | ||
+ | |||
+ | |||
+ | (The only exception is when | ||
+ | <math>u={\pi }/{2}\;\text{ }</math> | ||
+ | or | ||
+ | <math>u=3{\pi }/{2}\;\text{ }</math>, in which case | ||
+ | <math>u</math> | ||
+ | and | ||
+ | <math>\pi -u\text{ }</math> | ||
+ | correspond to the same direction and there is only one angle | ||
+ | <math>v</math> | ||
+ | which satisfies the equality.) | ||
+ | |||
+ | We obtain all the angles | ||
+ | <math>v</math> | ||
+ | which satisfy (*) by adding multiples of | ||
+ | <math>\text{2}\pi </math>, | ||
+ | |||
+ | |||
+ | <math>v=u+2n\pi </math> | ||
+ | and | ||
+ | <math>v=\pi -u+2n\pi </math> | ||
+ | |||
+ | |||
+ | where | ||
+ | <math>n\text{ }</math> | ||
+ | is an arbitrary integer. | ||
+ | |||
+ | If we now go back to our equation | ||
+ | |||
+ | |||
+ | <math>\sin 3x=\sin x</math> | ||
+ | |||
+ | |||
+ | the reasoning above shows that the equation is only satisfied when | ||
+ | |||
+ | |||
+ | <math>3x=x+2n\pi </math> | ||
+ | or | ||
+ | <math>3x=\pi -x+2n\pi </math> | ||
+ | |||
+ | |||
+ | If we make | ||
+ | <math>x</math> | ||
+ | the subject of each equation, we obtain the full solution to the equation: | ||
+ | |||
+ | |||
+ | <math>\left\{ \begin{array}{*{35}l} | ||
+ | x=0+n\pi \\ | ||
+ | x=\frac{\pi }{4}+\frac{1}{2}n\pi \\ | ||
+ | \end{array} \right.</math> |
Version vom 10:55, 1. Okt. 2008
If we consider for a moment the equality
\displaystyle \sin u=\sin v\quad \quad \quad (*)
where \displaystyle u has a fixed value, there are usually two angles \displaystyle v in the unit circle which ensure that the equality holds:
\displaystyle v=u
and
\displaystyle v=\pi -u
(The only exception is when
\displaystyle u={\pi }/{2}\;\text{ }
or
\displaystyle u=3{\pi }/{2}\;\text{ }, in which case
\displaystyle u
and
\displaystyle \pi -u\text{ }
correspond to the same direction and there is only one angle
\displaystyle v
which satisfies the equality.)
We obtain all the angles \displaystyle v which satisfy (*) by adding multiples of \displaystyle \text{2}\pi ,
\displaystyle v=u+2n\pi
and
\displaystyle v=\pi -u+2n\pi
where
\displaystyle n\text{ }
is an arbitrary integer.
If we now go back to our equation
\displaystyle \sin 3x=\sin x
the reasoning above shows that the equation is only satisfied when
\displaystyle 3x=x+2n\pi
or
\displaystyle 3x=\pi -x+2n\pi
If we make
\displaystyle x
the subject of each equation, we obtain the full solution to the equation:
\displaystyle \left\{ \begin{array}{*{35}l}
x=0+n\pi \\
x=\frac{\pi }{4}+\frac{1}{2}n\pi \\
\end{array} \right.