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Solution 4.4:5a

From Förberedande kurs i matematik 1

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m (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
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<center> [[Image:4_4_5a-1(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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{{NAVCONTENT_START}}
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<math>\sin u=\sin v\quad \quad \quad (*)</math>
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<center> [[Image:4_4_5a-2(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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 +
 
 +
where
 +
<math>u</math>
 +
has a fixed value, there are usually two angles
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<math>v</math>
 +
in the unit circle which ensure that the equality holds:
 +
 
 +
 
 +
<math>v=u</math>
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and
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<math>v=\pi -u</math>
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[[Image:4_4_5_a.gif]]
[[Image:4_4_5_a.gif]]
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 +
 +
(The only exception is when
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<math>u={\pi }/{2}\;\text{ }</math>
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or
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<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>,
 +
 +
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<math>v=u+2n\pi </math>
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and
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<math>v=\pi -u+2n\pi </math>
 +
 +
 +
where
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<math>n\text{ }</math>
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is an arbitrary integer.
 +
 +
If we now go back to our equation
 +
 +
 +
<math>\sin 3x=\sin x</math>
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 +
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the reasoning above shows that the equation is only satisfied when
 +
 +
 +
<math>3x=x+2n\pi </math>
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or
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<math>3x=\pi -x+2n\pi </math>
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 +
 +
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 \\
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x=\frac{\pi }{4}+\frac{1}{2}n\pi \\
 +
\end{array} \right.</math>

Revision as of 10:55, 1 October 2008

If we consider for a moment the equality


sinu=sinv()


where u has a fixed value, there are usually two angles v in the unit circle which ensure that the equality holds:


v=u and v=−u


Image:4_4_5_a.gif


(The only exception is when u=2 or u=32 , in which case u and −u correspond to the same direction and there is only one angle v which satisfies the equality.)

We obtain all the angles v which satisfy (*) by adding multiples of 2,


v=u+2n and v=−u+2n


where n is an arbitrary integer.

If we now go back to our equation


sin3x=sinx


the reasoning above shows that the equation is only satisfied when


3x=x+2n or 3x=−x+2n


If we make x the subject of each equation, we obtain the full solution to the equation:


x=0+nx=4+21n 

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