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Solution 4.4:7c

From Förberedande kurs i matematik 1

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m (Lösning 4.4:7c moved to Solution 4.4:7c: Robot: moved page)
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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
 +
 
 +
 
 +
<math>\cos u=\cos v</math>
 +
 
 +
 
 +
We know that for fixed
 +
<math>u</math>
 +
there are two angles
 +
<math>v=u\text{ }</math>
 +
and
 +
<math>v=-\text{u}</math>
 +
in the unit circle which have the cosine value
 +
<math>\cos u</math>, i.e. their
 +
<math>x</math>
 +
-coordinate is equal to
 +
<math>\cos u</math>.
 +
 
[[Image:4_4_7_c1.gif|center]]
[[Image:4_4_7_c1.gif|center]]
 +
 +
Imagine now that the whole unit circle is rotated anti-clockwise an angle
 +
<math>{\pi }/{2}\;</math>. The line
 +
<math>x=\cos u</math>
 +
will become the line
 +
<math>y=\cos u</math>
 +
and the angles
 +
<math>u</math>
 +
and
 +
<math>-u</math>
 +
are rotated to
 +
<math>u+{\pi }/{2}\;</math>
 +
and
 +
<math>-u+{\pi }/{2}\;</math>, respectively.
 +
[[Image:4_4_7_c2.gif|center]]
[[Image:4_4_7_c2.gif|center]]
 +
 +
The angles
 +
<math>u+{\pi }/{2}\;</math>
 +
and
 +
<math>-u+{\pi }/{2}\;</math>
 +
therefore have their
 +
<math>y</math>
 +
-coordinate, and hence sine value, equal to
 +
<math>\cos u</math>. In other words, the equality
 +
 +
 +
<math>\text{cos }u=\text{sin }v</math>
 +
 +
 +
holds for fixed
 +
<math>u</math>
 +
in the unit circle when
 +
<math>v=\pm u+{\pi }/{2}\;</math>, and more generally when
 +
 +
 +
<math>v=\pm u+\frac{\pi }{2}+2n\pi </math>
 +
(
 +
<math>n</math>
 +
an arbitrary integer).
 +
 +
For our equation
 +
<math>\text{cos 3}x=\text{sin 4}x</math>, this result means that
 +
<math>x\text{ }</math>
 +
must satisfy
 +
 +
 +
<math>4x=\pm 3x+\frac{\pi }{2}+2n\pi </math>
 +
 +
 +
This means that the solutions to the equation are
 +
 +
 +
<math>\left\{ \begin{array}{*{35}l}
 +
x=\frac{\pi }{2}+2n\pi \\
 +
x=\frac{\pi }{14}+\frac{2}{7}\pi n \\
 +
\end{array} \right.</math>
 +
(
 +
<math>n</math>
 +
an arbitrary integer)

Revision as of 13:20, 1 October 2008

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

So, we start by looking at the equality


cosu=cosv


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


Imagine now that the whole unit circle is rotated anti-clockwise an angle 2. The line x=cosu will become the line y=cosu and the angles u and −u are rotated to u+2 and −u+2, respectively.


The angles u+2 and −u+2 therefore have their y -coordinate, and hence sine value, equal to cosu. In other words, the equality


cos u=sin v


holds for fixed u in the unit circle when v=u+2, and more generally when


v=u+2+2n ( n an arbitrary integer).

For our equation cos 3x=sin 4x, this result means that x must satisfy


4x=3x+2+2n


This means that the solutions to the equation are


x=2+2nx=14+72n  ( \displaystyle n an arbitrary integer)

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