Solution 4.4:7c
From Förberedande kurs i matematik 1
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- | If we want to solve the equation | + | If we want to solve the equation <math>\cos 3x = \sin 4x</math>, we need an additional result which tells us for which values of ''u'' and ''v'' the equality |
- | <math>\ | + | <math>\cos u = \sin v</math> holds, but to get that we have to start with the equality <math>\cos u=\cos v</math>. |
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- | and | + | |
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- | the equality | + | |
- | <math>\ | + | |
- | holds, but to get that we have to start with the equality | + | |
- | <math>\cos u=\cos v</math>. | + | |
So, we start by looking at the equality | So, we start by looking at the equality | ||
+ | {{Displayed math||<math>\cos u=\cos v\,\textrm{.}</math>}} | ||
- | + | We know that for fixed ''u'' there are two angles <math>v=u</math> and <math>v=-u</math> in the unit circle which have the cosine value <math>\cos u</math>, i.e. their ''x''-coordinate is equal to <math>\cos u\,</math>. | |
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- | We know that for fixed | + | |
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- | there are two angles | + | |
- | <math>v=u | + | |
- | and | + | |
- | <math>v=- | + | |
- | in the unit circle which have the cosine value | + | |
- | <math>\cos u</math>, i.e. their | + | |
- | + | ||
- | -coordinate is equal to | + | |
- | <math>\cos u</math>. | + | |
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[[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 | + | 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 ''u'' and -''u'' are rotated to <math>u+\pi/2</math> and <math>-u+\pi/2</math>, respectively. |
- | <math> | + | |
- | <math>x=\cos u</math> | + | |
- | will become the line | + | |
- | <math>y=\cos u</math> | + | |
- | and the angles | + | |
- | + | ||
- | and | + | |
- | + | ||
- | are rotated to | + | |
- | <math>u+ | + | |
- | and | + | |
- | <math>-u+ | + | |
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[[Image:4_4_7_c2.gif|center]] | [[Image:4_4_7_c2.gif|center]] | ||
- | The angles | + | The angles <math>u+\pi/2</math> and <math>-u+\pi/2</math> therefore have their ''y''-coordinate, and hence sine value, equal to <math>\cos u</math>. In other words, the equality |
- | <math>u+ | + | |
- | and | + | |
- | <math>-u+ | + | |
- | therefore have their | + | |
- | + | ||
- | -coordinate, and hence sine value, equal to | + | |
- | <math>\cos u</math>. In other words, the equality | + | |
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- | <math> | + | {{Displayed math||<math>\cos u = \sin v</math>}} |
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- | + | holds for fixed ''u'' in the unit circle when <math>v = \pm u + \pi/2</math>, and more generally when | |
- | <math> | + | |
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+ | {{Displayed math||<math>v = \pm u + \frac{\pi}{2} + 2n\pi\qquad</math>(''n'' is an arbitrary integer).}} | ||
- | <math> | + | For our equation <math>\cos 3x = \sin 4x</math>, this result means that ''x'' must satisfy |
+ | {{Displayed math||<math>4x = \pm 3x + \frac{\pi}{2} + 2n\pi\,\textrm{.}</math>}} | ||
This means that the solutions to the equation are | This means that the solutions to the equation are | ||
+ | {{Displayed math||<math>\left\{\begin{align} | ||
+ | x &= \frac{\pi}{2} + 2n\pi\,,\\[5pt] | ||
+ | x &= \frac{\pi}{14} + \frac{2}{7}\pi n\,, | ||
+ | \end{align}\right.</math>}} | ||
- | + | where ''n'' is an arbitrary integer. | |
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- | an arbitrary integer | + |
Current revision
If we want to solve the equation
So, we start by looking at the equality
We know that for fixed u there are two angles
Imagine now that the whole unit circle is rotated anti-clockwise an angle 2
2
2
The angles 2
2
holds for fixed u in the unit circle when u+
2
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For our equation
![]() ![]() ![]() |
This means that the solutions to the equation are
![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
where n is an arbitrary integer.