Lösung 3.3:6
Aus Online Mathematik Brückenkurs 2
We take up the exercise's challenge and solve the equation both in polar form and in the form
Polar form
In polar form,
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and, using de Moivre's formula, the equation becomes
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If both sides are to be equal, their magnitudes must be equal and their arguments must be equal, other than for multiples of
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This gives
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which corresponds two solutions, because all even values of 8
8
Thus, in polar form, we have the solutions,
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One solution 42(cos(
8)+isin(
8)
42(cos(9
8)+isin(9
8))
Rectangular form
The alternative way to solve the equation is to put
If
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Because both sides' real and imaginary parts must equal each other we have that
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All the information we need for determining
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Therefore, we have in total three equations,
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If we add the first and the third equations,
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we get that
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If we subtract the first equation from the third equation,
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we obtain that
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All in all, this gives us four possible solutions
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although we have only taken account of the first and third equations.
The second equation says that the product
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Now, we know already that the equation has two solutions, so we can draw the conclusion that these are
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If we compare the solution in the first quadrant when it is expressed in polar and rectangular forms, we have
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and therefore we must have that
\displaystyle \begin{align}
\cos\frac{\pi}{8} &= \frac{1}{\sqrt[4]{2}}\sqrt{\frac{\sqrt{2}+1}{2}}\,,\\[5pt] \sin\frac{\pi}{8} &= \frac{1}{\sqrt[4]{2}}\sqrt{\frac{\sqrt{2}-1}{2}}\,\textrm{.} \end{align} |
Thus, we have
\displaystyle \tan\frac{\pi}{8} = \frac{\sin\dfrac{\pi}{8}}{\cos\dfrac{\pi}{8}} = \frac{\dfrac{1}{\sqrt[4]{2}}\sqrt{\dfrac{\sqrt{2}-1}{2}}}{\dfrac{1}{\sqrt[4]{2}}\sqrt{\dfrac{\sqrt{2}+1}{2}}} = \sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\,\textrm{.} |
We can simplify the expression under the square root sign by multiplying top and bottom by the conjugate of the denominator,
\displaystyle \begin{align}
\tan\frac{\pi}{8} &= \sqrt{\frac{(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}} = \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-1^2}}\\[5pt] &= \sqrt{\frac{(\sqrt{2}-1)^2}{2-1}} = \sqrt{(\sqrt{2}-1)^2} = \sqrt{2}-1\,\textrm{.} \end{align} |