Solution 4.4:1g
From Förberedande kurs i matematik 1
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- | A quick sketch of the unit circle and the line | + | A quick sketch of the unit circle and the line <math>y=(-1/\!\sqrt{3})x</math>, which corresponds to the tangent value <math>-1/\!\sqrt{3}</math> shows that there are two angles which satisfy <math>\tan v = -1/\!\sqrt{3}</math>. One of the angles lies in the fourth quadrant and the other is the opposite angle in the second quadrant. |
- | <math>y=-\ | + | |
- | <math>- | + | |
- | shows that there are two angles which satisfy | + | |
- | <math>\tan v=- | + | |
[[Image:4_4_1_g1.gif|center]] | [[Image:4_4_1_g1.gif|center]] | ||
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[[Image:4_4_1_g2.gif|center]] | [[Image:4_4_1_g2.gif|center]] | ||
- | If we call | + | If we call <math>x</math> the side adjacent to the angle <math>\alpha</math> then the fact that <math>\tan v=-1/\!\sqrt{3}</math> gives the length of the opposite side as <math>x/\!\sqrt{3}</math>. |
- | <math>x</math> | + | |
- | the side adjacent to the angle | + | |
- | <math>\alpha </math> | + | |
- | then the fact that | + | |
- | <math>\tan v=- | + | |
- | gives the length of the opposite side as | + | |
- | <math> | + | |
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[[Image:4_4_1_g3.gif|center]] | [[Image:4_4_1_g3.gif|center]] | ||
- | + | The Pythagorean theorem gives | |
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+ | {{Displayed math||<math>x^2 + \Bigl(\frac{x}{\sqrt{3}}\Bigr)^2 = 1^2</math>}} | ||
- | <math> | + | and this equation has the solution <math>x = \sqrt{3}/2</math>, which means that |
- | + | {{Displayed math||<math>\cos\alpha = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,,</math>}} | |
- | <math>\alpha ={\ | + | |
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+ | i.e. <math>\alpha = \pi/6</math>. Because the angle <math>v</math> in the fourth quadrant is the complement of <math>\alpha</math> the angle <math>v</math> is given by | ||
- | <math>v=2\pi -\alpha =2\pi -\frac{\pi }{6}=\frac{11\pi }{6} | + | {{Displayed math||<math>v = 2\pi - \alpha = 2\pi - \frac{\pi}{6} = \frac{11\pi }{6}\,\textrm{.}</math>}} |
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+ | If we subtract half a turn, <math>\pi</math>, we obtain the other angle | ||
- | <math>v=\frac{11\pi }{6}-\pi =\frac{5\pi }{6}</math> | + | {{Displayed math||<math>v = \frac{11\pi}{6} - \pi = \frac{5\pi}{6}\,\textrm{.}</math>}} |
Current revision
A quick sketch of the unit circle and the line 3)x
3
3
We can therefore limit ourselves to the fourth quadrant and draw an auxiliary triangle in order to determine the angle there.
If we call 3
3
The Pythagorean theorem gives
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and this equation has the solution 3
2
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i.e. =
6
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If we subtract half a turn,
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