Solution 4.4:1g

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Current revision (13:54, 10 October 2008) (edit) (undo)
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A quick sketch of the unit circle and the line
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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.
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<math>y=-\frac{1}{\sqrt{3}}x</math>, which corresponds to the tangent value
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<math>-{1}/{\sqrt{3}}\;</math>
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shows that there are two angles which satisfy
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<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.
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[[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]]
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If we call
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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>.
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<math>x</math>
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the side adjacent to the angle
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<math>\alpha </math>
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then the fact that
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<math>\tan v=-{1}/{\sqrt{3}}\;</math>
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gives the length of the opposite side as
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<math>{x}/{\sqrt{3}}\;</math>.
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[[Image:4_4_1_g3.gif|center]]
[[Image:4_4_1_g3.gif|center]]
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Pythagoras' theorem gives
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The Pythagorean theorem gives
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<math>x^{2}+\left( \frac{x}{\sqrt{3}} \right)^{2}=1^{2}</math>
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and this equation has the solution
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<math>x={\sqrt{3}}/{2}\;</math>, which means that
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{{Displayed math||<math>x^2 + \Bigl(\frac{x}{\sqrt{3}}\Bigr)^2 = 1^2</math>}}
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<math>\cos \alpha =\frac{{\sqrt{3}}/{2}\;}{1}=\frac{\sqrt{3}}{2}</math>
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and this equation has the solution <math>x = \sqrt{3}/2</math>, which means that
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i.e.
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{{Displayed math||<math>\cos\alpha = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,,</math>}}
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<math>\alpha ={\pi }/{6}\;</math>. Because the angle
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<math>v</math>
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in the fourth quadrant is the complement of
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<math>\alpha </math>
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<math>v</math>
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is given by
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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
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<math>v=2\pi -\alpha =2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>.
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{{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,
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<math>\pi </math>, we obtain the other angle
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If we subtract half a turn, <math>\pi</math>, we obtain the other angle
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<math>v=\frac{11\pi }{6}-\pi =\frac{5\pi }{6}</math>
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{{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 \displaystyle y=(-1/\!\sqrt{3})x, which corresponds to the tangent value \displaystyle -1/\!\sqrt{3} shows that there are two angles which satisfy \displaystyle \tan v = -1/\!\sqrt{3}. One of the angles lies in the fourth quadrant and the other is the opposite angle in the second quadrant.

We can therefore limit ourselves to the fourth quadrant and draw an auxiliary triangle in order to determine the angle there.

If we call \displaystyle x the side adjacent to the angle \displaystyle \alpha then the fact that \displaystyle \tan v=-1/\!\sqrt{3} gives the length of the opposite side as \displaystyle x/\!\sqrt{3}.

The Pythagorean theorem gives

\displaystyle x^2 + \Bigl(\frac{x}{\sqrt{3}}\Bigr)^2 = 1^2

and this equation has the solution \displaystyle x = \sqrt{3}/2, which means that

\displaystyle \cos\alpha = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,,

i.e. \displaystyle \alpha = \pi/6. Because the angle \displaystyle v in the fourth quadrant is the complement of \displaystyle \alpha the angle \displaystyle v is given by

\displaystyle v = 2\pi - \alpha = 2\pi - \frac{\pi}{6} = \frac{11\pi }{6}\,\textrm{.}

If we subtract half a turn, \displaystyle \pi, we obtain the other angle

\displaystyle v = \frac{11\pi}{6} - \pi = \frac{5\pi}{6}\,\textrm{.}
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