Lösung 4.3:6a

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Using the Pythagorean theorem, it is possible to determine the remaining side from
Using the Pythagorean theorem, it is possible to determine the remaining side from
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{{Displayed math||<math>b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2</math>}}
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{{Abgesetzte Formel||<math>b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2</math>}}
which gives that
which gives that
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{{Displayed math||<math>b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
Because the angle ''v'' belongs to the fourth quadrant, its ''y''-coordinate is negative and is therefore equal to <math>-b</math>, i.e.
Because the angle ''v'' belongs to the fourth quadrant, its ''y''-coordinate is negative and is therefore equal to <math>-b</math>, i.e.
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{{Displayed math||<math>\sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
Thus, we have directly that
Thus, we have directly that
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{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}</math>}}

Version vom 08:56, 22. Okt. 2008

If we think of the angle v as an angle in the unit circle, then v lies in the fourth quadrant and has x-coordinate 3/4.

If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to 1 and an opposite side equal to 3/4.

Using the Pythagorean theorem, it is possible to determine the remaining side from

\displaystyle b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2

which gives that

\displaystyle b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}

Because the angle v belongs to the fourth quadrant, its y-coordinate is negative and is therefore equal to \displaystyle -b, i.e.

\displaystyle \sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}

Thus, we have directly that

\displaystyle \tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:6a“