Lösung 4.3:6c
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Version vom 15:07, 22. Okt. 2008
Because the angle \displaystyle v satisfies \displaystyle \pi \le v\le 3\pi/2\,, \displaystyle v belongs to the third quadrant in the unit circle. Furthermore, \displaystyle \tan v = 3 gives that the line which corresponds to the angle \displaystyle v has slope 3.
In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is 1 and the sides have a 3:1 ratio.
If we now use the Pythagorean theorem on the triangle, we see that the horizontal side a satisfies
| \displaystyle a^2 + (3a)^2 = 1^2 |
which gives us that \displaystyle 10a^{2}=1 i.e. \displaystyle a = 1/\!\sqrt{10}\,\textrm{.}
Thus, the angle v's x-coordinate is \displaystyle -1/\!\sqrt{10} and y-coordinate is \displaystyle -3/\!\sqrt{10}, i.e.
| \displaystyle \begin{align}
\cos v &= -\frac{1}{\sqrt{10}}\,,\\[5pt] \sin v &= -\frac{3}{\sqrt{10}}\,\textrm{.} \end{align} |
