Aus Online Mathematik Brückenkurs 1

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Because the angle <math>v</math> satisfies <math>\pi \le v\le 3\pi/2\,</math>, <math>v</math> belongs to the third quadrant in the unit circle. Furthermore, <math>\tan v = 3</math> gives that the line which corresponds to the angle <math>v</math> has slope 3. [[Image:4_3_6_c1.gif|center]] 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. [[Image:4_3_6_c2.gif|center]] If we now use the Pythagorean theorem on the triangle, we see that the horizontal side ''a'' satisfies {{Displayed math||<math>a^2 + (3a)^2 = 1^2</math>}} which gives us that <math>10a^{2}=1</math> i.e. <math>a = 1/\!\sqrt{10}\,\textrm{.}</math> Thus, the angle ''v'''s ''x''-coordinate is <math>-1/\!\sqrt{10}</math> and ''y''-coordinate is <math>-3/\!\sqrt{10}</math>, i.e. {{Displayed math||<math>\begin{align} \cos v &= -\frac{1}{\sqrt{10}}\,,\\[5pt] \sin v &= -\frac{3}{\sqrt{10}}\,\textrm{.} \end{align}</math>}}

• 4.3.6c - Solution - The unit circle with angle v
• 4.3.6c - Solution - The unit circle with angle v in the third quadrant and an auxiliary triangle
• Vorlage:Abgesetzte Formel

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