Lösung 4.2:4c
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In exercise 4.2:3e, we studied the angle <math>3\pi/4</math> and found that | In exercise 4.2:3e, we studied the angle <math>3\pi/4</math> and found that | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}</math>}} |
Because <math>\tan x</math> is defined as <math>\frac{\sin x}{\cos x}</math>, we get immediately that | Because <math>\tan x</math> is defined as <math>\frac{\sin x}{\cos x}</math>, we get immediately that | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}} |
Version vom 08:52, 22. Okt. 2008
In exercise 4.2:3e, we studied the angle \displaystyle 3\pi/4 and found that
| \displaystyle \cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.} |
Because \displaystyle \tan x is defined as \displaystyle \frac{\sin x}{\cos x}, we get immediately that
| \displaystyle \tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.} |
