Lösung 4.2:5d
Aus Online Mathematik Brückenkurs 1
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Version vom 14:59, 22. Okt. 2008
By subtracting 360° from 495°, we do not change the value of the tangent,
| \displaystyle \tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.} |
We know from exercise a that \displaystyle \cos 135^{\circ} = -1/\!\sqrt{2} and \displaystyle \sin 135^{\circ} = 1/\!\sqrt{2}\,, which gives
| \displaystyle \tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.} |
