Lösung 4.2:5d

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
By subtracting 360° from 495°, we do not change the value of the tangent,
By subtracting 360° from 495°, we do not change the value of the tangent,
-
{{Displayed math||<math>\tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.}</math>}}
We know from exercise a that <math>\cos 135^{\circ} = -1/\!\sqrt{2}</math> and <math>\sin 135^{\circ} = 1/\!\sqrt{2}\,</math>, which gives
We know from exercise a that <math>\cos 135^{\circ} = -1/\!\sqrt{2}</math> and <math>\sin 135^{\circ} = 1/\!\sqrt{2}\,</math>, which gives
-
{{Displayed math||<math>\tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}}

Version vom 08:53, 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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:5d“