Lösung 4.2:5d

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
By subtracting
+
By subtracting 360° from 495°, we do not change the value of the tangent,
-
<math>360^{\circ }</math>
+
-
from
+
-
<math>\text{495}^{\circ }</math>, 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>}}
-
<math>\tan \text{495}^{\circ }=\tan \left( \text{495}^{\circ }-360^{\circ } \right)=\tan \text{135}^{\circ }</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
+
{{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>}}
-
<math>\cos 135^{\circ }=-\frac{1}{\sqrt{2}}</math>
+
-
and
+
-
<math>\sin 135^{\circ }=\frac{1}{\sqrt{2}}</math>, which gives
+
-
 
+
-
 
+
-
<math>\tan 135^{\circ }=\frac{\sin 135^{\circ }}{\cos 135^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1</math>
+

Version vom 11:15, 9. Okt. 2008

By subtracting 360° from 495°, we do not change the value of the tangent,

Vorlage:Displayed math

We know from exercise a that \displaystyle \cos 135^{\circ} = -1/\!\sqrt{2} and \displaystyle \sin 135^{\circ} = 1/\!\sqrt{2}\,, which gives

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:5d“