Lösung 4.2:5a

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Because
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Because <math>135^{\circ} = 90^{\circ} + 45^{\circ}</math>, <math>135^{\circ}</math> is an angle in the second quadrant which makes an angle of <math>45^{\circ}</math> with the positive ''y''-axis.
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<math>\text{135}^{\circ }\text{ }=\text{ 9}0^{\circ }\text{ }+\text{45}^{\circ }</math>,
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<math>\text{135}^{\circ }\text{ }</math>
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is an angle in the second quadrant which makes an angle of
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<math>\text{45}^{\circ }</math>
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with the positive
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<math>y</math>
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-axis.
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[[Image:4_2_5_a1.gif|center]]
[[Image:4_2_5_a1.gif|center]]
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We can determine the point on the unit circle which corresponds to
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We can determine the point on the unit circle which corresponds to <math>135^{\circ}</math> by introducing an auxiliary triangle and calculating its edges using trigonometry.
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<math>\text{135}^{\circ }\text{ }</math>
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by introducing an auxiliary triangle and calculating its edges using trigonometry.
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[[Image:4_2_5_a2.gif|center]]
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opposite<math>=1\centerdot \sin \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>
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adjacent
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<math>=1\centerdot \cos \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>
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{| width="100%"
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|width="50%" align="center"|[[Image:4_2_5_a2.gif|center]]
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|width="50%" align="left"|<math>\begin{align}\text{opposite} &= 1\cdot\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\\[5pt] \text{adjacent} &= 1\cdot\cos 45^{\circ} = \frac{1}{\sqrt{2}}\end{align}</math>
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|}
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The coordinates of the point are
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The coordinates of the point are <math>( -1/\!\sqrt{2}, 1/\!\sqrt{2})</math> and this shows that <math>\cos 135^{\circ} = -1/\!\sqrt{2}\,</math>.
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<math>\left( -\frac{1}{\sqrt{2}} \right.,\left. \frac{1}{\sqrt{2}} \right)</math>
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and this shows that
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<math>\text{cos135}^{\circ }=-\frac{1}{\sqrt{2}}</math>.
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Version vom 11:01, 9. Okt. 2008

Because \displaystyle 135^{\circ} = 90^{\circ} + 45^{\circ}, \displaystyle 135^{\circ} is an angle in the second quadrant which makes an angle of \displaystyle 45^{\circ} with the positive y-axis.

We can determine the point on the unit circle which corresponds to \displaystyle 135^{\circ} by introducing an auxiliary triangle and calculating its edges using trigonometry.

\displaystyle \begin{align}\text{opposite} &= 1\cdot\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\\[5pt] \text{adjacent} &= 1\cdot\cos 45^{\circ} = \frac{1}{\sqrt{2}}\end{align}

The coordinates of the point are \displaystyle ( -1/\!\sqrt{2}, 1/\!\sqrt{2}) and this shows that \displaystyle \cos 135^{\circ} = -1/\!\sqrt{2}\,.

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