From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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.
-	<center> [[Bild:4_2_5a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	[[Image:4_2_5_a1.gif|center]]
-	[[Bild:4_2_5_a1.gif|center]]	+
-	[[Bild:4_2_5_a2.gif|center]]	+	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.
		+
		+	{| width="100%"
		+	|width="50%" align="center"|[[Image:4_2_5_a2.gif|center]]
		+	|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>
		+	|}
		+
		+	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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Current revision

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}\,.