Aus Online Mathematik Brückenkurs 1

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}\,.