Aus Online Mathematik Brückenkurs 1

In exercise 4.2:3e, we studied the angle \displaystyle 3\pi/4 and found that

\displaystyle \cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}

Because \displaystyle \tan x is defined as \displaystyle \frac{\sin x}{\cos x}, we get immediately that

\displaystyle \tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}