Aus Online Mathematik Brückenkurs 1

In exercise e, we studied the angle \displaystyle \frac{3\pi }{4} and found that

\displaystyle \cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}} and \displaystyle \sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}

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

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