Aus Online Mathematik Brückenkurs 1

If we think of the angle v as an angle in the unit circle, then \displaystyle v lies in the fourth quadrant and has \displaystyle x -coordinate \displaystyle \frac{3}{4}.

If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to \displaystyle \text{1} and an opposite side equal to \displaystyle \frac{3}{4}.

Using Pythagoras' theorem, it is possible to determine the remaining side from

\displaystyle b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2}

which gives that

\displaystyle \begin{align} & b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2} \\ & b=\sqrt{1-\left( \frac{3}{4} \right)^{2}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \\ \end{align}

Because the angle \displaystyle v belongs to the fourth quadrant, its \displaystyle y -coordinate is negative and is therefore equal to \displaystyle -b, i.e.

\displaystyle \sin v=-\frac{\sqrt{7}}{4}

Thus, we have directly that

\displaystyle \tan v=\frac{\sin v}{\cos v}=\frac{-\frac{\sqrt{7}}{4}}{\frac{3}{4}}=-\frac{\sqrt{7}}{3}