Aus Online Mathematik Brückenkurs 1

We draw an angle \displaystyle v in the unit circle, and the fact that \displaystyle \text{sin }v\text{ }=\frac{3}{10} means that its \displaystyle y -coordinate equals \displaystyle \frac{3}{10}.

With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of \displaystyle \text{1} and a vertical side of length \displaystyle \frac{3}{10}.

We can determine the triangle's remaining side by using Pythagoras' theorem,

\displaystyle a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}

which gives that

\displaystyle a=\sqrt{1-\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}

This means that the angle's \displaystyle x -coordinate is \displaystyle -a, i.e. we have

\displaystyle \cos v=-\frac{\sqrt{91}}{10}

and thus

\displaystyle \tan v=\frac{\sin v}{\cos v}=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}