Lösung 4.3:4b
Aus Online Mathematik Brückenkurs 1
If we once again use the Pythagorean identity we get
\displaystyle \cos ^{2}v+\sin ^{2}v=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}
Because the angle v lies between
\displaystyle 0
and
\displaystyle \pi ,
\displaystyle \text{sin }v
is positive (an angle in the first and second quadrants has a positive
\displaystyle y
-coordinate) and therefore
\displaystyle \sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}