From Förberedande kurs i matematik 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}}