Aus Online Mathematik Brückenkurs 1

One could write \displaystyle \tan \frac{u}{2} as a quotient involving sine and cosine, and then continue with the formula for half-angles,

\displaystyle \tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...

but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.

We write \displaystyle u as \displaystyle 2\left( \frac{u}{2} \right) and use the formula for double angles (so as to end up with a right-hand side which has \displaystyle \frac{u}{2} as its argument)

\displaystyle \frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}

Writing the \displaystyle \text{1} in the denominator as \displaystyle \cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2} using the Pythagorean identity,

\displaystyle \begin{align} & \frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}=\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}} \\ & =\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{2\cos ^{2}\frac{u}{2}}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\tan \frac{u}{2} \\ \end{align}