Proof of $\sin^{2}\frac{A}{2}=\frac{1-\cos A}{2}$ and $\cos^{2}\frac{A}{2}=\frac{1+\cos A}{2}$

We proof by using the addition formulas：

$$\cos (A + B)=\cos A\cos B- \sin A\sin B$$

If we set $B=A$ in addition formulas, we obtain

\begin{eqnarray*}\cos (A + A)&=&\cos A\cos A- \sin A\sin A\\&=&\cos^{2}A-\sin^{2}A\ \ \ \ \ (\text{∵ }\sin^{2} A+\cos^{2} A=1)\\&=&1-2\sin^{2}A=2\cos^{2} A-1\end{eqnarray*}

That is, we obtain the following two alternate forms of $\cos 2A$：

$$\text{① }\cos 2A=1-2\sin^{2} A,\ \ \ \text{② }\cos 2A=2\cos^{2} A-1$$

Therefore we get $$\text{① }\sin^{2} A=\frac{1-\cos 2A}{2},\ \ \ \text{② }\cos^{2} A=\frac{1+\cos 2A}{2}$$

and if we substitute $\frac{A}{2}$ for $A$, we obtain half-angel formulas：

\begin{eqnarray*}\text{① }\sin^{2}\frac{A}{2}&=&\frac{1-\cos 2\cdot \frac{A}{2}}{2}=\frac{1-\cos A}{2}\\\text{② }\cos^{2} \frac{A}{2}&=&\frac{1+\cos 2\cdot \frac{A}{2}}{2}=\frac{1+\cos A}{2}\end{eqnarray*}

Proof of $\tan^{2}\frac{A}{2}=\frac{1-\cos A}{1+\cos A}$

Since $\tan A=\frac{\sin A}{\cos A}$, we get

$$\tan^{2} \frac{A}{2}=\frac{\sin^{2} \frac{A}{2}}{\cos^{2} \frac{A}{2}}=\frac{\frac{1}{2}(1-\cos A)}{\frac{1}{2}(1+\cos A)}=\frac{1-\cos A}{1+\cos A}$$

How to use the half-angle formulas？

We use the half-angle formula for $\tan^{2} \frac{A}{2}$：

$$\tan^{2} \frac{A}{2}=\frac{1-\cos A}{1+\cos A}$$

If we set $\frac{A}{2}=15^{\circ}$, we get

$$\tan^{2} 15^{\circ}=\frac{1-\cos 30^{\circ}}{1+\cos 30^{\circ}}=\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}=\frac{2-\sqrt{3}}{2+\sqrt{3}}=(2-\sqrt{3})^{2}$$

Now, since $\tan 15^{\circ} > 0$, we obtain $$\tan 15^{\circ}=2-\sqrt{3}$$

We use the half-angle formula of $\cos^{2} \frac{A}{2}$ : $$\cos^{2}\frac{A}{2}=\frac{1+\cos A}{2}$$,

If we set $\frac{A}{2}=x$, we get

\begin{eqnarray*}\int \cos^{2} x\ dx&=&\int \frac{1+\cos 2x}{2}\ dx\\&=&\int \frac{1}{2}\ dx+\frac{1}{2}\int \cos 2x\ dx\\&=&\frac{x}{2}+\frac{1}{4}\sin 2x+C\end{eqnarray*}