Unclear step in half-angle formula derivation (trigonometric identities) In deriving the half-angle formulas, my textbook first says: "Let's take the following identities:"
$$\cos^2\left(\frac a2\right)+\sin^2\left(\frac a2\right)=1;$$
$$\cos^2\left(\frac a2\right)-\sin^2\left(\frac a2\right)=\cos(a);$$
These identities I know. But then the texbook says "through addition and subtraction, we respectively arrive at:"
$$2\cos^2\left(\frac a2\right)=1+\cos(a)$$
$$2\sin^2\left(\frac a2\right)=1-\cos(a)$$
I failed to catch what exactly is added and what is substracted to arrive from the first two formulas to the second pair. Give me a hint, please.
 A: The "addition" means that 
$$\left(\cos^2\left(\frac a2\right)+\sin^2\left(\frac a2\right)\right)+\left(\cos^2\left(\frac a2\right)-\sin^2\left(\frac a2\right)\right)=1+\cos (a).$$
The "subtraction" means that 
$$\left(\cos^2\left(\frac a2\right)+\sin^2\left(\frac a2\right)\right)-\left(\cos^2\left(\frac a2\right)-\sin^2\left(\frac a2\right)\right)=1-\cos (a).$$
In general, if you have
$$A+B=C$$
$$D+E=F$$
then you can have
$$(A+B)+(D+E)=C+F$$
$$(A+B)-(D+E)=C-F.$$
A: Let $a, b$ be real numbers. Recall that
$$
e^{ia}=\cos a + i \sin a
$$ where $i^2=1.$
Then 
$$
e^{i(a+b)}=\cos (a+b) + i \sin (a+b)
$$
But
$$\begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\\\\&=(\cos a + i \sin a)(\cos b + i \sin b)
\\\\&=(\cos a\cos b-\sin a\sin b) + i (\sin a\cos b+\sin b\cos a).
\end{align}
$$
Consequently
$$
\begin{align}
\cos (a+b)&=\cos a\cos b-\sin a\sin b\\
\sin (a+b)&=\sin a\cos b+\sin b\cos a.
\end{align}
$$
Putting $a=b$ gives
$$
\begin{align}
\cos (2a)&=\cos^2 a-\sin^2 a \tag1\\
\cos (2a) &=2\cos^2 a-1 \tag2
\end{align}
$$ since
$$ \cos^2 a+\sin^2 a=1. \tag3$$
From $(2)$ we easily get $$\cos^2 a=\frac{1+\cos (2a)}{2} \tag4
$$
From $(1)$ you also have
$$\cos (2a) =1-2 \sin^2 a \tag5 $$ 
giving 
$$\sin^2 a=\frac{1-\cos (2a)}{2}.\tag6
$$
