Trigonometric identity proof $\cos(A) + \cos( B)$ What is a nice proof of $$\cos(A) + \cos( B)= 2\cos\Big(\frac{A+B}{2}\Big)\cos\Big(\frac{A+B}{2}\Big)$$? 
I can prove it starting with the RHS but i want to be able to quickly prove it starting on the LHS as I won't have access to a formula book when I need it.
 A: $$\cos(x+y)=\cos x\cos y- \sin x\sin y$$
$$\cos(x-y)=\cos x\cos y+\sin x\sin y$$
$$\cos(x+y)+\cos(x-y)=2\cos(x)\cos(y)$$
Now let $x+y=A$, $x-y=B$ and you'll get $$\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\left(\frac{A-B}{2}\right)$$
A: Here is a fast and efficient way forward.  We use Euler's Formula to write 
$$\cos A+\cos B=\text{Re}\left(e^{iA}+e^{iB}\right) \tag 1$$
Now, simply factor out $e^{i(A+B)/2}$ under the "Real Part" operator in $(1)$ to obtain
$$\begin{align}\text{Re}\left(e^{iA}+e^{iB}\right) &=\text{Re}\left(e^{i(A+B)/2}\left(e^{i(A-B)/2}+e^{-i(A-B)/2}\right)\right) \\\\
&=\text{Re}\left(2\cos\left(\frac{A-B}{2}\right)e^{i(A+B)/2}\right) \\\\
&=2\cos\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)
\end{align}$$
And we are done!
A: HINT: start from
$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$
$$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$
A: From LHS to RHS:
$$
\cos B + \cos A \\
\cos \frac {2 B} 2 + \cos \frac {2 A} 2 \\
\cos\left(\frac {A + B - A + B} 2 \right) + \cos\left(\frac {A + B + A - B} 2 \right) \\
\cos\left(\frac {A + B} 2 - \frac {A - B} 2 \right) + \cos\left(\frac {A + B} 2 + \dfrac {A - B} 2 \right) \\
\left[\cos\left(\frac{A + B} 2\right)\cos\left(\frac{A - B} 2 \right) + \sin\left(\frac{A + B} 2 \right)\sin\left(\frac{A - B} 2 \right)\right] + \left[\cos\left(\frac{A + B} 2 \right)\cos\left(\frac{A - B} 2 \right) - \sin\left(\frac{A + B} 2 \right)\sin\left(\frac{A - B} 2 \right)\right]\\
2 \cos\left(\frac {A + B} 2 \right) \cos\left(\frac {A - B} 2 \right) \\
$$
