Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ I've seen this identity on examsolutions, but I'm unsure on how to prove it.
$$\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
 A: This is a standard formula.
Start from the linearisation formula:
$$2\cos x\cos y=\cos(x+y)+\cos(x-y)$$
and solve the system:
$$\begin{cases}
x+y=A,\\x-y=B.
\end{cases}$$
A: Notice, 
$$RHS=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$
$$=2\cos \left(\frac{A}{2}+\frac{B}{2}\right)\cos \left(\frac{A}{2}-\frac{B}{2}\right)$$
$$=2 \left(\cos\frac{A}{2}\cos\frac{B}{2}-\sin\frac{A}{2}\sin\frac{B}{2}\right)\left(\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\right)$$
$$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-\sin^2\frac{A}{2}\sin^2\frac{B}{2}\right)$$
$$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-\left(1-\cos^2\frac{A}{2}\right)\left(1-\cos^2\frac{A}{2}\right)\right)$$
$$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-1+\cos^2\frac{A}{2}+\cos^2\frac{B}{2}-\cos^2\frac{A}{2}\cos^2\frac{B}{2}\right)$$
$$=2 \left(\cos^2\frac{A}{2}+\cos^2\frac{B}{2}-1\right)$$
$$=2\cos^2\frac{A}{2}+2\cos^2\frac{B}{2}-2$$
$$=\left(2\cos^2\frac{A}{2}-1\right)+\left(2\cos^2\frac{B}{2}-1\right)$$
$$=\cos A+\cos B=LHS$$
A: Substitute the following: $v := \frac{x+y}{2}$ and $w := \frac{x-y}{2}$
Hence: $$\cos(x) + \cos(y) = \cos(v + w) + \cos(v - w)$$
Using the addition theorem of the cosinus yields:
$$= \cos(v) \cdot \cos(w) - \sin(v) \cdot \sin(w) + \cos(v) \cdot \cos(w) + \sin(v) \cdot \sin(w)$$
$$= 2 \cdot \cos(v) \cdot \cos(w) = 2 \cdot \cos(\frac{x+y}{2}) \cdot \cos(\frac{x-y}{2})$$
A: Start with $$\cos (x+y)=\cos x\cos y-\sin x\sin y$$and $$\cos (x-y)=\cos x\cos y+\sin x\sin y$$(just by changing the sign of $y$ which changes only the sign of $\sin y$)
Now add these together to obtain $$\cos (x+y)+\cos (x-y)=2\cos x\cos y$$
Now set $A=x+y; B=x-y$
You should be able to see how you can subtract to find the formula for the difference of cosines (or the product of two sines, using it the other way around).
Starting with $\sin (x\pm y)$ you can get similar formulae involving the sum and difference of sines. It is worth working them out for yourself as they will then be easy to derive, if you forget them.
