Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Question

I've seen this identity on examsolutions, but I'm unsure on how to prove it. $$\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$

there is a classical variable change: $S=(A+B)/2, T=(A-B)/2$, i.e., $A=S+T, B=S-T$. Then develop the left part of the equality. — Fabrice NEYRET, Oct 24 '15 at 22:04

Bernard · Answer 1 · 2015-10-24T22:20:13.120

5

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: $$\begin{cases} x+y=A,\\x-y=B. \end{cases}$$

edited Oct 24 '15 at 22:20

answered Oct 24 '15 at 22:01

Bernard

175,478

nice and simple proof! – More water plz Oct 24 '15 at 22:16
Incidently, here's a trigonograph of the linearization formula(s). – Blue Oct 24 '15 at 22:27
Fine! In my opinion, the very last formula shouldn't be mentioned, as it is derived from the previous one by swapping $A$ and $B$. More importantly, everyone should know them by heart. – Bernard Oct 24 '15 at 22:33

score 0 · Answer 2 · answered Oct 24 '15 at 22:08

Notice, $$RHS=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$ $$=2\cos \left(\frac{A}{2}+\frac{B}{2}\right)\cos \left(\frac{A}{2}-\frac{B}{2}\right)$$

$$=2 \left(\cos\frac{A}{2}\cos\frac{B}{2}-\sin\frac{A}{2}\sin\frac{B}{2}\right)\left(\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\right)$$ $$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-\sin^2\frac{A}{2}\sin^2\frac{B}{2}\right)$$ $$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-\left(1-\cos^2\frac{A}{2}\right)\left(1-\cos^2\frac{A}{2}\right)\right)$$ $$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-1+\cos^2\frac{A}{2}+\cos^2\frac{B}{2}-\cos^2\frac{A}{2}\cos^2\frac{B}{2}\right)$$ $$=2 \left(\cos^2\frac{A}{2}+\cos^2\frac{B}{2}-1\right)$$ $$=2\cos^2\frac{A}{2}+2\cos^2\frac{B}{2}-2$$ $$=\left(2\cos^2\frac{A}{2}-1\right)+\left(2\cos^2\frac{B}{2}-1\right)$$ $$=\cos A+\cos B=LHS$$

score 0 · Answer 3 · answered Oct 24 '15 at 22:25

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: $$= \cos(v) \cdot \cos(w) - \sin(v) \cdot \sin(w) + \cos(v) \cdot \cos(w) + \sin(v) \cdot \sin(w)$$ $$= 2 \cdot \cos(v) \cdot \cos(w) = 2 \cdot \cos(\frac{x+y}{2}) \cdot \cos(\frac{x-y}{2})$$

score 0 · Answer 4 · answered Oct 24 '15 at 22:32

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.

Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

4 Answers4