Can you prove this? $2+2\cos(x-y)=4\cos^2\left(\frac{x-y}{2}\right)$

Question

Prove the following

$$2+2\cos(x-y)=4\cos^2\left(\frac{x-y}{2}\right)$$

Answer 1

You have to prove:

$1+ \cos(t)=2\cos^2(t/2).$

To this end use

$\cos(t)= \cos(t/2+t/2)= \cos^2(t/2)-\sin^2(t/2)$

and

$\cos^2 (t/2)+ \sin^2(t/2)=1.$

Answer 2

Hint: $\cos (2x)=2\cdot\cos^2 x -1$, and $x-y=2\cdot\dfrac {x-y}2$.

Answer 3

We have $$\begin{align}\bigl(1+ \cos(t)-2\cos^2(t/2)\bigr)'&=-\sin(t)-4\cos(t/2)(-\sin(t/2))\cdot\frac12\\ &=-2\sin(t/2)\cos(t/2)+2\cos(t/2)\sin(t/2)\\ &=0, \end{align}$$ hence $1+ \cos(t)-2\cos^2(t/2)$ is constant. Now plug in $t=0$.

Answer 4

Let $t=\dfrac{x-y}{2}$,

$\cos \left(2t\right)=2\cos^2t-1 \\4\cos^2t=2\cos\left(2t\right)+2\\4\cos^2\left(\dfrac{x-y}{2}\right)=2\cos\left(x-y\right)+2$