The problem:
If the angle between $\vec a$ and $\vec b$ is $60^\circ $, find the angle between $2\vec a$ and $-2\vec b$.
My Attempt
The angle between $\vec a$ and $\vec b$ is given by: $$\cos \theta =\dfrac {\vec a\times \vec b}{ |\vec a|\times |\vec b|}.$$
But, how to solve this?
On
Let $\bar{\theta}$ be the angle between $2\vec a$ and $-2\vec b$. So $$\cos \bar{\theta} =\dfrac {\vec 2a\times \vec -2b}{ |2\vec a|\times |-2\vec b|} = \dfrac {-4(\vec a\times \vec b)}{ 4(|\vec a|\times |\vec b|)} = -\dfrac {\vec a\times \vec b}{ |\vec a|\times |\vec b|} = -\cos(\theta) = -\frac{1}{2},$$
what implies
$$\bar{\theta}=\arccos(-1/2)=120^\circ.$$
$\angle (2\vec{a},-2\vec{b})=\angle (\vec{a},-\vec{b})=180-\angle(\vec{a},\vec{b})$
The explanation why each equality holds:
For the first equality: $$ \cos{\angle (2\vec{a},-2\vec{b})}=\frac{4\vec{a}\vec{b}}{2||\vec{a}||2||\vec{b}||}= \cos{\angle (\vec{a},-\vec{b})} $$ For the second equality: $$ \cos{\angle (\vec{a},-\vec{b})}=-\frac{\vec{a} \vec{b}}{||\vec{a}||||\vec{b}||}=-\cos{\angle (\vec{a},\vec{b})}\Rightarrow \angle (\vec{a},-\vec{b})=180-\angle (\vec{a},\vec{b})=120 $$ Note that when we refer to the angle of two vectors we consider the angle that is between $0$ and $180$ so when we found $\cos{\theta}=-\cos{60}=-\frac 1 2$ we have that $\theta=120$.