So I have proved that the sequence $(a_n)$ converges (to $a$ say) where $a_{n+1}=\cos \frac{a_n}{2} $.
Now I have also shown that $a$ solves the equation $x=\cos \frac{x}{2} $, but I am now trying to show that this solution - which is $a$ - is unique.
How can I do this, I started by supposing that another solution $b$ existed which would give $b=\cos \frac{b}{2} $ and the goal is to show that $a=b$ but I'm not sure how I would do this.
Any ideas or help would be appreciated.
$f(x)=x-\cos(\frac x2)$ has the derivative $f'(x)=1+\frac12\sin\frac x2\ge\frac12>0$ so that you have a strongly monotonous function.
$g(x)=\cos\frac x2$ has derivative $|g'(x)|=\frac12|\sin\frac12|\le\frac12$ which means that the fixed point iteration $x_+=g(x)$ is contractive on $\Bbb R$ with Lipschitz constant $L=\frac12$ guaranteeing the convergence to the unique fixed point by the Banach fixed point theorem.
Related to the BFT proof, replace the mean value theorem by trigonometric identities $$\cos(A+B)-\cos(A-B)=-2\sin A\sin B.$$ If there were two fixed points $x,y$, then $$ |x-y|=|g(x)-g(y)|=2\sin\frac{|x-y|}4\sin\frac{|x+y|}4\le\frac12|x-y| $$ which is only possible for $x=y$.