Maximum Value of trig

So first take $\prod_{i=m}^n$ (cos {A m})=x

And from the restriction, you have $\prod_{i=m}^n$ (cos {A m})= $\prod_{i=m}^n$ (sin {A m})

take sin(2a)=2 sin(a) cos(a)

And finally, you get $(x^{2})$ * $(2^{n})$ = $\prod_{i=m}^n$ (sin2 {A m))

Now, think about the maximum possible value of $\prod_{i=m}^n$ (sin2 {A m)), and you can see that the answer is 1, when each angle will be pi/4.

and you may also notice that the maximum value of x can be obtained from the maximum value of $\prod_{i=m}^n$ (sin2 {A m))

And thus because the above solution also satisfies the given restriction $\prod_{i=m}^n$ (cot {A m)) =1,

We get $(x^{2})$ * $(2^{n})$ =1

Therefore x = $\frac{1}{2^{\frac{n}{2}}}$