Prove $\prod_{r=0}^n (\cos 2^r A + \cos 2^r B)$ w/ $A,B$ Trig III

[sbhatnagar]

Challenge Problem
If $A=\dfrac{\pi}{2^{n+1}}$ and $B=\dfrac{\pi}{2^{n+2}}$, prove that

\[\prod_{r=0}^n (\cos 2^r A + \cos 2^r B)=\frac{1}{2^{n+1}\left( \cos \frac{\pi}{2^{n+2}}-\cos \frac{\pi}{2^{n+1}}\right)}\]

 

[Opalg]

Challenge Problem
If $A=\dfrac{\pi}{2^{2^{n+1}}}$ and $B=\dfrac{\pi}{2^{2^{n+2}}}$, prove that

\[\prod_{r=0}^n (\cos 2^r A + \cos 2^r B)=\frac{1}{2^{n+1}\left( \cos \frac{\pi}{2^{n+2}}-\cos \frac{\pi}{2^{n+1}}\right)}\]

Something wrong here: I think there is one exponent too many in $A$ and $B$, it should be $A=\dfrac{\pi}{2^{n+1}}$ and $B=\dfrac{\pi}{2^{n+2}}$.

 

[Amer]

Challenge Problem
If $A=\dfrac{\pi}{2^{2^{n+1}}}$ and $B=\dfrac{\pi}{2^{2^{n+2}}}$, prove that

\[\prod_{r=0}^n (\cos 2^r A + \cos 2^r B)=\frac{1}{2^{n+1}\left( \cos \frac{\pi}{2^{n+2}}-\cos \frac{\pi}{2^{n+1}}\right)}\]

try for n = 0
A = \frac{\pi}{4} , \; B = \frac{\pi}{16}

Trying to prove
\prod_{r = 0 }^n (\cos 2^r A + \cos 2^r B ) = \frac{1}{2^{1}} \left( \cos \frac{\pi}{2^{2}} - \cos \frac{\pi}{2^{1}}\right)
By induction on r
when r = 0
\cos A + \cos B = \frac{1}{2(\cos \frac{\pi}{4} - \cos \frac{\pi}{2} ) }
the right hand side \frac{1}{\sqrt{2}}
which is not euqal to the left hand side

 

[sbhatnagar]

Opalg is right. It should have been $A=\frac{\pi}{2^{n+1}}$ and $B=\frac{\pi}{2^{n+2}}$. I am extremely sorry for this blunder.

 

[Opalg]

If $A=\dfrac{\pi}{2^{n+1}}$ and $B=\dfrac{\pi}{2^{n+2}}$, prove that

\[\prod_{r=0}^n (\cos 2^r A + \cos 2^r B)=\frac{1}{2^{n+1}\left( \cos \frac{\pi}{2^{n+2}}-\cos \frac{\pi}{2^{n+1}}\right)}\]

You want to show that $\displaystyle \bigl( \cos \tfrac{\pi}{2^{n+2}}-\cos \tfrac{\pi}{2^{n+1}}\bigr) \prod_{r=0}^n \bigl(\cos \tfrac{2^r\pi}{2^{n+2}} + \cos \tfrac{2^r\pi}{2^{n+1}}\bigr) = \tfrac1{2^{n+1}}.$ The left side of that is $$ \bigl( \cos \tfrac{\pi}{2^{n+2}}-\cos \tfrac{\pi}{2^{n+1}}\bigr) \bigl(\cos \tfrac{\pi}{2^{n+2}} + \cos \tfrac{\pi}{2^{n+1}}\bigr) \bigl(\cos \tfrac{\pi}{2^{n+1}} + \cos \tfrac{\pi}{2^{n}}\bigr) \bigl(\cos \tfrac{\pi}{2^{n}} + \cos \tfrac{\pi}{2^{n-1}}\bigr) \bigl(\cos \tfrac{\pi}{2^{n-1}} + \cos \tfrac{\pi}{2^{n-2}}\bigr) \cdots \bigl(\cos \tfrac{\pi}{2} + \cos \tfrac{\pi}{4}\bigr).\qquad(**) $$

Use the trig identity $\cos^2\theta = \tfrac12(\cos2\theta+1)$ to write the product of the first two factors in (**) as $$ \bigl( \cos \tfrac{\pi}{2^{n+2}}-\cos \tfrac{\pi}{2^{n+1}}\bigr) \bigl(\cos \tfrac{\pi}{2^{n+2}} + \cos \tfrac{\pi}{2^{n+1}}\bigr) = \cos^2\tfrac{\pi}{2^{n+2}}-\cos^2 \tfrac{\pi}{2^{n+1}} = \tfrac12\bigl( \cos \tfrac{\pi}{2^{n+1}}-\cos \tfrac{\pi}{2^{n}}\bigr).$$ Substitute that into (**) and then repeat the process of combining the first two factors. Each time you do that, it will introduce a factor of 1/2 and decrease by one the number of factors in the product. After doing this $n+1$ times you will be left with $\frac1{2^{n+1}}\bigl( \cos \tfrac{\pi}{2}-\cos \pi\bigr) = \frac1{2^{n+1}}.$