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Trigonometry III

sbhatnagar

Active member
Jan 27, 2012
95
Challenge Problem
If $A=\dfrac{\pi}{2^{n+1}}$ and $B=\dfrac{\pi}{2^{n+2}}$, prove that

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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

Active member
Mar 1, 2012
275
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
[tex] A = \frac{\pi}{4} , \; B = \frac{\pi}{16} [/tex]

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

sbhatnagar

Active member
Jan 27, 2012
95
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 $\frac1{2^{n+1}}\bigl( \cos \tfrac{\pi}{2}-\cos \pi\bigr) = \frac1{2^{n+1}}.$