Sigma Notation Question/Trig Identity

AI Thread Summary
The discussion revolves around deducing a second formula from a first using trigonometric identities. The user initially struggles to find the correct identity but later realizes that the "Sum to product identity" applies. They clarify the formulas involved, specifically adjusting the notation for accuracy. The conversation highlights the importance of correctly identifying variables A and B in the context of the identity. Ultimately, the user expresses gratitude for the clarification on the identity's name.
Lamoid
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[SOLVED] Sigma Notation Question/Trig Identity

I posted this elsewhere but I think I put it in the wrong place so I'm going to post my question again here.


Basically I have to deduce the second formula from the first. Both equations are the same except for the top of the right side, which makes me think it is just a simple matter of a trig identity. Unfortunately, I can't find an identity that would work. Can anyone help?
 

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i do not think those formulas are correctly written, i mean the second one. the last part should read like this i guess:

\frac{sin(\frac{nx}{2})cos(\frac{1}{2}(n+1)x)}{sin\frac{x}{2}}

to deduce this one from the first one, you need to apply this rule

sin(A)-sin(B)=2cos(\frac{A+B}{2})sin(\frac{A-B}{2})
 
OK, but looking at the first formula, wouldn't A be nx? and B be x/2?

Edit: Ah nvm. OK thank you very much! What is that formula called?
 
Last edited:
Lamoid said:
OK, but looking at the first formula, wouldn't A be nx? and B be x/2?

Edit: Ah nvm. OK thank you very much! What is that formula called?

A=(n+\frac{1}{2})x

B=\frac{x}{2}
 
It is a "Sum to product identity".
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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