Find the value of the trigonometric sum

AI Thread Summary
The discussion revolves around solving a trigonometric sum involving angles related to π/7. Participants suggest using the sine function and product-to-sum identities to simplify the problem, leading to significant cancellations. One participant successfully solves the problem by choosing α as π/7, ultimately concluding that the sum equals -1/2. Another approach involves analyzing the unit circle with symmetrical vectors, demonstrating that the sum of the cosine components also results in -1/2 due to symmetry. Both methods confirm the same final result, showcasing different techniques to arrive at the solution.
Amlan mihir
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Homework Statement



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i have no idea how to proceed
 

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Must be the hint that throws you off. What can it be aiming at ? Pick some angle ##\alpha## and multiply with ##\sin\alpha## to see what you get.
You know that ##sin(n\pi)=0## so maybe there's something.

Or else perhaps make a cosine sketch marking the given angles ?
 
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Here's another hint: Pick some number ##\alpha##, multiply ##S## by ##sin(\alpha)##, and then use the fact that ##sin(\alpha)cos(\beta) = \frac{1}{2} (sin(\alpha + \beta) + sin(\alpha - \beta))##. If you pick ##\alpha## cleverly, you get some amazing cancellations to get a very simple result.
 
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solved it! , taking α as π/7 and using product to sum conversion for trig identities , it boils down to -[1][/2]
 
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Congratulation that you solved the problem. Here is an other way:
See the unit circle with 7 symmetrical vectors, with angle θ=2pi/7 between them. Their sum is zero, and so is the sum of the x components:cos(0) + cos(θ) +cos(2θ)+cos(3θ)+cos(4θ)+cos(5θ) +cos (6θ)=0. Because of symmetry, cos(θ)=cos (6θ) , cos(2θ)=cos(5θ) and cos(3θ)=cos(4θ). Therefore, 2(cos(θ) +cos(2θ)+cos(3θ))+1=0, cos(θ) +cos(2θ)+cos(3θ)=-1/2

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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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