Find the value of the trigonometric sum

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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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