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2080 Points
3 years ago
well, the question is dayumn puzzling tbh. but not if u know the steps in the proof of summation of sine and cos series when the arguments are in AP. very few students bother to understand the proof, rather just memorise the formula as it is:
sina+sin(a+b)+sin(a+2b)+.....+sin(a+(n-1)b)=sin(a+(n-1)b/2)*sin(nb/2)/sin(b/2).......(1)
now in your question, let S be the reqd sum. multiply both sides by 2sinθ/2
2sinθ/2*S=S’=Σ r(sin(r+1/2)θ-sin(r-1/2)θ)
=Σr(sin(r+1/2)θ-(r-1)sin(r-1/2)θ – Σsin(r-1/2)θ
now, r(sin(r+1/2)θ-(r-1)sin(r-1/2)θ easily telescopes, and sin(r-1/2)θ is easily obtained because the argument inside [(r-1/2)θ] forms an AP. so simply use the result from (1).
the first sum turns out to be
n*sin(n+1/2)θ
while the latter will be
sin(nθ/2)*sin(θ/2+(n-1)*θ/2)/sinθ/2=sin^2(nθ/2)/sin(θ/2)