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S-k-1-n-sin-k-k-1-n-cos-k-where-k-k-1-n-is-an-arithmetic-progression-show-that-S-tan-where-1-n-k-1-n-k-is-the-arithmetic-mean-of-

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Question Number 153758 by yeti123 last updated on 10/Sep/21
S = ((Σ_(k = 1) ^n sin(θ_k ))/(Σ_(k = 1) ^n cos(θ_k ))); where (θ_k )_(k = 1) ^n is an arithmetic progression. show that S = tan(θ^� ) where θ^� = (1/n)Σ_(k = 1) ^n θ_k is the arithmetic mean of (θ_k )
$${S}\:=\:\frac{\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}\left(\theta_{{k}} \right)}{\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\left(\theta_{{k}} \right)};\:\mathrm{where}\:\left(\theta_{{k}} \right)_{{k}\:=\:\mathrm{1}} ^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}. \\ $$$$\mathrm{show}\:\mathrm{that}\:{S}\:=\:\mathrm{tan}\left(\bar {\theta}\right) \\ $$$$\mathrm{where}\:\bar {\theta}\:=\:\frac{\mathrm{1}}{{n}}\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\theta_{{k}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{arithmetic}\:\mathrm{mean}\:\mathrm{of}\:\left(\theta_{{k}} \right) \\ $$
Answered by mindispower last updated on 10/Sep/21
θ_k =ak+b Σ_(k=1) ^n e^(iθ_k ) =e^(ib) .e^(ia) ((1−(e^(ia) )^n )/(1−e^(ia) ))=S =e^(i(a+b)) .((e^(i(((na)/2)−(a/2))) cos(((na)/2)))/(cos((a/2)))) =e^(i((a/2)(n+1)+b)) ((cos(n(a/2)))/(cos((a/2))))n Σsin(θ_k )=((cos(((na)/2)))/(cos((a/2))))sin((1/n)(((n(n+1)a)/2)+bn)) Σcos(θ_k )=((cos(((na)/2)))/(cos((a/2))))cos(((1/n)(((n(n+1)a)/2)+bn)) Σθ_k =Σ_(k=1) ^n (ak+b)=nb+(a/2)(n(n+1) ((Σsin(θ_k ))/(Σcos(θ_k )))=((((cos(((na)/(2)))))/(cos((a/2))))sin((1/n)(n(n+1)(a/2)+nb)))/(((cos(((na)/2)))/(cos((a/2))))cos((1/n)((((n+1)/2))na+nb))) =tg((1/n)(Σθ_k ))=tg(θ_k ^− )
$$\theta_{{k}} ={ak}+{b} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{i}\theta_{{k}} } ={e}^{{ib}} .{e}^{{ia}} \frac{\mathrm{1}−\left({e}^{{ia}} \right)^{{n}} }{\mathrm{1}−{e}^{{ia}} }={S} \\ $$$$={e}^{{i}\left({a}+{b}\right)} .\frac{{e}^{{i}\left(\frac{{na}}{\mathrm{2}}−\frac{{a}}{\mathrm{2}}\right)} {cos}\left(\frac{{na}}{\mathrm{2}}\right)}{{cos}\left(\frac{{a}}{\mathrm{2}}\right)} \\ $$$$={e}^{{i}\left(\frac{{a}}{\mathrm{2}}\left({n}+\mathrm{1}\right)+{b}\right)} \frac{{cos}\left({n}\frac{{a}}{\mathrm{2}}\right)}{{cos}\left(\frac{{a}}{\mathrm{2}}\right)}{n} \\ $$$$\Sigma{sin}\left(\theta_{{k}} \right)=\frac{{cos}\left(\frac{{na}}{\mathrm{2}}\right)}{{cos}\left(\frac{{a}}{\mathrm{2}}\right)}{sin}\left(\frac{\mathrm{1}}{{n}}\left(\frac{{n}\left({n}+\mathrm{1}\right){a}}{\mathrm{2}}+{bn}\right)\right) \\ $$$$\Sigma{cos}\left(\theta_{{k}} \right)=\frac{{cos}\left(\frac{{na}}{\mathrm{2}}\right)}{{cos}\left(\frac{{a}}{\mathrm{2}}\right)}{cos}\left(\left(\frac{\mathrm{1}}{{n}}\left(\frac{{n}\left({n}+\mathrm{1}\right){a}}{\mathrm{2}}+{bn}\right)\right)\right. \\ $$$$\Sigma\theta_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({ak}+{b}\right)={nb}+\frac{{a}}{\mathrm{2}}\left({n}\left({n}+\mathrm{1}\right)\right. \\ $$$$\frac{\Sigma{sin}\left(\theta_{{k}} \right)}{\Sigma{cos}\left(\theta_{{k}} \right)}=\frac{\frac{{cos}\left(\frac{{na}}{\left.\mathrm{2}\right)}\right)}{{cos}\left(\frac{{a}}{\mathrm{2}}\right)}{sin}\left(\frac{\mathrm{1}}{{n}}\left({n}\left({n}+\mathrm{1}\right)\frac{{a}}{\mathrm{2}}+{nb}\right)\right)}{\frac{{cos}\left(\frac{{na}}{\mathrm{2}}\right)}{{cos}\left(\frac{{a}}{\mathrm{2}}\right)}{cos}\left(\frac{\mathrm{1}}{{n}}\left(\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right){na}+{nb}\right)\right.} \\ $$$$={tg}\left(\frac{\mathrm{1}}{{n}}\left(\Sigma\theta_{{k}} \right)\right)={tg}\left(\overset{−} {\theta}_{{k}} \right) \\ $$$$ \\ $$

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