Question Number 199226 by mr W last updated on 29/Oct/23
![find ((2 sin 2°+4 sin 4°+...+180 sin 180°)/(90))=? [an unsolved old question]](Q199226.png)
$${find} \\ $$$$\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}°+\mathrm{4}\:\mathrm{sin}\:\mathrm{4}°+...+\mathrm{180}\:\mathrm{sin}\:\mathrm{180}°}{\mathrm{90}}=? \\ $$$$ \\ $$$$\left[{an}\:{unsolved}\:{old}\:{question}\right] \\ $$
Answered by mr W last updated on 29/Oct/23
![let′s find at first S=Σ_(k=1) ^n kq^k =q+2q^2 +3q^3 +...+nq^n qS=q^2 +2q^3 +3q^4 +...+(n−1)q^n +nq^(n+1) (1−q)S=q+q^2 +q^3 +...+q^n −nq^(n+1) (1−q)S=((q−q^(n+1) )/(1−q))−nq^(n+1) =((q−(n+1)q^(n+1) +nq^(n+2) )/(1−q)) ⇒S=((q−(n+1)q^(n+1) +nq^(n+2) )/((1−q)^2 )) say A=Σ_(k=1) ^n k cos kα B=Σ_(k=1) ^n k sin kα A+Bi=Σ_(k=1) ^n k(cos kα+i sin kα)=Σ_(k=1) ^n ke^(kαi) A+Bi=Σ_(k=1) ^n k(e^(αi) )^k =((e^(αi) −(n+1)e^((n+1)αi) +ne^((n+2)αi) )/((1−e^(αi) )^2 )) A+Bi=((e^(αi) −(n+1)e^((n+1)αi) +ne^((n+2)αi) )/(1+e^(2αi) −2e^(αi) )) A+Bi=((cos α+i sin α−(n+1)cos (n+1)α−i (n+1)sin (n+1)α+n cos (n+2)α+i n sin (n+2)α)/(1+cos 2α+i sin 2α−2 cos α−i 2 sin α)) A+Bi=(([cos α−(n+1)cos (n+1)α+n cos (n+2)α]+i [sin α−(n+1)sin (n+1)α+n sin (n+2)α])/([1+cos 2α−2 cos α]+i [sin 2α−2 sin α])) A+Bi=((p+iq)/(u+iv)) A+Bi=(((p+iq)(u−iv))/((u+iv)(u−iv)))=(((pu+qv)+(qu−pv)i)/(u^2 +v^2 )) ⇒A=((pu+qv)/(u^2 +v^2 )) =(([cos α−(n+1)cos (n+1)α+n cos (n+2)α][1+cos 2α−2 cos α]+[sin α−(n+1)sin (n+1)α+n sin (n+2)α][sin 2α−2 sin α])/((1+cos 2α−2 cos α)^2 +(sin 2α−2 sin α)^2 )) =((cos α [cos α−(n+1)cos (n+1)α+n cos (n+2)α]+sin α [sin α−(n+1)sin (n+1)α+n sin (n+2)α])/(2(cos α−1))) =(((n+1)cos nα−n cos (n+1)α−1)/(2(1−cos α))) ⇒B=((qu−pv)/(u^2 +v^2 )) =(([sin α−(n+1)sin (n+1)α+n sin (n+2)α][1+cos 2α−2 cos α]−[cos α−(n+1)cos (n+1)α+n cos (n+2)α][sin 2α−2 sin α])/((1+cos 2α−2 cos α)^2 +(sin 2α−2 sin α)^2 )) =((cos α [(n+1)sin (n+1)α−n sin (n+2)α]+sin α [n cos (n+2)α−(n+1)cos (n+1)α])/(2(1−cos α))) =(((n+1) sin nα−n sin (n+1)α)/(2(1−cos α))) ((2 sin 2°+4 sin 4°+...×180 sin 180°)/(90)) =((Σ_(k=1) ^(90) (2k) sin (2k)°)/(90))=((2Σ_(k=1) ^(90) k sin (2k)°)/(90)) =(2/(90))×B_(α=2°, n=90) =(2/(90))×((91 sin (90×2°)−90 sin (91×2°))/(2(1−cos 2°))) =((sin 2°)/(1−cos 2°))=((2 sin 1° cos 1°)/(1−1+2 sin^2 1°))=(1/(tan 1°)) btw. we get ((2 cos 2°+4 cos 4°+...+180 cos 180°)/(90)) =(2/(90))×A_(α=2°, n=90) =(2/(90))×((91 cos (90×2°)−90 cos (91×2°)−1)/(2(1−cos 2°))) =−1−(1/(45(1−cos 2°)))](Q199227.png)
$${let}'{s}\:{find}\:{at}\:{first} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kq}^{{k}} ={q}+\mathrm{2}{q}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{3}} +...+{nq}^{{n}} \\ $$$${qS}={q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} +\mathrm{3}{q}^{\mathrm{4}} +...+\left({n}−\mathrm{1}\right){q}^{{n}} +{nq}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{q}\right){S}={q}+{q}^{\mathrm{2}} +{q}^{\mathrm{3}} +...+{q}^{{n}} −{nq}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{q}\right){S}=\frac{{q}−{q}^{{n}+\mathrm{1}} }{\mathrm{1}−{q}}−{nq}^{{n}+\mathrm{1}} =\frac{{q}−\left({n}+\mathrm{1}\right){q}^{{n}+\mathrm{1}} +{nq}^{{n}+\mathrm{2}} }{\mathrm{1}−{q}} \\ $$$$\Rightarrow{S}=\frac{{q}−\left({n}+\mathrm{1}\right){q}^{{n}+\mathrm{1}} +{nq}^{{n}+\mathrm{2}} }{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${say} \\ $$$${A}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\:\mathrm{cos}\:{k}\alpha \\ $$$${B}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\:\mathrm{sin}\:{k}\alpha \\ $$$${A}+{Bi}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left(\mathrm{cos}\:{k}\alpha+{i}\:\mathrm{sin}\:{k}\alpha\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ke}^{{k}\alpha{i}} \\ $$$${A}+{Bi}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({e}^{\alpha{i}} \right)^{{k}} =\frac{{e}^{\alpha{i}} −\left({n}+\mathrm{1}\right){e}^{\left({n}+\mathrm{1}\right)\alpha{i}} +{ne}^{\left({n}+\mathrm{2}\right)\alpha{i}} }{\left(\mathrm{1}−{e}^{\alpha{i}} \right)^{\mathrm{2}} } \\ $$$${A}+{Bi}=\frac{{e}^{\alpha{i}} −\left({n}+\mathrm{1}\right){e}^{\left({n}+\mathrm{1}\right)\alpha{i}} +{ne}^{\left({n}+\mathrm{2}\right)\alpha{i}} }{\mathrm{1}+{e}^{\mathrm{2}\alpha{i}} −\mathrm{2}{e}^{\alpha{i}} } \\ $$$${A}+{Bi}=\frac{\mathrm{cos}\:\alpha+{i}\:\mathrm{sin}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha−{i}\:\left({n}+\mathrm{1}\right)\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{cos}\:\left({n}+\mathrm{2}\right)\alpha+{i}\:{n}\:\mathrm{sin}\:\left({n}+\mathrm{2}\right)\alpha}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha+{i}\:\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{cos}\:\alpha−{i}\:\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$${A}+{Bi}=\frac{\left[\mathrm{cos}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{cos}\:\left({n}+\mathrm{2}\right)\alpha\right]+{i}\:\left[\mathrm{sin}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{sin}\:\left({n}+\mathrm{2}\right)\alpha\right]}{\left[\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{cos}\:\alpha\right]+{i}\:\left[\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{sin}\:\alpha\right]} \\ $$$${A}+{Bi}=\frac{{p}+{iq}}{{u}+{iv}} \\ $$$${A}+{Bi}=\frac{\left({p}+{iq}\right)\left({u}−{iv}\right)}{\left({u}+{iv}\right)\left({u}−{iv}\right)}=\frac{\left({pu}+{qv}\right)+\left({qu}−{pv}\right){i}}{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}=\frac{{pu}+{qv}}{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} } \\ $$$$\:\:=\frac{\left[\mathrm{cos}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{cos}\:\left({n}+\mathrm{2}\right)\alpha\right]\left[\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{cos}\:\alpha\right]+\left[\mathrm{sin}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{sin}\:\left({n}+\mathrm{2}\right)\alpha\right]\left[\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{sin}\:\alpha\right]}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left(\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{cos}\:\alpha\:\left[\mathrm{cos}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{cos}\:\left({n}+\mathrm{2}\right)\alpha\right]+\mathrm{sin}\:\alpha\:\left[\mathrm{sin}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{sin}\:\left({n}+\mathrm{2}\right)\alpha\right]}{\mathrm{2}\left(\mathrm{cos}\:\alpha−\mathrm{1}\right)} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\mathrm{cos}\:{n}\alpha−{n}\:\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)} \\ $$$$\Rightarrow{B}=\frac{{qu}−{pv}}{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} } \\ $$$$\:\:=\frac{\left[\mathrm{sin}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{sin}\:\left({n}+\mathrm{2}\right)\alpha\right]\left[\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{cos}\:\alpha\right]−\left[\mathrm{cos}\:\alpha−\left({n}+\mathrm{1}\right)\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha+{n}\:\mathrm{cos}\:\left({n}+\mathrm{2}\right)\alpha\right]\left[\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{sin}\:\alpha\right]}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left(\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{cos}\:\alpha\:\left[\left({n}+\mathrm{1}\right)\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha−{n}\:\mathrm{sin}\:\left({n}+\mathrm{2}\right)\alpha\right]+\mathrm{sin}\:\alpha\:\left[{n}\:\mathrm{cos}\:\left({n}+\mathrm{2}\right)\alpha−\left({n}+\mathrm{1}\right)\mathrm{cos}\:\left({n}+\mathrm{1}\right)\alpha\right]}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\:\mathrm{sin}\:{n}\alpha−{n}\:\mathrm{sin}\:\left({n}+\mathrm{1}\right)\alpha}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)} \\ $$$$ \\ $$$$\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}°+\mathrm{4}\:\mathrm{sin}\:\mathrm{4}°+...×\mathrm{180}\:\mathrm{sin}\:\mathrm{180}°}{\mathrm{90}} \\ $$$$=\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{90}} {\sum}}\left(\mathrm{2}{k}\right)\:\mathrm{sin}\:\left(\mathrm{2}{k}\right)°}{\mathrm{90}}=\frac{\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{90}} {\sum}}{k}\:\mathrm{sin}\:\left(\mathrm{2}{k}\right)°}{\mathrm{90}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{90}}×{B}_{\alpha=\mathrm{2}°,\:{n}=\mathrm{90}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{90}}×\frac{\mathrm{91}\:\mathrm{sin}\:\left(\mathrm{90}×\mathrm{2}°\right)−\mathrm{90}\:\mathrm{sin}\:\left(\mathrm{91}×\mathrm{2}°\right)}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}°\right)} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{2}°}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}°}=\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{1}°\:\mathrm{cos}\:\mathrm{1}°}{\mathrm{1}−\mathrm{1}+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{1}°} \\ $$$$ \\ $$$${btw}.\:{we}\:{get} \\ $$$$\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}°+\mathrm{4}\:\mathrm{cos}\:\mathrm{4}°+...+\mathrm{180}\:\mathrm{cos}\:\mathrm{180}°}{\mathrm{90}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{90}}×{A}_{\alpha=\mathrm{2}°,\:{n}=\mathrm{90}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{90}}×\frac{\mathrm{91}\:\mathrm{cos}\:\left(\mathrm{90}×\mathrm{2}°\right)−\mathrm{90}\:\mathrm{cos}\:\left(\mathrm{91}×\mathrm{2}°\right)−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}°\right)} \\ $$$$=−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{45}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}°\right)} \\ $$
Commented by hardmath last updated on 29/Oct/23
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{are} \\ $$$$\mathrm{great}... \\ $$