Question Number 207065 by manxsol last updated on 05/May/24
![f(x)=[cos2x+cos3x][cos4x+cos6x][[cosx+cos5x] evaluar f(((2π)/(13)))](Q207065.png)
$$ \\ $$$$\:\:\:{f}\left({x}\right)=\left[{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x}\right]\left[{cos}\mathrm{4}{x}+{cos}\mathrm{6}{x}\right]\left[\left[{cosx}+{cos}\mathrm{5}{x}\right]\right. \\ $$$${evaluar}\:\:\:{f}\left(\frac{\mathrm{2}\pi}{\mathrm{13}}\right)\:\: \\ $$
Answered by Berbere last updated on 06/May/24
![4cos(x)cos(5x)cos(2x)cos(3x).[cos(2x)+cos( 3x)]]a2( 8cos((π/(13)))cos(((5π)/(13)))cos(((2π)/(13)))cos(((4π)/(13)))cos(((6π)/(13)))cos(((10π)/(13))) =f((π/(12)))−8Π_(k=1) ^6 cos(((kπ)/(13)))=−8Σ_(k=7) ^(12) cos(k(π/(13))) ⇒(Π_(k=1) ^(12) (cos(k(π/(13)))))^(1/2) =(Π_(k=1) ^6 cos(((kπ)/(13)))) =p(z)=Z^(13) −1=Π_(k=0) ^(12) (z−e^((2ikπ)/(13)) )=p(−1)=−2 =−Π_(k=0) ^(12) (1+e^((2ikπ)/(13)) )=−2^(13) Π_(k=0) ^(12) cos(((kπ)/(13))).e^(i12π) =−2 =Π_(k=0) ^(12) cos(k(π/(13)))=(1/2^(12) )⇒f((π/(12)))=−2^3 .(1/2^6 )=−(1/8)](Q207097.png)
$$\left.\mathrm{4}{cos}\left({x}\right){cos}\left(\mathrm{5}{x}\right){cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right).\left[{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\:\mathrm{3}{x}\right)\right]\right]{a}\mathrm{2}\left(\right. \\ $$$$\mathrm{8}{cos}\left(\frac{\pi}{\mathrm{13}}\right){cos}\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{13}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{13}}\right){cos}\left(\frac{\mathrm{6}\pi}{\mathrm{13}}\right){cos}\left(\frac{\mathrm{10}\pi}{\mathrm{13}}\right) \\ $$$$={f}\left(\frac{\pi}{\mathrm{12}}\right)−\mathrm{8}\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{cos}\left(\frac{{k}\pi}{\mathrm{13}}\right)=−\mathrm{8}\underset{{k}=\mathrm{7}} {\overset{\mathrm{12}} {\sum}}{cos}\left({k}\frac{\pi}{\mathrm{13}}\right) \\ $$$$\Rightarrow\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\prod}}\left({cos}\left({k}\frac{\pi}{\mathrm{13}}\right)\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{cos}\left(\frac{{k}\pi}{\mathrm{13}}\right)\right) \\ $$$$={p}\left({z}\right)={Z}^{\mathrm{13}} −\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{12}} {\prod}}\left({z}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{13}}} \right)={p}\left(−\mathrm{1}\right)=−\mathrm{2} \\ $$$$=−\underset{{k}=\mathrm{0}} {\overset{\mathrm{12}} {\prod}}\left(\mathrm{1}+{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{13}}} \right)=−\mathrm{2}^{\mathrm{13}} \underset{{k}=\mathrm{0}} {\overset{\mathrm{12}} {\prod}}{cos}\left(\frac{{k}\pi}{\mathrm{13}}\right).{e}^{{i}\mathrm{12}\pi} =−\mathrm{2} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{12}} {\prod}}{cos}\left({k}\frac{\pi}{\mathrm{13}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{12}} }\Rightarrow{f}\left(\frac{\pi}{\mathrm{12}}\right)=−\mathrm{2}^{\mathrm{3}} .\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$ \\ $$
Commented by manxsol last updated on 07/May/24
$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{complicatedt} \\ $$$$\mathrm{wha}\:\mathrm{theory}\:\mathrm{do}\:\mathrm{you}\:\mathrm{need}\:\mathrm{to}\:\mathrm{study}\:\mathrm{tou} \\ $$$$\mathrm{nderstand}\:\mathrm{it} \\ $$