Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 176603 by a.lgnaoui last updated on 23/Sep/22

look the anser

$${look}\:{the}\:{anser} \\ $$

Commented by a.lgnaoui last updated on 23/Sep/22

Commented by mr W last updated on 23/Sep/22

x=±11, ±((11)/3), ±((11)/5), ±((11)/7), ±((11)/9)

$${x}=\pm\mathrm{11},\:\pm\frac{\mathrm{11}}{\mathrm{3}},\:\pm\frac{\mathrm{11}}{\mathrm{5}},\:\pm\frac{\mathrm{11}}{\mathrm{7}},\:\pm\frac{\mathrm{11}}{\mathrm{9}} \\ $$

Commented by Shrinava last updated on 23/Sep/22

How dear professor please

$$\mathrm{How}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{please} \\ $$

Commented by MJS_new last updated on 23/Sep/22

Σ_(k=1) ^n (−1)^(k+1) cos ((kπ)/(2n+1)) =(1/2) for n=5 Σ_(k=1) ^5 (−1)^(k+1) cos ((kπ)/(11)) =(1/2) cos (π/(11)) −cos ((2π)/(11)) +cos ((3π)/(11)) −cos ((4π)/(11)) +cos ((5π)/(11)) =(1/2) cos (π/(11)) +cos ((3π)/(11)) +cos ((5π)/(11)) =(1/2)+cos ((2π)/(11)) +cos ((4π)/(11))

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \mathrm{cos}\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{for}\:{n}=\mathrm{5} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \mathrm{cos}\:\frac{{k}\pi}{\mathrm{11}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{11}}\:−\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{11}}\:+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{11}}\:−\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{11}}\:+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{11}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{11}}\:+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{11}}\:+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{11}}\:=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{11}}\:+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{11}} \\ $$

Commented by MJS_new last updated on 23/Sep/22

maybe this helps to see... cos py =((e^(ipy) +e^(−ipy) )/2)=((e^(2ipy) +1)/(2e^(ipy) )) let e^(iy) =t cos py =((t^(2p) +1)/(2t^p )) cos y −cos 2y +cos 3y −cos 4y +cos 5y =(1/2) ((t^2 +1)/(2t))−((t^4 +1)/(2t^2 ))+((t^6 +1)/(2t^3 ))−((t^8 +1)/(2t^4 ))+((t^(10) +1)/(2t^5 ))=(1/2) ((t^(10) −t^9 +t^8 −t^7 +t^6 +t^4 −t^3 +t^2 −t+1)/(2t^5 ))=(1/2) t^(10) −t^9 +t^8 −t^7 +t^6 −t^5 +t^4 −t^3 +t^2 −t+1=0 (t+1)(t^(10) −t^9 +t^8 −t^7 +t^6 −t^5 +t^4 −t^3 +t^2 −t+1)=0 t^(11) +1=0

$$\mathrm{maybe}\:\mathrm{this}\:\mathrm{helps}\:\mathrm{to}\:\mathrm{see}... \\ $$$$\mathrm{cos}\:{py}\:=\frac{\mathrm{e}^{\mathrm{i}{py}} +\mathrm{e}^{−\mathrm{i}{py}} }{\mathrm{2}}=\frac{\mathrm{e}^{\mathrm{2i}{py}} +\mathrm{1}}{\mathrm{2e}^{\mathrm{i}{py}} } \\ $$$$\mathrm{let}\:\mathrm{e}^{\mathrm{i}{y}} ={t} \\ $$$$\mathrm{cos}\:{py}\:=\frac{{t}^{\mathrm{2}{p}} +\mathrm{1}}{\mathrm{2}{t}^{{p}} } \\ $$$$\mathrm{cos}\:{y}\:−\mathrm{cos}\:\mathrm{2}{y}\:+\mathrm{cos}\:\mathrm{3}{y}\:−\mathrm{cos}\:\mathrm{4}{y}\:+\mathrm{cos}\:\mathrm{5}{y}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}}−\frac{{t}^{\mathrm{4}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }+\frac{{t}^{\mathrm{6}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{3}} }−\frac{{t}^{\mathrm{8}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{4}} }+\frac{{t}^{\mathrm{10}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{t}^{\mathrm{10}} −{t}^{\mathrm{9}} +{t}^{\mathrm{8}} −{t}^{\mathrm{7}} +{t}^{\mathrm{6}} +{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{\mathrm{2}{t}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}^{\mathrm{10}} −{t}^{\mathrm{9}} +{t}^{\mathrm{8}} −{t}^{\mathrm{7}} +{t}^{\mathrm{6}} −{t}^{\mathrm{5}} +{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{10}} −{t}^{\mathrm{9}} +{t}^{\mathrm{8}} −{t}^{\mathrm{7}} +{t}^{\mathrm{6}} −{t}^{\mathrm{5}} +{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$${t}^{\mathrm{11}} +\mathrm{1}=\mathrm{0} \\ $$

Commented by Tawa11 last updated on 23/Sep/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Answered by a.lgnaoui last updated on 23/Sep/22

Answered by a.lgnaoui last updated on 23/Sep/22

Terms of Service

Privacy Policy

Contact: info@tinkutara.com