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

$$looktheanser$$

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

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

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

Commented by Shrinava last updated on 23/Sep/22

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

Commented by MJS_new last updated on 23/Sep/22

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

Commented by MJS_new last updated on 23/Sep/22

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

Commented by Tawa11 last updated on 23/Sep/22

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

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

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

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