trigonometrisk identitet

[Janhaa]

Vis at:

[tex]\tan({\pi\over 7})\cdot \tan({2\pi\over 7})\cdot \tan({3\pi\over 7})=\sqrt7[/tex]

[olalia]

[tex]x=tan(\frac{\pi}{7})tan(\frac{2\pi}{7})tan(\frac{3\pi}{7})=i\frac{e^{\frac{6\pi i}{7}}-e^{\frac{2\pi i}{7}}+e^{\frac{10\pi i}{7}}-e^{\frac{4\pi i}{7}}+e^{\frac{12\pi i}{7}}-e^{\frac{8\pi i}{7}}}{2+e^{\frac{6\pi i}{7}}+e^{\frac{2\pi i}{7}}+e^{\frac{10\pi i}{7}}+e^{\frac{4\pi i}{7}}+e^{\frac{12\pi i}{7}}+e^{\frac{8\pi i}{7}}}>0[/tex]

[tex] 2+e^{\frac{6\pi i}{7}}+e^{\frac{2\pi i}{7}}+e^{\frac{3\pi i}{7}}+e^{\frac{4\pi i}{7}}+e^{\frac{5\pi i}{7}}+e^{\frac{\pi i}{7}}=1+\sum \{x : x^7=1\}=1[/tex]
bruker en dette i nevneren og finner konjugerte par i telleren får en:

[tex]x=2(sin(\frac{2\pi}{7})+sin(\frac{4\pi}{7})-sin(\frac{6\pi}{7})[/tex]
Bruker videre at
[tex]sin(a)sin(b)=\frac{1}{2}(cos(a-b)-cos(a+b))[/tex]
[tex]x^2=4(sin^2(\frac{2\pi}{7})+sin^2(\frac{4\pi}{7})+sin^2(\frac{6\pi}{7})+2sin(\frac{2\pi}{7})sin(\frac{4\pi}{7})- 2sin(\frac{2\pi}{7})sin(\frac{6\pi}{7})-2sin(\frac{4\pi}{7})sin(\frac{6\pi}{7}))[/tex]
Til slutt bruker en samme likning som for nevneren øverst, men bare for realdelen.
[tex]x^2=6-2cos(\frac{2\pi}{7})-2cos(\frac{4\pi}{7})-2cos(\frac{6\pi}{7})=6-\sum_{n=1}^6 cos(\frac{\2\pi}{n})= 7-\sum_{n=0}^6 cos(\frac{\2\pi}{n})=7 [/tex]

[tex] x>0,x^2=7 \Leftrightarrow x=\underline{\underline{\sqrt{7}}} [/tex]

[Charlatan]

Oppfølger:

Finn [tex]\prod^{n}_{k=1} \tan(\frac{\pi k}{2n+1})[/tex].

[Janhaa]

[tex]x=tan(\frac{\pi}{7})tan(\frac{2\pi}{7})tan(\frac{3\pi}{7})=i\frac{e^{\frac{6\pi i}{7}}-e^{\frac{2\pi i}{7}}+e^{\frac{10\pi i}{7}}-e^{\frac{4\pi i}{7}}+e^{\frac{12\pi i}{7}}-e^{\frac{8\pi i}{7}}}{2+e^{\frac{6\pi i}{7}}+e^{\frac{2\pi i}{7}}+e^{\frac{10\pi i}{7}}+e^{\frac{4\pi i}{7}}+e^{\frac{12\pi i}{7}}+e^{\frac{8\pi i}{7}}}>0[/tex]
[tex] 2+e^{\frac{6\pi i}{7}}+e^{\frac{2\pi i}{7}}+e^{\frac{3\pi i}{7}}+e^{\frac{4\pi i}{7}}+e^{\frac{5\pi i}{7}}+e^{\frac{\pi i}{7}}=1+\sum \{x : x^7=1\}=1[/tex]
bruker en dette i nevneren og finner konjugerte par i telleren får en:
[tex]x=2(sin(\frac{2\pi}{7})+sin(\frac{4\pi}{7})-sin(\frac{6\pi}{7})[/tex]
Bruker videre at
[tex]sin(a)sin(b)=\frac{1}{2}(cos(a-b)-cos(a+b))[/tex]
[tex]x^2=4(sin^2(\frac{2\pi}{7})+sin^2(\frac{4\pi}{7})+sin^2(\frac{6\pi}{7})+2sin(\frac{2\pi}{7})sin(\frac{4\pi}{7})- 2sin(\frac{2\pi}{7})sin(\frac{6\pi}{7})-2sin(\frac{4\pi}{7})sin(\frac{6\pi}{7}))[/tex]
Til slutt bruker en samme likning som for nevneren øverst, men bare for realdelen.
[tex]x^2=6-2cos(\frac{2\pi}{7})-2cos(\frac{4\pi}{7})-2cos(\frac{6\pi}{7})=6-\sum_{n=1}^6 cos(\frac{\2\pi}{n})= 7-\sum_{n=0}^6 cos(\frac{\2\pi}{n})=7 [/tex]
[tex] x>0,x^2=7 \Leftrightarrow x=\underline{\underline{\sqrt{7}}} [/tex]

nå er jeg ikke så godt bevandra i komplekse tall sine anvendelser til trigonometri, men dette så finfint ut!

(husker jeg løste den på "vanlig" måte).