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andu_flavius95
maestru (V)
Pe: 2011-09-23T15:28:27+03:00 In: In:

Inductie cu sin

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	\begin{array}{l}
	 Sa\;se\;verifice\;prin\;inductie\;matematica\;relatia]
	

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Multumesc mult! 🙂

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16 raspunsuri

  1. PhantomR
    expert (VI)

    Cu riscul de a parea aiurea deja cu astfel de intrebari, ai certitudinea ca exista o solutie prin inductie? (spun asta din cauza numitorului care la pasul de inductie va deveni n+1).

  2. andu_flavius95
    maestru (V)

    Asa se cere in manual.. 🙂

  4. andu_flavius95
    maestru (V)

    Poate este mai usoara aceasta, tot prin inductie se cere:

        \[ \cos 3x + \cos 5x + ... + \cos (2n + 1)x = \frac{{\cos (n + 2)x \cdot \sin nx}}{{\sin x}}\quad ,n \in N^{*.} \]

    Multumesc mult!

  5. Anonymous

    Super tare exercitiul e exact genu` de exercitiul de care-mi place mie…iti scriu imediat si rezolvarea…dar nu prin inductie

  6. Anonymous

    \rm {Notam cu A=cos(\frac{\pi}{n})+cos(\frac{2*\pi}{n})+...+cos(\frac{(n-1)*\pi}{n}) iar cu B expresia noastra. A+i*B=(cos(\frac{\pi}{n})+i*sin(\frac{\pi}{n}))+...+(cos(\frac{(n-1)*\pi}{n})+i*sin(\frac{(n-1)*\pi}{n})) \\=>A+i*B=\frac{1-cos(\frac{n*\pi}{n})-i*sin(\frac{n*\pi}{n})}{1-cos(\frac{\pi}{n})-i*sin(\frac{\pi}{n})}*(cos(\frac{\pi}{n})+i*sin(\frac{\pi}{n})) \\=> notam \frac{\pi}{n}=x => 2*\frac{[cos(x)+i*sin(x)]*[1-cos(x)+i*sin(x)]}{1-2*cos(x)+(cos(x))^2+(sin(x))^2} \\=> A+i*B=\frac{cos(x)-(cos(x))^2+i*sin(x)*cos(x)+i*sin(x)-i*sin(x)*cos(x)-(sin(x))^2}{2*(sin(\frac{x}{2}))^2}=\frac{cos(x)-1+i*sin(x)}{2*(sin(\frac{x}{2}))^2} \\=> B=\frac{sin(x)}{2*sin(\frac{x}{2})^2} =>B=\frac{2*sin(\frac{x}{2})*cos(\frac{x}{2})}{2*sin(\frac{x}{2})^2} \\=>B=\frac{cos(\frac{x}{2})}{sin(\frac{x}{2})} = ctg(\frac{\pi}{2*n})

  8. andu_flavius95
    maestru (V)

    Frumoasa metoda de rezolvare. Am o intrebare: i la ce se refera in context, ce fel de numar este?

  9. Anonymous

    i=sqrt{-1}

  10. andu_flavius95
    maestru (V)

    Multumesc mult! ma gandeam la complexitate

  12. Anonymous

    Cu celalalt te descurci? Ash vrea sa postezi rezolvarea ta prin metoda „mea” daca poti si ai timp! (ca sa-mi demonstrezi ca ai inteles) 😀

  13. andu_flavius95
    maestru (V)

    Exista o problema, acestea sunt de clasa a 9 iar eu nu am facut inca numerele complexe.Scuze ca nu am mentionam mai inainte.

  14. Anonymous

    Ah, atunci nu-ti ramane alta solutie decat inductia zic eu. Dar arata naspa. Cand ajungi in a 11-a nu mai ai scuze…tb sa faci exercitiul asta cum ti-am aratat 😀 (banuiesc ca in a 11-a se fac numerele complexe).

  16. andu_flavius95
    maestru (V)

    In clasa a 10 se fac numerele complexe, urmeaza chiar anul acesta la mine.

  17. red12dog34
    veteran (III)

    andu_flavius95 wrote: 

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 Sa\;se\;verifice\;prin\;inductie\;matematica\;relatia]
	

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Emergency stop.

    Multumesc mult! 🙂

    Prin inductie se poate demonstra identitatea:

    \sin\alpha+\sin 2\alpha+\ldots+\sin (n-1)\alpha=\frac{\sin\frac{(n-1)\alpha}{2}\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}, \ \forall n\geq 2, \ \alpha\neq 2k\pi

    Pentru \alpha=\frac{\pi}{n} se obtine identitatea ceruta.

  18. red12dog34
    veteran (III)

    andu_flavius95 wrote: 

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 Sa\;se\;verifice\;prin\;inductie\;matematica\;relatia]
	

*** Error message:
\begin{array} on input line 10 ended by \end{document}.
leading text: \end{document}
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leading text: \end{document}
Missing $ inserted.
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Missing } inserted.
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Emergency stop.

    Multumesc mult! 🙂

    Fara inductie: se inmulteste egalitatea cu 2\sin\frac{\pi}{2n}:

    Atunci trebuie demonstrata identitatea

    2\sin\frac{\pi}{2n}\sin\frac{\pi}{n}+\2\sin\frac{\pi}{2n}\sin\frac{2\pi}{n}+\ldots+\2\sin\frac{\pi}{2n}\sin\frac{(n-1)\pi}{n}=2\cos\frac{\pi}{2n}

    Folosind formula 2\sin a\sin b=\cos(a-b)-\cos(a+b) membrul stang devine

    \cos\frac{\pi}{2n}-\cos\frac{3\pi}{2n}+\cos\frac{3\pi}{2n}-\cos\frac{5\pi}{2n}+\ldots+\cos\frac{(2n-3)\pi}{2n}-\cos\frac{(2n-1)\pi}{2n}=\\ =\cos\frac{\pi}{2n}-\cos\frac{(2n-1)\pi}{2n}=2\sin\frac{(n-1)\pi}{2n}\sin\frac{\pi}{2}=2\cos\left[\frac{\pi}{2}-\frac{(n-1)\pi}{2n}\right]=2\cos\frac{\pi}{2n}

  20. red12dog34
    veteran (III)

    andu_flavius95 wrote: Poate este mai usoara aceasta, tot prin inductie se cere:

        \[ \cos 3x + \cos 5x + ... + \cos (2n + 1)x = \frac{{\cos (n + 2)x \cdot \sin nx}}{{\sin x}}\quad ,n \in N^{*.} \]

    Multumesc mult!

    Fara inductie:

    Fie S=\cos 3x+\cos 5x+\ldots+\cos(2n+1)x
    Se inmultesc ambii membri cu 2\sin x.
    Se transforma produsele in sume (de fapt in diferente de sinusi, care se reduc doi cate doi).
    Mai ramane o diferenta de sinusi care se transforma inapoi in produs, apoi se imparte la loc cu ce am inmultit la inceput.

  21. andu_flavius95
    maestru (V)

    Multumesc mult, red_dog

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