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dana1101
Pe: 2012-12-18T13:08:01+02:00 In: In:

Exercitiu

Sa se demonstreze egalitatile:
a)cos(π/n)+cos([2π/n)+…+cos[(n-1)π/n]=0, n≥2

b)sin(π/n)+sin(2π/n)+…+sin[(n-1)π/n]=ctg(π/2n), n≥2.

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2 raspunsuri

  1. gunty
    maestru (V)

    a)

    Se stie ca

        \[ \cos \left( {\frac{{t\pi }}{n}} \right) = - \cos \left( {\pi - \frac{{t\pi }}{n}} \right) = - \cos \left( {\frac{{\left( {n - t} \right)\pi }}{n}} \right) \]

    Deci egalitatea ta este echivalenta cu urmatoarea:

        \[ - \cos \left( {\frac{{\left( {n - 1} \right)\pi }}{n}} \right) - \cos \left( {\frac{{\left( {n - 2} \right)\pi }}{n}} \right) - ... - \cos \left( {\frac{{\left[ {\frac{n}{2}} \right]\pi }}{n}} \right) + \cos \left( {\frac{{\left( {\left[ {\frac{n}{2}} \right] + 1} \right)\pi }}{n}} \right) + ... + \cos \left( {\frac{{(n - 1)\pi }}{n}} \right) = 0 \]

    1. Daca n este numar par, ramane doar

        \[ - \cos \left( {\frac{{\left[ {\frac{n}{2}} \right]\pi }}{n}} \right) = 0 \Leftrightarrow \cos \left( {\frac{\pi }{2}} \right) = 0_{(adevarat)} \]

    2. Daca n este impar, toti termenii „dispar”, deci 0=0 este adevarat.

    [k] reprezinta partea intreaga a numarului real k.

  2. George_Gaumont

    Doua metode clasice.

    Attached files

    sume_trigonometrice_202.pdf (50.9 KB) 

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