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grapefruit
maestru (V)
Pe: 2015-06-14T13:03:48+03:00 In: In:

polinom

Determinati polinom de gradul 4 cu coef intregi care admite urmatoarele solutii:ctg(pi/16) ,ctg(5pi/16) ,ctg(9pi/16),ctg(13pi/16)

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2 raspunsuri

  1. ghioknt
    profesor

    Se ştie că dacă notăm x_1+x_2=s_1;\;x_1x_2=p_1;\;x_3+x_4=s_2;x_3x_4=p_2, atunci polinomul poate fi
    f=X^4-S_1X^3+S_2X^2-S_3X+S_4,\;cu\;S_1=s_1+s_2,\;S_2=s_1s_2+p_1+p_2,\;S_3=p_1s_2+p_2s_1,\;S_4=p_1p_2.
    Aleg x_1=ctg \frac{\pi }{16},\;x_2=ctg \frac{9\pi }{16}=-tg \frac{\pi }{16}\;(c)\;x_3=ctg \frac{5\pi }{16}=tg \frac{3\pi }{16}\;(c)\;\\x_4=ctg \frac{13\pi }{16}=-ctg \frac{3\pi }{16}\;(s), unde (c) înseamnă unghiuri complementare, (s) – suplementare.
    In continuare voi folosi identitatea ctg x-tg x=2ctg(2x). Obţin:
    s_1=2ctg \frac{\pi}{8},\;s_2=-2ctg \frac{3\pi}{8}=-2tg \frac{\pi}{8}\;(c)\;p_1=p_2=-1\;apoi\;\\S_1=2\left ( ctg \frac{\pi}{8}-tg \frac{\pi}{8} \right )=4ctg \frac{\pi}{4}=4,\;S_2=-6,\;S_3=-S_1=-4,\;S_4=1.

    f=X^4-4X^3-6X^2+4X+1.

  2. grapefruit
    maestru (V)

    Cum demonstrez acea identitate?

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