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andrei_93
Pe: 2010-08-29T08:26:35+03:00 In: In:

trigonometrie

Am un exercitiu la care am miscat ceva dar nu sa ajung la rezultat.
Mi se cere sa demonstrez egalitatea:
cos(arcctg(x))=\frac{x}{\sqrt{1+x^{2}}}
Ca sa va spun si ce am reusit si eu sa fac:
Am atribuit membrului stang un y si ma scris apoi ca arcctg(x)=arccos(cos(y)). Si de aici am afltat x=ctg(arccos(y)). Dar nu ajung la ce mi se cere.

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6 raspunsuri

  1. sindex

    formula e foarte usor de demonstrat, avem nevoie de un triunghi dreptunghic, sa fi atent la explicatii:

    1.

        \[{\text{Notezi }}arcctg(x) = \alpha \]

    2.

        \[arcctg(x) = \alpha \Leftrightarrow ctg\alpha = x\]

    3.

        $$ctg\alpha = {{AB} \over {AC}} = x \Rightarrow \left\{ \matrix{ AB = x \hfill \cr AC = 1 \hfill \cr} \right.$$

    4.Cu Pitagora, aflii ipotenuza =>

        $$BC = \sqrt {1 + {x^2}} $$

    5.Prin urmare,

        $$\left. {\left\{ \matrix{ \cos \alpha = {{AB} \over {BC}} = {x \over {\sqrt {1 + {x^2}} }} \hfill \cr \cos (arcctg(x)) = \cos \alpha \hfill \cr} \right.} \right| \Rightarrow \cos (arcctg(x)) = {x \over {\sqrt {1 + {x^2}} }}$$

    Metoda este foarte usoara si te ajuta cand astfel de exercitii apar in probleme gen (sin(arctg(x)) + tg(arccos(x)) = 2010 afla pe x) Daca ai nelamuriri, nu ezita sa intrebi.

  2. andrei_93

    am incercat sa rezolv exercitiul.
    sin(arctg(x))+tg(arccos(x))=2010
    \frac{x}{\sqrt{1+x^{2}}}+\frac{\sqrt{1-x^{2}}}{x}=2010

    \frac{x}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+x^{2}}}.\frac{1}{x}=2010

    \frac{x^{2}+1}{x\sqrt{1+x^{2}}}=2010

    x^{2}+1=2010x\sqrt{1+x^{2}}

    (x^{2}+1)^{2}=2010^{2}x^{2}(1+x^{2})^2

    x^{4}+2x^{2}+1=2010^{2}x^{2}+2010^{2}x^{4}

    2010^{2}x^{4}-x^{4}+2010^{2}x^{2}-2x^{2}-1=0

    x^{4}(2010^{2}-1)+x^{2}(2010^{2}-2)-1=0

    notez x^{2}=a si o sa am o ecuatie de gardul al 2 – lea
    Este corect asa?
    Un ajutor la subiectul functii inverse, postat anterior?

  4. sindex

    da, e corect. felicitari 🙂

  5. VMD

    2) Sa se arate ca :
    sin^2(1*) + sin^2(2*) + … + sin^2(90*) = 91/2

    🙁

  6. sindex

        \[\begin{array}{l} \underbrace {{{\sin }^2}({0^o})}_{ + {{\cos }^2}({0^o})} + \underbrace {{{\sin }^2}({1^o})}_{ + {{\cos }^2}({1^o})} + \underbrace {{{\sin }^2}({2^o})}_{ + {{\cos }^2}({2^o})} + ... + \underbrace {{{\sin }^2}({{90}^o})}_{ + {{\cos }^2}({{90}^o})} = \frac{{91}}{2} + \underbrace {{{\cos }^2}({0^o}) + {{\cos }^2}({1^o}) + {{\cos }^2}({2^o}) + ... + {{\cos }^2}({{90}^o})}_{\underbrace { = {{\sin }^2}({0^o}) + {{\sin }^2}({1^o}) + ... + {{\sin }^2}({{90}^2})}_{ = 0 + \frac{{91}}{2}}}[1] \\ \Leftrightarrow \underbrace {{{\sin }^2}({0^o})}_{ + {{\cos }^2}({0^2})} + \underbrace {1 + 1 + 1 + ... + 1}_{90} = \frac{{91}}{2} + \frac{{91}}{2} \\ \Leftrightarrow \underbrace {1 + 1 + ... + 1}_{91} = 91 \Leftrightarrow 91 = 91(A) \\ {\rm{In stanga am folosit }}{\sin ^2}x + {\cos ^2}x = 1,{\rm{ cosinusi pe care i - am reintrodus in dreapta ptr a < balansa > egalitatea}}{\rm{.}} \\ {\rm{Am schimbat putin egalitatea introducand pe si}}{{\rm{n}}^2}(0){\rm{ si apoi perechea sa co}}{{\rm{s}}^2}(0).{\rm{ Acest lucru nu schimba egalitatea pentru ca:}} \\ \left\{ \begin{array}{l} \sin 0 = 0 \Rightarrow {\sin ^2}0 = 0 \\ \cos 90 = 0 \Rightarrow {\cos ^2}90 = 0 \\ \end{array} \right. \\ {\rm{[1] M - am folosit de : co}}{{\rm{s}}^2}{\rm{(0) + co}}{{\rm{s}}^2}{\rm{(1) + co}}{{\rm{s}}^2}{\rm{(2) + }}...{\rm{ + co}}{{\rm{s}}^2}{\rm{(90) = si}}{{\rm{n}}^2}{\rm{(0) + si}}{{\rm{n}}^2}{\rm{(1) + si}}{{\rm{n}}^2}{\rm{(2) + }}...{\rm{ + si}}{{\rm{n}}^2}{\rm{(90) }} \\ {\rm{Demonstratie: }} \\ {\rm{S = co}}{{\rm{s}}^2}{\rm{(0) + co}}{{\rm{s}}^2}(1) + {\cos ^2}(2) + ...{\cos ^2}90 - {\sin ^2}90 - {\sin ^2}89 - ... - {\sin ^2}(1) - {\sin ^2}0 \\ Dar\left. {:\left\{ \begin{array}{l} \cos 0 = \sin 90 \Rightarrow {\cos ^2}0 = {\sin ^2}90 \Leftrightarrow {\cos ^2}0 - {\sin ^2}90 = 0 \\ \cos 1 = \sin 89 \Rightarrow \cos 1 - \sin 89 = 0 \Leftrightarrow {\cos ^2}1 - {\sin ^2}89 = 0 \\ s.a.m.d \\ \end{array} \right.} \right| \Rightarrow S = \underbrace {0 + 0 + 0 + ... + 0}_{{\rm{de 91 ori}}} = 0 \\ S = 0 \Rightarrow {\rm{co}}{{\rm{s}}^2}{\rm{(0) + co}}{{\rm{s}}^2}{\rm{(1) + co}}{{\rm{s}}^2}{\rm{(2) + }}...{\rm{ + co}}{{\rm{s}}^2}{\rm{(90) = si}}{{\rm{n}}^2}{\rm{(0) + si}}{{\rm{n}}^2}{\rm{(1) + si}}{{\rm{n}}^2}{\rm{(2) + }}...{\rm{ + si}}{{\rm{n}}^2}{\rm{(90)}} \\ \end{array}\]

    De altfel, problema se poate reduce la a demonstra doar ca cos^2(0)+…cos^2(90)=sin^2(0)+…+sin^2(90), pentru ca observi ca in prima faza m-am folosit de acest lucru si am egalat pe cos^2(0)+cos^2(1)+….cos^2(90) cu 91/2 presupunand ca sin^2(0)+sin^2(1)+…+sin^2(90)=91/2

  8. VMD

    ms 😀 … am inteles

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