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kdfgh
Pe: 2013-09-30T12:55:08+03:00 In: In:

trigonometrie

de rezolvat:
arcsin x+ arccos x=pi\2
x apartine [-1 , 1]

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1 raspuns

  1. gunty
    maestru (V)

    Vrei sa spui „de demonstrat”, caci este adevarat pentru oricare x apartine [-1,1]. Uite de ce:

    In primul rand, voi demonstra alte doua identitati:

        \[ \begin{array}{l} \left. \begin{array}{l} \sin ^2 \left( {\arccos x} \right) + \cos ^2 \left( {\arccos x} \right) = 1 \Leftrightarrow \sin ^2 \left( {\arccos x} \right) = 1 - x^2 \Rightarrow \sin \left( {\arccos x} \right) = \sqrt {1 - x^2 } \vee \sin \left( {\arccos x} \right) = - \sqrt {1 - x^2 } \\ \arccos x \in \left[ {0,\pi } \right] \\ \end{array} \right\} \Rightarrow \sin \left( {\arccos x} \right) = \sqrt {1 - x^2 } . \\ \\ \left. \begin{array}{l} \sin ^2 \left( {\arcsin x} \right) + \cos ^2 \left( {\arcsin x} \right) = 1 \Leftrightarrow \cos ^2 \left( {\arcsin x} \right) = 1 - x^2 \Rightarrow \cos \left( {\arcsin x} \right) = \sqrt {1 - x^2 } \vee \cos \left( {\arcsin x} \right) = - \sqrt {1 - x^2 } \\ \arcsin x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \\ \end{array} \right\} \Rightarrow \cos \left( {\arcsin x} \right) = \sqrt {1 - x^2 } . \\ \end{array} \]

    Folosindu-ma de ceea ce am demonstrat, primesc:

        \[ \begin{array}{l} \sin \left( {\arcsin x + \arccos x} \right) = \sin \left( {\arcsin x} \right) \cdot \cos \left( {\arccos x} \right) + \sin \left( {\arccos x} \right)\cos \left( {\arcsin x} \right) = x^2 + \sqrt {1 - x^2 } \cdot \sqrt {1 - x^2 } = 1. \\ Daca\;sinusul\;unei\;valori \in \left[ { - \frac{\pi }{2},\frac{{3\pi }}{2}} \right]\;este\;1,\;atunci\;acea\;valoare\;este\;\frac{\pi }{2}.\;In\;concluzie,\;\arcsin x + \arccos x = \frac{\pi }{2}. \\ \end{array} \]

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