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andrei_93
Pe: 2009-08-24T09:14:55+03:00 In: In:

trigonometrie>egalitati

Am terminat zilele astea temele de vacanta la algebra si am inceput la geometrie. Primul pas facut primul obstacol! Am dat peste un exercitiu de trigonometrie care nu il pot rezolva.
Se cere sa se verifica egalitatile
a) \frac{cos12 + sqrt3{sin12}}{{cos3 - sin3}} = sqrt2

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11 raspunsuri

  1. ali
    maestru (V)

    Rezolvare:

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 \frac{{\cos 12 + \sqrt 3 \sin 12}}{{\cos 3 - \sin 3}} = \frac{{\cos 12 + ctg30 \cdot \sin 12}}{{\cos 3 - \sin 3}} =  \\
	 \frac{{\sin 30 \cdot \cos 12 + \cos 30\sin 12}}{{\sin 30\left( {\cos 3 - \sin 3} \right)}} = \frac{{\sin (30 + 12)}}{{\sin 30\left( {\cos 3 - \sin 3} \right)}} \\
	  = \frac{{\sin 42}}{{\sin 30\left( {\cos 3 - \sin 3} \right)}} \\
	 {\rm Stiim ca]
	

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  2. andrei_93

    Ce as putea face cu puterile exercitiului urmator. Face parte din aceeasi tema cu precedentul.
    sin^4\frac{pi}{{8}} + sin^4\frac{3pi}{{8}} + sin^4\frac{5pi}{{8}} + sin^4\frac{7pi}{{8}} = \frac{3}{{2}}
    Nu stiu cum sa scriu „pi” in tex.

  4. ali
    maestru (V)

    Indicatie::
    Observam ca:

        \[ \sin \frac{\pi }{8} = \sin \left( {22.5} \right) = \sin \left( {\frac{{45}}{2}} \right)\]

    Folosim formula:

        \[\sin \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{{1 - \cos x}}{2}} \]

    fiecare ‘termen’ al egalitatii se poate scrie sub forma indicata, de exemplu:

        \[ \begin{array}{l} \sin ^4 \frac{\pi }{8} = \left( {\sin \frac{\pi }{8}} \right)^4 = \left[ {\sin \left( {22.5} \right)} \right]^4 = \left[ {\sin \left( {\frac{{45}}{2}} \right)} \right]^4 = \left[ { \pm \sqrt {\frac{{1 - \cos 45}}{2}} } \right]^4 = ... = \\ = \frac{1}{8}\left( {3 - 2\sqrt 2 } \right) \\ \end{array} \]

    Continund la fel, se obtine rezultatul dorit.

  5. andrei_93

    Nu ajung la rezultatul cerut. Ca de exemplu sin^4\frac{3pi}{{8}} = +-sqrt{\frac{1 - cos 3ori45}{{2}}}. Pe acesta il pot rezolva cu formula cos3x. Dar la celelalte nu ma nici o idee.
    Pe sin^4\frac{5pi}{{8}} = +-sqrt{\frac{1 - cos 5ori45}{{2}}} nu am idee cum sa il scriu.Multumesc!

  6. ali
    maestru (V)

    Prin transformare din radian in grade, se obtine

        \[ \sin \frac{{5\pi }}{8} = \sin (112.5) = \sin \left( {\frac{\pi }{2} + 22.5} \right) = \cos (22.5) = \cos \left( {\frac{{45}}{2}} \right) \]

    Dupa care aplici formula pentru

        \[\cos \left( {\frac{x}{2}} \right)\]

    De asemena:

        \[ \sin \frac{{3\pi }}{8} = \sin (67.5) = \sin \left( {\frac{\pi }{2} - 22.5} \right) = \cos (22.5) = \cos \left( {\frac{{45}}{2}} \right) \]

  8. andrei_93

    M-am cam grabit si am uitat materia. Nu prea imi dau seama unde am gresit.Ori am gresit eu ori e gresit rezultatul dat de ei. Am rezolvat asa:

    Attached files

  9. ali
    maestru (V)

    Defapt:

        \[ \cos \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{{1 + \cos x}}{2}} \]

    Deci:

        \[ \begin{array}{l} \left[ {\sin \left( {\frac{{5\pi }}{8}} \right)} \right]^4 = \left[ {\cos (22.5)} \right]^4 = \left[ {\cos \left( {\frac{{45}}{2}} \right)} \right]^4 = ... = \frac{1}{{16}}\left( {2 + \sqrt 2 } \right)^2 \\ \left[ {\sin \left( {\frac{{7\pi }}{8}} \right)} \right]^4 = ... = \frac{1}{8}\left( {3 - 2\sqrt 2 } \right) \\ \left[ {\sin \left( {\frac{{3\pi }}{8}} \right)} \right]^4 = .... = \frac{1}{{16}}\left( {2 + \sqrt 2 } \right)^2 \\ \end{array} \]

  10. andrei_93

    la formula aceea cu cos(\frac{a}{{2}} ma uitam, dar aveam impresia ca socotesc sin(\frac{a}{{2}} .Multumesc! Greseala mea. 🙁
    E ok.Oricine greseste! 😉
    Rezolvarea corecta:

    Attached files

  12. andrei_93

    Ce pot face la egalitatea aceasta.Am rezolvat pana la un moment dat, pana cand nu am mai putut continua. Nu imi da nici o putere de 8 in rezolvare.Multumesc!
    (7 + cos4x)^2 = 32(1 + sin^8x + cos^8x) Multumesc!

  13. ali
    maestru (V)

    Indicatie:

        \[ \begin{array}{l} {\rm Notam E} = \left( {7 + \cos 4x} \right)^2 = 49 + 14\cos \left( {4x} \right) + \cos ^2 \left( {4x} \right) \\ \cos \left( {4x} \right) = - 1 + 2\cos ^2 2x \\ \cos ^2 \left( {4x} \right) = \left( { - 1 + 2\cos ^2 2x} \right)^2 \\ \Rightarrow E = 49 - 14 + 28\cos ^2 \left( {2x} \right) + 1 - 4\cos ^2 \left( {2x} \right) + 4\cos ^4 \left( {2x} \right) \\ = 36 + 24\cos ^2 \left( {2x} \right) + 4\cos ^4 \left( {2x} \right) \\ = ... = 32\left( {1 + \cos ^8 x + \sin ^8 x} \right) \\ \end{array} \]

    Ultima egalitate se obtine usor stiind ca:

        \[ \cos (2x) = \cos ^2 x - \sin ^2 x \]

  14. RazzvY

    O rezolvare frumoasa pentru al doilea exercitiu: se reduce expresia la 2*( \sin^4{ \frac{\pi}{8}} + \cos^4{ \frac{\pi}{8}})
    Apoi se foloseste \sin^4{x} + \cos^4{x} = 1 - 2*\sin^2{x}*\cos^2{x}, si apoi 2*\sin^2{x}*cos^2{x} = \sin^2{2x} / 2.

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