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Matrix20122
Pe: 2012-06-07T19:06:34+03:00 In: In:

Demonstratie

Puteti sa ma ajutati sa demonstrez si eu doua relatii intr-un triunghi ABC??? 😀 Vaa rog foarte muult ..Imi trebuie pentru nota si nu am nici o idee …

1)cos A+cosB+cos C=1+r/R

2) sin A+sinB+sinC<=3sqrt3/2

Va multumesc anticipat 🙂 ! 😥

Similare

7 raspunsuri

  1. PhantomR
    expert (VI)

    b) Functia \sin x este concava pe \left[0;\pi \right ], iar cum A,B,C\in\left[ 0; \pi] (valorile extreme se pot atinge pentru un triunghi degenerat) din inegalitatea lui Jensen rezulta \frac{\sin A + \sin B + \sin C}{3}\leq \sin\left( \frac{A+B+C}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}, de unde \sin A + \sin B + \sin C \leq \frac{3\sqrt 3}{2}.

    Am folosit faptul ca A+B+C=\pi.

  2. PhantomR
    expert (VI)

    O alta relatie (voi incerca sa o fac si pe cealalta):

    \sin A +\sin B + \sin C = 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C = 2\sin \frac{\pi-C}{2}\cos \frac{A-B}{2}+\cos C = 2\sin \left( \frac{\pi}{2} - \frac{C}{2}\right)\cos \frac{A-B}{2}+\cos C=2 \cos \frac{C}{2} \cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2}=2\cos \frac{C}{2} \left( \cos \frac{A-B}{2}+\sin \frac{C}{2}\right)=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2} + \sin \frac{\pi-A-B}{2}\right)=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2} + \sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right) =
    =2\cos \frac{C}{2}\left( \cos \frac{A-B}{2} + \cos \frac{A+B}{2}\right)=2\cos \frac{C}{2}\cdot 2\cos \frac{\frac{A-B+(A+B)}{2}}{2}\cos \frac{\frac{A-B-(A+B)}{2}}{2}=4\cos \frac{C}{2}\cos\frac{A}{2} \left( -\frac{B}{2}\right) si cum functia \cos e para, adica \cos (-x)=\cos x ,\forall x\in\mathbb{R}, rezulta \cos A + \cos B +\cos C = 4\cos \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}=4\cdot \sqrt{\frac{p(p-a)}{bc}\cdot \frac{p(p-b)}{ca}\cdot \frac{p(p-c)}{ab}}=
    =\sqrt{\left( \frac{p}{abc}\right)^2\cdot p(p-a)(p-b)(p-c}=\sqrt{\left( \frac{p}{abc}\right)^2\cdot S^2}=4\cdot \frac{p}{abc}\cdot S=\frac{p}{R}

  4. PhantomR
    expert (VI)

    \cos A + \cos B + \cos C = 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+1-2\sin^2\frac{C}{2}=2\cos \frac{\pi-C}{2}\cos \frac{A-B}{2}-2\sin^2 \frac{C}{2}+1=2\cos \left(\frac{\pi}{2}-\frac{C}{2}\right)\cos \frac{A-B}{2}-2\sin^2 \frac{C}{2}+1=2\sin \frac{C}{2}\cos \frac{A-B}{2}-2\sin^2\frac{C}{2}+1=2\sin \frac{C}{2}\left( \cos \frac{A-B}{2}-\sin \frac{C}{2}\right)+1=2\sin \frac{C}{2}\left( \cos \frac{A-B}{2}-\sin \frac{\pi-(A+B)}{2}\right)+1=
    =2\sin \frac{C}{2}\left( \cos \frac{A-B}{2}-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right)+1=2\sin \frac{C}{2}\left( \cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right)+1=2\sin \frac{C}{2} \cdot (-2) \sin \frac{\frac{A-B+(A+B)}{2}}{2}\sin \frac{\frac{A-B-(A+B)}{2}}{2}+1=-4\sin \frac{C}{2}\sin \frac{A}{2}\sin \left( -\frac{B}{2}\right)+1=-4 \sin \frac{C}{2} \sin \frac{A}{2} \cdot (-1) \sin \frac{B}{2}+1=4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}+1=4\sqrt{\frac{(p-b)(p-c)}{bc} \cdot \frac{(p-c)(p-a)}{ca}\cdot \frac{(p-a)(p-b)}{ab}}+1=
    =4 \sqrt{\frac{[(p-a)(p-b)(p-c)]^2}{(abc)^2}}+1=4\cdot \frac{(p-a)(p-b)(p-c)}{abc}+1=4\frac{p(p-a)(p-b)(p-c)}{p\cdot abc}+1=4\frac{S^2}{p \cdot abc}+1=\frac{S}{p} \cdot \frac{4S}{abc}+1=r \cdot \frac{1}{R}+1=1+\frac{r}{R}, ceea ce trebuia demonstrat.

  5. Zeus
    veteran (III)

    Foarte buna rezolvarea. Sunt mandru de tine 😀. Jos palaria!

  6. PhantomR
    expert (VI)

    Va multumesc , dar cred ca am vazut-o candva (anul trecut daca nu ma insel)!

  8. Integrator
    maestru (V)

    PhantomR wrote: O alta relatie (voi incerca sa o fac si pe cealalta):

    \sin A +\sin B + \sin C = 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C = 2\sin \frac{\pi-C}{2}\cos \frac{A-B}{2}+\cos C = 2\sin \left( \frac{\pi}{2} - \frac{C}{2}\right)\cos \frac{A-B}{2}+\cos C=2 \cos \frac{C}{2} \cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2}=2\cos \frac{C}{2} \left( \cos \frac{A-B}{2}+\sin \frac{C}{2}\right)=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2} + \sin \frac{\pi-A-B}{2}\right)=2\cos \frac{C}{2}\left( \cos \frac{A-B}{2} + \sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right) =
    =2\cos \frac{C}{2}\left( \cos \frac{A-B}{2} + \cos \frac{A+B}{2}\right)=2\cos \frac{C}{2}\cdot 2\cos \frac{\frac{A-B+(A+B)}{2}}{2}\cos \frac{\frac{A-B-(A+B)}{2}}{2}=4\cos \frac{C}{2}\cos\frac{A}{2} \left( -\frac{B}{2}\right) si cum functia \cos e para, adica \cos (-x)=\cos x ,\forall x\in\mathbb{R}, rezulta \cos A + \cos B +\cos C = 4\cos \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}=4\cdot \sqrt{\frac{p(p-a)}{bc}\cdot \frac{p(p-b)}{ca}\cdot \frac{p(p-c)}{ab}}=
    =\sqrt{\left( \frac{p}{abc}\right)^2\cdot p(p-a)(p-b)(p-c}=\sqrt{\left( \frac{p}{abc}\right)^2\cdot S^2}=4\cdot \frac{p}{abc}\cdot S=\frac{p}{R}


    Demonstratia relatiei sinA+sinB+sinC=\frac{p}{R} rezulta mult mai repede folosind teorema sinusurilor aplicand regula ca un sir de rapoarte egale este egal cu raportul dintre suma numaratorilor acelor rapoarte si suma numitorilor acelorasi rapoarte.

  9. PhantomR
    expert (VI)

    Aveti foarte mare dreptate si va multumesc pentru remarca! Acea relatie a iesit de fapt deoarece am aplicat gresit formula pentru transformarea unei sume de cosinusi in produs. Pana la urma mi-am dat seama si am schimbat relatia cu cos in cea cu sin.

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