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Prove that psquared +qsquared +rsquared +2pqr =1

  1. Jul 23, 2005 #1
    if arc cos[p] +arc cos[q] +arc cos[r] =180,
    prove that psquared +qsquared +rsquared +2pqr =1
    i need a comprehensive solution
     
  2. jcsd
  3. Jul 23, 2005 #2
    Let us get the ball rolling. What have you done towards it?

    The Bob (2004 ©)
     
  4. Jul 23, 2005 #3

    HallsofIvy

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    Looks like the cosine law to me.
     
  5. Jul 23, 2005 #4

    Zurtex

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    To me it quite clearly reads as:

    [tex]p^2 + q^2 + r^2 + 2pqr = 1[/tex]
     
  6. Jul 23, 2005 #5

    VietDao29

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    Dear me, how can I made such a stupid mistake??? :mad:
    Viet Dao,
     
  7. Jul 23, 2005 #6

    HallsofIvy

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    If θ= arc cos[p], φ= arccos[q], ψ= arccos[r], then
    saying that "arc cos[p] +arc cos[q] +arc cos[r] =180" is simply saying that θ+φ+ψ= 180 or that θ, φ, ψ are angles in a triangle.
     
  8. Jul 23, 2005 #7
    acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?
     
  9. Jul 23, 2005 #8
    ϖ radians
     
  10. Jul 23, 2005 #9
    Π radians
     
  11. Jul 23, 2005 #10
    3.1415... radians
     
  12. Jul 23, 2005 #11

    HallsofIvy

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    Yes, but the original post said "arc cos[p] +arc cos[q] +arc cos[r] =180" which only makes sense if the angles are given in degrees.
     
  13. Jul 23, 2005 #12
    if the angles are given in degrees then
    p*p + q*q + r*r +2*p*q*r != 1
     
  14. Jul 23, 2005 #13

    Gokul43201

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    And why is that ? p, q, r are NOT angles !
     
  15. Jul 23, 2005 #14

    Gokul43201

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    "arc cos" is simply the inverse cosine or "cos -1"
     
  16. Jul 24, 2005 #15
    no one is helpin,all i said is that i need a proof on how psquared +qsquared +rsquared +2pqr=1,p,q,r,are angles in degrees.my friend abia ubong has a solution of this problem,but wont show me.so i need it urgently please
     
  17. Jul 24, 2005 #16

    VietDao29

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    No-one's gonna do the homework for you. Here's a hint:
    [tex]\alpha = \arccos p \Rightarrow \cos \alpha = p[/tex]
    [tex]\beta = \arccos q \Rightarrow \cos \beta = q[/tex]
    [tex]\zeta = \arccos r \Rightarrow \cos \zeta = r[/tex]
    You have [tex]\alpha + \beta + \zeta = 180[/tex]
    [tex]p^2 + q^2 + r^2 + 2pqr= \cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \zeta + 2 \cos \alpha \cos \beta \cos \zeta= ?[/tex]
    Viet Dao,
     
    Last edited: Jul 24, 2005
  18. Jul 24, 2005 #17
    You could start by doing:
    [tex]\cos ((arc cos p + arc cos q) + arc cos r)=-1[/tex]
    [tex]\cos (arc cosp + arc cosq)cos(arc cosr) - \sin (arc cosp+arc cosq) \sin (arc cosr)=-1[/tex]
    [tex]rpq-r\sin (arc cosp) \sin (arc cosq))-q\sin (arc cosp) \sin (arc cosr) - p\sin (arc cos q)\sin (arc cosr) = -1[/tex]
    You can then use identities like [tex]\sin x =\pm \sqrt{1-cos^{2} x}[/tex] to give [tex]\sin (arc cosp) =\pm \sqrt{1-p^{2}}[/tex] and so tidy your expression to give the desired one.
     
  19. Jul 26, 2005 #18
    let arccosp = A, arccosq = B, arccosr = C, then A+B+C = 180 => A + B = 180 - C
    => cos(A+B) = cos(180 - c) => - cosC = cosa.cosb - sina.sinb. squaring both sides,
    change all sin square to cos square. take things common, you'll get the answer. dont forget to use...... cosA = p, cosB = q, cosC = r
     
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