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

  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. Let us get the ball rolling. What have you done towards it?

    The Bob (2004 ©)
     
  4. HallsofIvy

    HallsofIvy 40,314
    Staff Emeritus
    Science Advisor

    Looks like the cosine law to me.
     
  5. Zurtex

    Zurtex 1,123
    Science Advisor
    Homework Helper

    To me it quite clearly reads as:

    [tex]p^2 + q^2 + r^2 + 2pqr = 1[/tex]
     
  6. VietDao29

    VietDao29 1,422
    Homework Helper

    Dear me, how can I made such a stupid mistake??? :mad:
    Viet Dao,
     
  7. HallsofIvy

    HallsofIvy 40,314
    Staff Emeritus
    Science Advisor

    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. acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?
     
  9. ϖ radians
     
  10. Π radians
     
  11. 3.1415... radians
     
  12. HallsofIvy

    HallsofIvy 40,314
    Staff Emeritus
    Science Advisor

    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. if the angles are given in degrees then
    p*p + q*q + r*r +2*p*q*r != 1
     
  14. Gokul43201

    Gokul43201 11,141
    Staff Emeritus
    Science Advisor
    Gold Member

    And why is that ? p, q, r are NOT angles !
     
  15. Gokul43201

    Gokul43201 11,141
    Staff Emeritus
    Science Advisor
    Gold Member

    "arc cos" is simply the inverse cosine or "cos -1"
     
  16. 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. VietDao29

    VietDao29 1,422
    Homework Helper

    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. 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. 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
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?