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

  • Thread starter mathelord
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  • #1
mathelord
if arc cos[p] +arc cos[q] +arc cos[r] =180,
prove that psquared +qsquared +rsquared +2pqr =1
i need a comprehensive solution
 

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  • #2
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Let us get the ball rolling. What have you done towards it?

The Bob (2004 ©)
 
  • #3
HallsofIvy
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Looks like the cosine law to me.
 
  • #4
Zurtex
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VietDao29 said:
Do you mean:
if [tex]\arccos p + \arccos q + \arccos r = 180[/tex] then [tex]\sqrt{p} + \sqrt{q} + \sqrt{r} + 2pqr = 1[/tex]?
Maybe I am wrong, but if you mean that, it's a wrong problem.
p = 0.573576
q = 0.422618
r = 0.5
Am I missing something?
Viet Dao,
To me it quite clearly reads as:

[tex]p^2 + q^2 + r^2 + 2pqr = 1[/tex]
 
  • #5
VietDao29
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Dear me, how can I made such a stupid mistake??? :mad:
Viet Dao,
 
  • #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.
 
  • #7
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acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?
 
  • #8
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ϖ radians
 
  • #9
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Π radians
 
  • #10
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3.1415... radians
 
  • #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.
 
  • #12
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if the angles are given in degrees then
p*p + q*q + r*r +2*p*q*r != 1
 
  • #13
Gokul43201
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bao_ho said:
if the angles are given in degrees then
p*p + q*q + r*r +2*p*q*r != 1
And why is that ? p, q, r are NOT angles !
 
  • #14
Gokul43201
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"arc cos" is simply the inverse cosine or "cos -1"
 
  • #15
mathelord
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
 
  • #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:
  • #17
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mathelord said:
if arc cos[p] +arc cos[q] +arc cos[r] =180,
prove that psquared +qsquared +rsquared +2pqr =1
i need a comprehensive solution
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.
 
  • #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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