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

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Discussion Overview

The discussion revolves around proving the equation \( p^2 + q^2 + r^2 + 2pqr = 1 \) under the condition that \( \arccos[p] + \arccos[q] + \arccos[r] = 180^\circ \). Participants explore the implications of this condition and seek a comprehensive solution, engaging in various mathematical reasoning and interpretations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the problem relates to the cosine law.
  • Another participant questions the interpretation of the original equation and proposes an alternative formulation, expressing confusion about the problem's validity.
  • Several participants clarify that the angles represented by \( p, q, r \) are not angles themselves but rather the cosines of angles.
  • There is a discussion on whether the angles should be considered in degrees or radians, with differing opinions on the implications of this choice.
  • One participant provides a hint involving the use of trigonometric identities and the relationship between the angles and their cosines.
  • Another participant outlines a method involving the cosine of the sum of angles to approach the proof.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the problem or the validity of the proposed formulations. Multiple competing views remain regarding the correct approach to proving the equation.

Contextual Notes

There are unresolved assumptions regarding the definitions of \( p, q, r \) and their relationship to angles. The discussion also highlights the dependence on whether angles are considered in degrees or radians, which affects the validity of the claims made.

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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The Bob (2004 ©)
 
Looks like the cosine law to me.
 
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]
 
Dear me, how can I made such a stupid mistake? :mad:
Viet Dao,
 
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.
 
acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?
 
ϖ radians
 
Π radians
 
  • #10
3.1415... radians
 
  • #11
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
if the angles are given in degrees then
p*p + q*q + r*r +2*p*q*r != 1
 
  • #13
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
"arc cos" is simply the inverse cosine or "cos -1"
 
  • #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 won't show me.so i need it urgently please
 
  • #16
No-one's going to 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
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. don't forget to use... cosA = p, cosB = q, cosC = r
 

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