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

by mathelord
Tags: solved, trig
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mathelord
#1
Jul23-05, 03:50 AM
P: n/a
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
#2
Jul23-05, 04:35 AM
P: 1,116
Let us get the ball rolling. What have you done towards it?

The Bob (2004 )
HallsofIvy
#3
Jul23-05, 06:03 AM
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PF Gold
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Looks like the cosine law to me.

Zurtex
#4
Jul23-05, 08:15 AM
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Prove that psquared +qsquared +rsquared +2pqr =1

Quote Quote by VietDao29
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]
VietDao29
#5
Jul23-05, 08:31 AM
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Dear me, how can I made such a stupid mistake???
Viet Dao,
HallsofIvy
#6
Jul23-05, 11:30 AM
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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.
bao_ho
#7
Jul23-05, 02:36 PM
P: 42
acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?
bao_ho
#8
Jul23-05, 02:36 PM
P: 42
ϖ radians
bao_ho
#9
Jul23-05, 02:37 PM
P: 42
Π radians
bao_ho
#10
Jul23-05, 02:38 PM
P: 42
3.1415... radians
HallsofIvy
#11
Jul23-05, 04:50 PM
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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.
bao_ho
#12
Jul23-05, 05:57 PM
P: 42
if the angles are given in degrees then
p*p + q*q + r*r +2*p*q*r != 1
Gokul43201
#13
Jul23-05, 06:02 PM
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Quote Quote by bao_ho
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 !
Gokul43201
#14
Jul23-05, 06:04 PM
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"arc cos" is simply the inverse cosine or "cos -1"
mathelord
#15
Jul24-05, 10:20 AM
P: n/a
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
VietDao29
#16
Jul24-05, 10:25 AM
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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,
Gaz031
#17
Jul24-05, 12:48 PM
P: 51
Quote Quote by mathelord
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.
geniusprahar_21
#18
Jul26-05, 12:03 PM
P: 29
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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