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[SOLVED] trig

 
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Jul23-05, 03:50 AM   #1
 

[SOLVED] trig


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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Jul23-05, 04:35 AM   #2
 
Let us get the ball rolling. What have you done towards it?

The Bob (2004 ©)
Jul23-05, 06:03 AM   #3
 
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Looks like the cosine law to me.
Jul23-05, 08:15 AM   #4
 
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[SOLVED] trig


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