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Prove that psquared +qsquared +rsquared +2pqr =1 
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#1
Jul2305, 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 


#2
Jul2305, 04:35 AM

P: 1,116

Let us get the ball rolling. What have you done towards it?
The Bob (2004 ©) 


#3
Jul2305, 06:03 AM

Math
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PF Gold
P: 39,533

Looks like the cosine law to me.



#4
Jul2305, 08:15 AM

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P: 1,123

Prove that psquared +qsquared +rsquared +2pqr =1
[tex]p^2 + q^2 + r^2 + 2pqr = 1[/tex] 


#6
Jul2305, 11:30 AM

Math
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PF Gold
P: 39,533

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
Jul2305, 02:36 PM

P: 42

acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?



#8
Jul2305, 02:36 PM

P: 42

ϖ radians



#9
Jul2305, 02:37 PM

P: 42

Π radians



#10
Jul2305, 02:38 PM

P: 42

3.1415... radians



#11
Jul2305, 04:50 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,533

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
Jul2305, 05:57 PM

P: 42

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


#13
Jul2305, 06:02 PM

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PF Gold
P: 11,155




#14
Jul2305, 06:04 PM

Emeritus
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PF Gold
P: 11,155

"arc cos" is simply the inverse cosine or "cos ^{1}"



#15
Jul2405, 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



#16
Jul2405, 10:25 AM

HW Helper
P: 1,422

Noone'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, 


#17
Jul2405, 12:48 PM

P: 51

[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]rpqr\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{1cos^{2} x}[/tex] to give [tex]\sin (arc cosp) =\pm \sqrt{1p^{2}}[/tex] and so tidy your expression to give the desired one. 


#18
Jul2605, 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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