#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
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

Looks like the cosine law to me.




#4
Jul2305, 08:15 AM

Sci Advisor
HW Helper
P: 1,123

[SOLVED] trig[tex]p^2 + q^2 + r^2 + 2pqr = 1[/tex] 



#6
Jul2305, 11:30 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

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: 38,900

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

Emeritus
Sci Advisor
PF Gold
P: 11,154





#14
Jul2305, 06:04 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

"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 


Register to reply 
Related Discussions  
[SOLVED] Trig  Precalculus Mathematics Homework  6  
[SOLVED] Trig Question  Precalculus Mathematics Homework  1  
[SOLVED] More trig substitution help...  Calculus & Beyond Homework  6  
[SOLVED] Trig question  Precalculus Mathematics Homework  8  
[SOLVED] Trig...  General Math  3 