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

[mathelord]

if arc cos[p] +arc cos[q] +arc cos[r] =180,
prove that psquared +qsquared +rsquared +2pqr =1
i need a comprehensive solution

 

[The Bob]

Let us get the ball rolling. What have you done towards it?

The Bob (2004 ©)

 

[HallsofIvy]

Looks like the cosine law to me.

 

[Zurtex]

Do you mean:
if $$\arccos p + \arccos q + \arccos r = 180$$ then $$\sqrt{p} + \sqrt{q} + \sqrt{r} + 2pqr = 1$$?
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:

$$p^2 + q^2 + r^2 + 2pqr = 1$$

 

[VietDao29]

Dear me, how can I made such a stupid mistake?
Viet Dao,

 

[HallsofIvy]

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]

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

 

[bao_ho]

ϖ radians

 

[bao_ho]

Π radians

 

[bao_ho]

3.1415... radians

 

[HallsofIvy]

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]

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

 

[Gokul43201]

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]

"arc cos" is simply the inverse cosine or "cos ^-1"

 

[mathelord]

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

 

[VietDao29]

No-one's going to do the homework for you. Here's a hint:
$$\alpha = \arccos p \Rightarrow \cos \alpha = p$$
$$\beta = \arccos q \Rightarrow \cos \beta = q$$
$$\zeta = \arccos r \Rightarrow \cos \zeta = r$$
You have $$\alpha + \beta + \zeta = 180$$
$$p^2 + q^2 + r^2 + 2pqr= \cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \zeta + 2 \cos \alpha \cos \beta \cos \zeta= ?$$
Viet Dao,

 

[Gaz031]

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:
$$\cos ((arc cos p + arc cos q) + arc cos r)=-1$$
$$\cos (arc cosp + arc cosq)cos(arc cosr) - \sin (arc cosp+arc cosq) \sin (arc cosr)=-1$$
$$rpq-r\sin (arc cosp) \sin (arc cosq))-q\sin (arc cosp) \sin (arc cosr) - p\sin (arc cos q)\sin (arc cosr) = -1$$
You can then use identities like $$\sin x =\pm \sqrt{1-cos^{2} x}$$ to give $$\sin (arc cosp) =\pm \sqrt{1-p^{2}}$$ and so tidy your expression to give the desired one.

 

[geniusprahar_21]

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