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

M

#### 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

Science Advisor
Homework Helper
Looks like the cosine law to me.

#### Zurtex

Science Advisor
Homework Helper
VietDao29 said:
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

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

#### HallsofIvy

Science Advisor
Homework Helper
If &theta;= arc cos[p], &phi;= arccos[q], &psi;= arccos[r], then
saying that "arc cos[p] +arc cos[q] +arc cos[r] =180" is simply saying that &theta;+&phi;+&psi;= 180 or that &theta;, &phi;, &psi; are angles in a triangle.

#### bao_ho

acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = &pi; radians?

&piv; radians

&Pi; radians

#### bao_ho

3.1415... radians

#### HallsofIvy

Science Advisor
Homework Helper
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

Staff Emeritus
Science Advisor
Gold Member
bao_ho said:
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

Staff Emeritus
Science Advisor
Gold Member
"arc cos" is simply the inverse cosine or "cos -1"

M

#### 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 wont show me.so i need it urgently please

#### VietDao29

Homework Helper
No-one's gonna 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,

Last edited:

#### Gaz031

mathelord said:
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. dont forget to use...... cosA = p, cosB = q, cosC = r

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving