[SOLVED] trig

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

To me it quite clearly reads as:

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

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

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 wont show me.so i need it urgently please

VietDao29

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,

Gaz031

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.