# [SOLVED] trig

by mathelord
Tags: solved, trig
 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
 P: 1,116 Let us get the ball rolling. What have you done towards it? The Bob (2004 ©)
 PF Patron Sci Advisor Thanks Emeritus P: 38,400 Looks like the cosine law to me.
HW Helper
P: 1,123

## [SOLVED] trig

 Quote by VietDao29 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:

$$p^2 + q^2 + r^2 + 2pqr = 1$$
 HW Helper P: 1,422 Dear me, how can I made such a stupid mistake??? Viet Dao,
 PF Patron Sci Advisor Thanks Emeritus P: 38,400 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.
 P: 42 ϖ radians
 P: 42 Π radians
 P: 42 3.1415... radians
 PF Patron Sci Advisor Thanks Emeritus P: 38,400 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.
 P: 42 if the angles are given in degrees then p*p + q*q + r*r +2*p*q*r != 1
PF Patron
Emeritus
P: 11,137
 Quote by bao_ho 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 !
 PF Patron Sci Advisor Emeritus P: 11,137 "arc cos" is simply the inverse cosine or "cos -1"
 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
 HW Helper P: 1,422 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,
P: 51
 Quote by mathelord 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.
 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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