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