Question

How would you solve this?

Answer 1

\[\cos^{-1} p + \cos^{-1} q+\cos^{-1} r=\pi \]

Answer 2

correction : What is the value of \[p^2 + q^2 + r^2 + 2pqr\]

Answer 3

This might work but try moving one of those arccos over with the pi
and the taking cosine of both sides
That might help
I haven't tried it yet

Answer 4

I got the answer guys! thanks for trying to help though! :) :D

Answer 5

so here is a question
there are an infinite number of triples that add up to \(\pi\)
so there should be an infinite number of answers
but since each could be \(\frac{\pi}{3}\) so it it is always the same, wouldn't it be \(\cos^{-1}(p)=\cos^{-1}(q)=\cos^{-1}(r)=\frac{\pi}{3}\) making \(p=q=r=\frac{1}{2}\) ?

Answer 6

i guess i mean if \[p^2 + q^2 + r^2 + 2pqr\] is a constant, then it would be that constant if \(p=q=r=\frac{1}{2}\)

Answer 7

or for that matter \(p=q=0,r=1\)

Answer 8

hmm looks like \(p^2 + q^2 + r^2 + 2pqr=1\) doesn't it
i wonder why...

Answer 9

\[\cos^{-1}(p)+\cos^{-1}(q)=\pi-\cos^{-1}(r)\]
\[\cos(\cos^{-1}(p)+\cos^{-1}(q))=\cos(\pi-\cos^{-1}(r))\]
\[\cos(\cos^{-1}(p))\cos(\cos^{-1}(q))-\sin(\cos^{-1}(p))\sin(\cos^{-1}(q))=\cos(\pi)\cos(\cos^{-1}(r))+0\]
\[pq-\sqrt{1-p^2}\sqrt{1-q^2}=-r\]
\[\sqrt{1-p^2}\sqrt{1-q^2}=-r-pq\]
\[(1-p^2)(1-q^2)=r^2+2rpq+p^2q^2 \]
\[1-q^2-p^2+p^2q^2=r^2+2rpq+p^2q^2 \]
\[r^2+p^2+q^2+2rpq=-p^2q^2+p^2q^2+1\]

so 1

Answer 10

wow!

Answer 11

i got rid of that negative too soon but i squared both sides afterwards so it's okay i guess

Answer 12

damn i was trying to copy it but it is all different lines of latex
at least i got the 1

Answer 13

I don't know how to do a new line.

Answer 14

\[x+y=2\\
x-y=1\]

Answer 15

\[\cos^{-1}(p)+\cos^{-1}(q)=\pi-\cos^{-1}(r) \\ \cos(\cos^{-1}(p)+\cos^{-1}(q))=\cos(\pi-\cos^{-1}(r)) \\ \cos(\cos^{-1}(p))\cos(\cos^{-1}(q))-\sin(\cos^{-1}(p))\sin(\cos^{-1}(q))= \\ \cos(\pi)\cos(\cos^{-1}(r))+\sin(\pi)\sin(\cos^{-1}(r)) \\ pq-\sqrt{1-p^2}\sqrt{1-q^2}=-r \\ -\sqrt{1-p^2}\sqrt{1-q^2}=-r-pq \\ (1-p^2)(1-q^2)=r^2+2rpq+p^2q^2 \\ 1-p^2-q^2+p^2q^2=r^2+2rpq+p^2q^2 \\ r^2+p^2+q^2+2rpq=p^2q^2-p^2q^2+1 \]

Answer 16

ok I just need a //

Answer 17

oops \\

Answer 18

Your way is much easier.

Answer 19

like this a lot
given that it is a constant, it has to be 1, but i had not idea why it had to be constant

Answer 20

Well I guess we proved it is a constant but I don't know if I can visualize it right now.

Answer 21

not me

Answer 22

guess it has something to do with the interior angles adding to \(\pi\)

Answer 23

oh no nvm, it is that weird expression that is a constant
i was thinking along the lines of \(\cos^{-1}(x)+\sin^{-1}(x)=1\) but this is different completely

Answer 24

i mean \(\pi\) doh

Answer 25

or maybe i mean \(\frac{\pi}{2}\) i must be tired

Answer 26

hmm...sleepy. no more thinking.

Answer 27

g'night

Answer 28

night