Question

can someone help with a trigonometry problem?

Answer 1

\[Cos^2 A -\cos^2 B -\cos^2 C = -1\]
If this equation then show that ABC is a rigt triangle

Answer 2

use cosine rule for each of the angles..
2 steps really..
\[\cos A=\frac{b^2+c^2-a^2}{2ab}\]
and others

Answer 3

I have tried for quite some time to reason it out this way can you jelp walk me throug it?

Answer 4

sorry I meant help not jelp

Answer 5

ok i'll start you up
\[
\frac{b^2+c^2-a^2}{2bc}-\frac{c^2+a^2-b^2}{2ca}-\frac{a^2+b^2-c^2}{2ab}=-1
\]
take the common deno

Answer 6

ok I will try thank you @electrokid

Answer 7

I found the common denominator and it turns out to be\[\frac{ a^3-a^2 b +ab^2 +b^3 -b^2 c-bc^2 +c^3 +c^2 a-ca^2}{ 2abc }=-1\]
and I can't figure out what to do to prove they are equivalent from here @electrokid

Answer 8

oh wait...
my bad....
I did not see the cos^2
the law of cosines is for "cos"

Answer 9

now go ahead.
you will need Pythagorean thm too