This is a follow-up to my previous question about an identity with the sides and angles of a triangle.
Can $a \cos A + b \cos B + c \cos C$ equal $4R \sin A \sin B \sin C$?
I'm not sure if this is possible. Is it?
R is the circumradius.
Asked
Active
Viewed 83 times
0
Mathy Person
- 1,565
-
2what are $a,b,c$ and $A,B,C$ ? – r9m Jan 13 '15 at 05:24
-
You need a circumradius factor on the right-hand side. – Blue Jan 13 '15 at 05:30
-
@Blue: Oops, I forgot to add that. I will fix that. But how would I prove that? – Mathy Person Jan 13 '15 at 05:34
-
@r9m: a, b, and c are sides of a triangle. A, B, and C are the angles corresponding to the sides. – Mathy Person Jan 13 '15 at 05:50
-
2Once $a$, $b$, $c$ are written as $2 R \sin A$, $2 R\sin B$, $2 R\sin C$, so that the left-hand side equals $R(\sin 2A + \sin 2B+\sin 2C)$, this question becomes a duplicate. – Blue Jan 13 '15 at 05:53
-
@Blue: How can $a \cos A + b \cos B + c \ cos C = R(\sin 2A + \sin 2B + \sin 2C)$, though? I understand the second part, where $R(\sin 2A + \sin 2B + \sin 2C) = 4R \sin A \sin B \sin C$. – Mathy Person Jan 13 '15 at 05:57
-
2$a \cos A = 2 R\sin A \cos A = R ( 2\sin A\cos A) = R \sin 2A$. – Blue Jan 13 '15 at 05:58
-
@Blue: Got it now. Thanks. – Mathy Person Jan 13 '15 at 06:15
1 Answers
1
$$a,b,c,A,B,C=0$$ works. Probably not the solution you were looking for though.
Teoc
- 8,700
-
@MathNoob: No, not really, haha. I was asking it for this problem: http://math.stackexchange.com/questions/1102050/prove-fraca-sin-ab-sin-bc-sin-ca-cos-ab-cos-bc-cos-c-r-left-fraca2/1102070?noredirect=1#comment2247087_1102070 , where one of the answers mentioned that those two were equal. – Mathy Person Jan 13 '15 at 05:28