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If $A+B+C=\pi$ does it imply that $$\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$$

If yes, how?

Julien
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chndn
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    Duplicate: http://math.stackexchange.com/q/154505/409 . (I think there are also other duplicates, as well, but this was the first I found.) – Blue Apr 17 '13 at 03:32

1 Answers1

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I think this has been asked here before.

One way is to chase ordinary high-school trigonometric identities.

Another is to consider a triangle in a circle of diameter (not radius) $1$. Partition it into three triangles, each with a vertex at the center of the circle. Their areas are $\frac18\sin2A$, $\frac18\sin2B$, and $\frac18\sin2C$. The sum of their areas is the area of the original triangle. If the original triangle does not have the center of the circle in its interior, then one of those is negative, and that's appropriate as you'll see if you draw the picture.

Then you need to show that the original triangle has area $\frac12\sin A\sin B\sin C$. Just use the fact that the lengths of the sides are equal to the sines of the opposite angles (figure out how to prove that if you've never done it before) and the area of any triangle is half the product of the lengths of any two adjacent sides times the sine of the angle between them.

Michael Hardy
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  • I need the first way. Thanks ... +1 – chndn Apr 17 '13 at 03:25
  • Do you mean that this is an identity. So can you prove it? – chndn Apr 17 '13 at 03:32
  • What I've written above is a sketch of a way to prove it. You can also write $\sin2A$ as $2\sin A\cos A$, and similarly for $B$, and $\sin(2C)=\sin(-2(A+B))$ and then apply the identity for the sine of a sum, and then sine and cosine of a double angle, etc. OK, maybe I'll write this out here tomorrow if no one else does it first. – Michael Hardy Apr 17 '13 at 03:35