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Prove that any triangle, $b^2\sin(2\gamma)+c^2\sin(2\beta)=2ac\sin(\beta)$.

Hello. I am very stuck on this problem. How could I go? Expand the double-angle sine but don't get to anything simpler. Also use that $\alpha+\beta+\gamma = 180$ to use identities of the type $\sin(2\pi-x) = -\sin(x)$ but also don't get to something simpler. Some hint?

eraldcoil
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    It may be helpful for you if you show your work. – Wolgwang Dec 16 '21 at 17:00
  • "Expand the double-angle sine but don't get to anything simpler." Don't you? Writing $\sin\gamma$ in terms of side lengths and the area (whatever symbol you use for it), and $\cos\gamma$ in terms of side lengths with the cosine rule, should be very helpful. – J.G. Dec 16 '21 at 17:07

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Hint (use double-angle formula to expand the left side first):
$1$. By law of sines, we have $b \sin(\gamma)=c \sin(\beta)$;
$2$. Construct the height of edge $a$ and prove $a=b \cos(\gamma)+c \cos(\beta)$.

Zerox
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    Thanks.

    $b^2\sin(2\gamma)+c^2\sin(2\beta)=2ac\sin(\beta)\ (b\sin\gamma)(b\cos(\gamma)+(c\sin\beta)(c\cos\beta)=ac\sin\beta\ c\sin\beta)(b\cos\gamma+c\cos\beta)=ac\sin(\beta)\ b\cos\gamma+c\cos\beta=a$

     – eraldcoil Dec 16 '21 at 17:17
  • @eraldcoil Well done, although it's best to write proofs forwards, so to speak. – J.G. Dec 16 '21 at 17:18