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Identity

Posted July 14th, 2009 by alpha
in

If ABC are the angles of a triangle , then prove that
$ \frac{sin 2A+sin 2B+sin2C}{cos A+cosB+cosC-1}=8cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2} $

$$\sin 2A+\sin 2B+\sin 2C=2\sin (A+B)\cos (A-B)-2\sin (A+B)\cos (A+B)=2 \sin (A+B)(\cos (A-B)-\cos (A+B))=4\sin C \sin A\sin B$$

as

$$ \sin 2C=2 sin C\cos C=2\sin (\pi-A-B)\cos (\pi-A-B)=-2\sin (A+B)\cos (A+B)$$
$$\cos A+\cos B+\cos C-1=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-1-\cos (A+B)=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-2\cos^2\frac{A+B}{2}=$$
$$=2\cos \frac{A+B}{2}(\cos \frac{A-B}{2}-\cos \frac{A+B}{2})=4\cos \frac{A+B}{2}\sin \frac {A}{2}\sin \frac{B}{2}=4 \sin \frac{A}{2}\sin\frac{B}{2}\sin \frac{C}{2}$$

Then use

$$\sin A=2 \sin\frac{A}{2}\cos \frac{A}{2}$$
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Comments

answer

On July 14th, 2009 Structure says:
$$\sin 2A+\sin 2B+\sin 2C=2\sin (A+B)\cos (A-B)-2\sin (A+B)\cos (A+B)=2 \sin (A+B)(\cos (A-B)-\cos (A+B))=4\sin C \sin A\sin B$$

as

$$ \sin 2C=2 sin C\cos C=2\sin (\pi-A-B)\cos (\pi-A-B)=-2\sin (A+B)\cos (A+B)$$
$$\cos A+\cos B+\cos C-1=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-1-\cos (A+B)=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-2\cos^2\frac{A+B}{2}=$$
$$=2\cos \frac{A+B}{2}(\cos \frac{A-B}{2}-\cos \frac{A+B}{2})=4\cos \frac{A+B}{2}\sin \frac {A}{2}\sin \frac{B}{2}=4 \sin \frac{A}{2}\sin\frac{B}{2}\sin \frac{C}{2}$$

Then use

$$\sin A=2 \sin\frac{A}{2}\cos \frac{A}{2}$$

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