In triangle ABC,find the value of frac{sin 2A | vedclass

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Question

(A) We know that in $	riangle ABC$,$A+B+C = \pi$. The numerator is $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$. The denominator is $\cos A

Vedclass · Accepted Answer

$2[\sin A + \sin B + \sin C]$

Vedclass · Answer

(A) We know that in $	riangle ABC$,$A+B+C = \pi$. The numerator is $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$. The denominator is $\cos A + \cos B + \cos C - 1 = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. Thus,the expression becomes: $\frac{4 \sin A \sin B \sin C}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = \frac{4 (2 \sin \frac{A}{2} \cos \frac{A}{2}) (2 \sin \frac{B}{2} \cos \frac{B}{2}) (2 \sin \frac{C}{2} \cos \frac{C}{2})}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$. We also know that $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$. Therefore,$8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = 2 [4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}] = 2 [\sin A + \sin B + \sin C]$.

In $\triangle ABC$,find the value of $\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1}$.

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