Question

In a $△ABC(\frac{a^2}{\sin A}+\frac{b^2}{\sin B}+\frac{c^2}{\sin C})\cdot \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ simplifies to

• A. $2△$
• B. $△$
• C. $\frac{△}{2}$
• D. $\frac{△}{4}$

Answer 1

Answer: (B)

$△$

$△ABC(\frac{a^2}{\sin A}+\frac{b^2}{\sin B}+\frac{c^2}{\sin C})\cdot \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$=(\frac{4R^2{\sin }^{2}A}{\sin A}+\frac{4R^2{\sin }^{2}B}{\sin B}+\frac{4R^2{\sin }^{2}C}{\sin C})\cdot \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$=4R^2(\sin A+\sin B+\sin C)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$\sin A+\sin B+\sin C=\sin A+\sin B+\sin (A+B)$
$=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A+B}{2}\right)$
$=2\sin \left(\frac{A+B}{2}\right)(\cos \left(\frac{A+B}{2}\right)+\cos \left(\frac{A-B}{2}\right))$
$=2\sin \left(\frac{180-C}{2}\right)(2\cos \left(\frac{\frac{A+B}{2}+\frac{A-B}{2}}{2}\right)+\cos \left(\frac{\frac{A+B}{2}+\frac{A+B}{2}}{2}\right))$
$=2\cos \frac{C}{2}\cdot 2\cdot \cos \frac{A}{2}\cos \frac{B}{2}$
$\therefore =4R^2\cdot \left(4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\right)\cdot \left(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)$
$=2R^2\sin A+\sin B+\sin C=△$