# prove that a^2sin(A-C)=

• Dec 10th 2009, 11:07 PM
bigwave
prove that a^2sin(A-C)=
prove that

$\displaystyle a^2 \sin{\left(B-C\right)} = \left(b^2- c^2\right)\sin{A}$

where
$\displaystyle a,b,c$ are sides of any triangle
$\displaystyle A,B,C$ are the corresponding angles
• Dec 10th 2009, 11:18 PM
alexmahone
Quote:

Originally Posted by bigwave
prove that

$\displaystyle a^2 \sin{\left(B-C\right)} = \left(b^2- c^2\right)\sin{A}$

where
$\displaystyle a,b,c$ are sides of any triangle
$\displaystyle A,B,C$ are the corresponding angles

$\displaystyle \frac{a^2}{b^2-c^2}=\frac{sin^2A}{sin^2B-sin^2C}$

= $\displaystyle \frac{sin^2A}{(sinB+sinC)(sinB-sinC)}$

= $\displaystyle \frac{sin^2A}{2sin(\frac{B+C}{2})cos(\frac{B-C}{2}).2cos(\frac{B+C}{2})sin(\frac{B-C}{2})}$

= $\displaystyle \frac{sin^2A}{2sin(\frac{B+C}{2})cos(\frac{B+C}{2} ).2sin(\frac{B-C}{2})cos(\frac{B-C}{2})}$

= $\displaystyle \frac{sin^2A}{sin(B+C)sin(B-C)}$

= $\displaystyle \frac{sin^2A}{sin(\pi-A)sin(B-C)}$

= $\displaystyle \frac{sin^2A}{sinAsin(B-C)}$

= $\displaystyle \frac{sinA}{sin(B-C)}$

Thus, $\displaystyle a^2sin(B-C)=(b^2-c^2)sinA$