Question

prove that b^2-c^2/a^2 sin2A+c^2-a^2/b^2 sin2b+b^2-b^2/b^2 sin2c=0

Answer 1

Answer:

We know that in a triangle a=k\sin A,b=k\sin B,c=k\sin Ca=ksinA,b=ksinB,c=ksinC .

So

\frac{\sin 2A}{a^2} =\frac{2}{a^2} \sin A \cos A=\frac{2a}{ka^2} \frac{b^2+c^2-a^2}{2bc} =\frac{b^2+c^2-a^2}{2kabc}

a

2

sin2A

=

a

2

2

sinAcosA=

ka

2

2a

2bc

b

2

+c

2

−a

2

=

2kabc

b

2

+c

2

−a

2

.

Similarly,

$$\begin{lgathered}\frac{\sin 2B}{b^2} =\frac{c^2+a^2-b^2}{2kabc} \\ \frac{\sin 2C}{c^2} =\frac{a^2+b^2-c^2}{2kabc}\end{lgathered}$$ .

Thus,

$$\begin{lgathered}LHS=(b^2 -c^2)\frac{\sin 2A}{a^2} +(c^2 -b^2)\frac{\sin 2B}{b^2}+(a^2 -c^2)\frac{\sin 2C}{c^2}\\\end{lgathered}$$

$$\begin{lgathered}LHS=(b^2 -c^2)\frac{b^2+c^2-a^2}{2kabc} +(c^2 -b^2)\frac{c^2+a^2-b^2}{2kabc}+(a^2 -b^2)\frac{a^2+b^2-c^2}{2kabc}\\ LHS=\frac{b^4-c^4-a^2(b^2 -c^2)}{2kabc} +\frac{c^4-a^4-b^2(c^2 -b^2)}{2kabc}+\frac{a^4-b^4-c^2(a^2 -b^2)}{2kabc}\\ LHS=\frac{b^4-c^4-a^2(b^2 -c^2)+c^4-a^4-b^2(c^2 -b^2)+a^4-b^4-c^2(a^2 -b^2)}{2kabc} \\ LHS=\frac{0}{2kabc} =0=RHS\end{lgathered}$$ .

The proof is complete.