Question

If $a^2+c^2=2b^2$ where a, b, c are the sides of a triangle, then prove that $\frac{sin3B}{sinB}=[\frac{a^2-c^2}{2ac}{]}^2$

Answer 1

L.H.S.$\frac{sin3B}{sinB}$= $3-4si{n}^{2}B=-1+4co{s}^{2}B$ ...... (1)

From the given relation, we have

$a^2+c^2-b^2=b^2$ or $2accosB=b^2$

$\therefore 4co{s}^{2}B=(\frac{b^2}{ac}{)}^2=(\frac{a^2+c^2}{2ac}{)}^2$ Put in (1) etc.

$\therefore L.H.S.=\frac{(a^2+c^2{)}^2-4a^2{c}^2}{4a^2{c}^2}=[\frac{a^2-c^2}{2ac}{]}^2$