Question

If $a,b,c$ are in $H.P.$, prove that
${\sin }^2\frac{1}{2}A,{\sin }^2\frac{1}{2}B,{\sin }^2\frac{1}{2}C$ are also in $H.P.$

Answer 1

$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$
$\Rightarrow \frac{(\sin A-\sin B)}{\sin A\sin B}=\frac{\sin B-\sin C}{\sin B\sin C}$
or $\frac{2\sin \frac{A-B}{2}\cos \frac{A+B}{2}}{2\sin \left(A/2\right)\cos \left(A/2\right)}$
$=\frac{2\sin \frac{B-C}{2}\cos \frac{B+C}{2}}{2\sin \left(C/2\right)\cos \left(C/2\right)}$
or ${\sin }^2\frac{C}{2}\cos \frac{C}{2}\sin \frac{A-B}{2}$
$={\sin }^2\frac{A}{2}\cos \frac{A}{2}\sin \frac{B-C}{2}$
or $={\sin }^2\frac{A}{2}\cos \frac{B+C}{2}\sin \frac{B-C}{2}$
or ${\sin }^2\frac{C}{2}({\sin }^2\frac{A}{2}-{\sin }^2\frac{B}{2})$
$={\sin }^2\frac{A}{2}({\sin }^2\frac{B}{2}-{\sin }^2\frac{C}{2})$
Divide by ${\sin }^2\left(A/2\right){\sin }^2\left(B/2\right){\sin }^2\left(C/2\right)$
$\therefore \frac{1}{{\sin }^2\left(B/2\right)}-\frac{1}{{\sin }^2\left(A/2\right)}$
$=\frac{1}{{\sin }^2\left(C/2\right)}-\frac{1}{{\sin }^2\left(B/2\right)}$
$=\frac{1}{{\sin }^2\left(C/2\right)}-\frac{1}{{\sin }^2\left(B/2\right)}$
$\therefore \frac{1}{{\sin }^2\left(A/2\right)},\frac{1}{{\sin }^2\left(B/2\right)},\frac{1}{{\sin }^2\left(C/2\right)}$ are in $A.P.$
or ${\sin }^2\left(A/2\right),{\sin }^2\left(B/2\right),{\sin }^2\left(C/2\right)$ are in $H.P.$