Question

If the sides a, b and c of ∆ABC are in H.P., prove that ${\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2}\text{and}{\sin }^2\frac{C}{2}$ are in H.P.

Answer 1

$$\begin{gathered} {\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2}\mathrm{and}{\sin }^2\frac{C}{2}\mathrm{is}\mathrm{a}\mathrm{H}.\mathrm{P}. \\ \iff \frac{1}{{\sin }^2{\displaystyle \frac{A}{2}}},\frac{1}{{\sin }^2{\displaystyle \frac{B}{2}}}\mathrm{and}\frac{1}{{\sin }^2{\displaystyle \frac{C}{2}}}\mathrm{is}\mathrm{an}\mathrm{A}.\mathrm{P}. \\ \iff \frac{1}{{\sin }^2{\displaystyle \frac{B}{2}}}-\frac{1}{{\sin }^2{\displaystyle \frac{A}{2}}}=\frac{1}{{\sin }^2{\displaystyle \frac{C}{2}}}-\frac{1}{{\sin }^2{\displaystyle \frac{B}{2}}} \\ \iff \frac{{\sin }^2{\displaystyle \frac{A}{2}}-{\sin }^2{\displaystyle \frac{B}{2}}}{{\sin }^2{\displaystyle \frac{A}{2}}{\sin }^2{\displaystyle \frac{B}{2}}}=\frac{{\sin }^2{\displaystyle \frac{B}{2}}-{\sin }^2{\displaystyle \frac{C}{2}}}{{\sin }^2{\displaystyle \frac{B}{2}}{\sin }^2{\displaystyle \frac{C}{2}}} \\ \iff \frac{\sin \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)}{{\sin }^2{\displaystyle \frac{A}{2}}}=\frac{\sin {\displaystyle \left(\frac{B+C}{2}\right)}\sin \left({\displaystyle \frac{B-C}{2}}\right)}{{\sin }^2{\displaystyle \frac{C}{2}}} \\ \iff \frac{\cos \left({\displaystyle \frac{C}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)}{{\sin }^2{\displaystyle \frac{A}{2}}}=\frac{\cos {\displaystyle \left(\frac{\mathrm{A}}{2}\right)}\sin \left({\displaystyle \frac{B-C}{2}}\right)}{{\sin }^2{\displaystyle \frac{C}{2}}}\left[\mathrm{As},A+B+C=\mathrm{\pi }\right] \\ \iff {\sin }^2\frac{C}{2}\cos \left(\frac{C}{2}\right)\sin \left(\frac{A-B}{2}\right)={\sin }^2\frac{A}{2}\cos \left(\frac{A}{\mathit{2}}\right)\sin \left(\frac{B-C}{2}\right) \\ \iff 2\sin \frac{C}{2}\sin \frac{C}{2}\cos \left(\frac{C}{2}\right)\sin \left(\frac{A-B}{2}\right)=2\sin \frac{A}{2}\sin \frac{A}{2}\cos \left(\frac{A}{\mathit{2}}\right)\sin \left(\frac{B-C}{2}\right) \\ \iff \sin \frac{C}{2}\sin C\sin \left(\frac{A-B}{2}\right)=\sin \frac{A}{2}\sin A\sin \left(\frac{B-C}{2}\right)\left[\because \sin 2\mathrm{\theta }=2\mathrm{sin\theta cos\theta }\right] \\ \iff \sin C\mathit{}\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)=\sin A\cos \left(\frac{B+C}{2}\right)\sin \left(\frac{B-C}{2}\right)\left[\mathrm{As},A+B+C=\mathrm{\pi }\right] \\ \iff \sin C\frac{\left(\sin A-\sin B\right)}{2}=\sin A\frac{\left(\sin B-\sin C\right)}{2}\left[\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)\right] \\ \iff \sin C\left(\sin A-\sin B\right)=\sin A\left(\sin B-\sin C\right) \\ \iff ck\left(ak\mathit{-}bk\right)=ak\left(bk\mathit{-}ck\right)\left(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\left(\mathrm{say}\right)\right) \\ \iff ca\mathit{-}cb=ab\mathit{-}ac \\ \iff 2ac=ab+bc \\ \iff \frac{2}{b}=\frac{1}{c}+\frac{1}{a}\left[\mathrm{multiplying}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{by}abc\right] \\ \iff a\mathit{,}\mathit{}b\mathit{,}\mathit{}c\mathrm{are}\mathrm{in}\mathrm{H}.\mathrm{P}. \end{gathered}$$