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Question
If the sides a, b and c of ∆ABC are in H.P., prove that \[\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2} \text{ and } \sin^2 \frac{C}{2}\]
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Solution
\[\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2} \text{ and } \sin^2 \frac{C}{2} \text{ is a H . P }. \]
\[ \Leftrightarrow \frac{1}{\sin^2 \frac{A}{2}}, \frac{1}{\sin^2 \frac{B}{2}} \text{ and } \frac{1}{\sin^2 \frac{C}{2}} \text{ is an A . P } . \]
\[ \Leftrightarrow \frac{1}{\sin^2 \frac{B}{2}} - \frac{1}{\sin^2 \frac{A}{2}} = \frac{1}{\sin^2 \frac{C}{2}} - \frac{1}{\sin^2 \frac{B}{2}}\]
\[ \Leftrightarrow \frac{\sin^2 \frac{A}{2} - \sin^2 \frac{B}{2}}{\sin^2 \frac{A}{2} \sin^2 \frac{B}{2}} = \frac{\sin^2 \frac{B}{2} - \sin^2 \frac{C}{2}}{\sin^2 \frac{B}{2} \sin^2 \frac{C}{2}}\]
\[ \Leftrightarrow \frac{\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)}{\sin^2 \frac{A}{2}} = \frac{\sin\left( \frac{B + C}{2} \right)\sin\left( \frac{B - C}{2} \right)}{\sin^2 \frac{C}{2}}\]
\[ \Leftrightarrow \frac{\cos\left( \frac{C}{2} \right)\sin\left( \frac{A - B}{2} \right)}{\sin^2 \frac{A}{2}} = \frac{\cos\left( \frac{A}{2} \right)\sin\left( \frac{B - C}{2} \right)}{\sin^2 \frac{C}{2}} \left[ As, A + B + C = \pi \right]\]
\[ \Leftrightarrow \sin^2 \frac{C}{2}\cos\left( \frac{C}{2} \right)\sin\left( \frac{A - B}{2} \right) = \sin^2 \frac{A}{2}\cos\left( \frac{A}{2} \right)\sin\left( \frac{B - C}{2} \right)\]
\[ \Leftrightarrow 2\sin\frac{C}{2}\sin\frac{C}{2}\cos\left( \frac{C}{2} \right)\sin\left( \frac{A - B}{2} \right) = 2\sin\frac{A}{2}\sin\frac{A}{2}\cos\left( \frac{A}{2} \right)\sin\left( \frac{B - C}{2} \right)\]
\[ \Leftrightarrow \sin\frac{C}{2}\sin C \sin\left( \frac{A - B}{2} \right) = \sin\frac{A}{2}\sin A\sin\left( \frac{B - C}{2} \right) \left[ \because \sin2\theta = 2sin\thetacos\theta \right]\]
\[ \Leftrightarrow \sin C \cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right) = \sin A \cos\left( \frac{B + C}{2} \right) \sin\left( \frac{B - C}{2} \right) \left[ As, A + B + C = \pi \right]\]
\[ \Leftrightarrow \sin C\frac{\left( \sin A - \sin B \right)}{2} = \sin A\frac{\left( \sin B - \sin C \right)}{2} \left[ \sin C - \sin D = 2\cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ \Leftrightarrow \sin C\left( \sin A - \sin B \right) = \sin A\left( \sin B - \sin C \right)\]
\[ \Leftrightarrow ck\left( ak - bk \right) = ak\left( bk - ck \right) \left( \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \left( say \right) \right)\]
\[ \Leftrightarrow ca - cb = ab - ac\]
\[ \Leftrightarrow 2ac = ab + bc\]
\[ \Leftrightarrow \frac{2}{b} = \frac{1}{c} + \frac{1}{a} \left[ \text{ multiplying both the sides by abc } \right]\]
\[ \Leftrightarrow \text{ a, b, c are in H . P } . \]
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