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Question Number 16041 by Tinkutara last updated on 17/Jun/17

If sides a, b, c of ΔABC are in H.P., prove that sin^2 ((A/2)), sin^2 ((B/2)), sin^2 ((C/2)) are in H.P.

$$\mathrm{If}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{of}\:\Delta{ABC}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}., \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{A}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{B}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$

Answered by Tinkutara last updated on 06/Jul/17

Let sin^2 ((A/2)), sin^2 ((B/2)), sin^2 ((C/2)) be in H.P. and we have to prove a, b, c are in H.P. ⇒ (1/(sin^2 (B/2))) − (1/(sin^2 (A/2))) = (1/(sin^2 (C/2))) − (1/(sin^2 (B/2))) sin^2 (C/2) (sin^2 (A/2) − sin^2 (B/2)) = sin^2 (A/2) (sin^2 (B/2) − sin^2 (C/2)) sin^2 (C/2) sin ((A + B)/2) sin ((A − B)/2) = sin^2 (A/2) sin ((B + C)/2) sin ((B − C)/2) 2 sin^2 (C/2) cos (C/2) sin ((A − B)/2) = 2 sin^2 (A/2) cos (A/2) sin ((B − C)/2) sin (C/2) sin C sin ((A − B)/2) = sin (A/2) sin A sin ((B − C)/2) ((sin A)/(sin C)) = ((2 sin (C/2) sin ((A − B)/2))/(2 sin (A/2) sin ((B − C)/2))) (a/c) = ((sin A − sin B)/(sin B − sin C)) = ((a − b)/(b − c)) ∴ (2/b) = (1/a) + (1/c)

$$\mathrm{Let}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{A}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{B}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{C}}{\mathrm{2}}\right)\:\mathrm{be}\:\mathrm{in} \\ $$$$\mathrm{H}.\mathrm{P}.\:\mathrm{and}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{in} \\ $$$$\mathrm{H}.\mathrm{P}. \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{{B}}{\mathrm{2}}}\:−\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{{C}}{\mathrm{2}}}\:−\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{{B}}{\mathrm{2}}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{{C}}{\mathrm{2}}\:\left(\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:−\:\mathrm{sin}^{\mathrm{2}} \:\frac{{B}}{\mathrm{2}}\right)\:=\:\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:\left(\mathrm{sin}^{\mathrm{2}} \:\frac{{B}}{\mathrm{2}}\:−\:\mathrm{sin}^{\mathrm{2}} \:\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{A}\:+\:{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{A}\:−\:{B}}{\mathrm{2}}\:=\:\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}\:+\:{C}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}\:−\:{C}}{\mathrm{2}} \\ $$$$\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{C}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{A}\:−\:{B}}{\mathrm{2}}\:=\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}\:−\:{C}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\:{C}\:\mathrm{sin}\:\frac{{A}\:−\:{B}}{\mathrm{2}}\:=\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:{A}\:\mathrm{sin}\:\frac{{B}\:−\:{C}}{\mathrm{2}} \\ $$$$\frac{\mathrm{sin}\:{A}}{\mathrm{sin}\:{C}}\:=\:\frac{\mathrm{2}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{A}\:−\:{B}}{\mathrm{2}}}{\mathrm{2}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}\:−\:{C}}{\mathrm{2}}} \\ $$$$\frac{{a}}{{c}}\:=\:\frac{\mathrm{sin}\:{A}\:−\:\mathrm{sin}\:{B}}{\mathrm{sin}\:{B}\:−\:\mathrm{sin}\:{C}}\:=\:\frac{{a}\:−\:{b}}{{b}\:−\:{c}} \\ $$$$\therefore\:\frac{\mathrm{2}}{{b}}\:=\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{c}} \\ $$

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