In $\Delta ABC$ if $a,b,c$ are in Harmonic Progression

Question

In $\Delta ABC$ if $a,b,c$ are in Harmonic Progression Then Prove that

$$\sin ^2\left(\frac{A}{2}\right),\sin ^2\left(\frac{B}{2}\right),\sin ^2\left(\frac{C}{2}\right)$$ are in Harmonic Progression

My Try:

we have

$$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$$ Then

$$\frac{a-b}{a}=\frac{b-c}{c}$$ and by Sine Rule

$$ \frac{\sin A-\sin B}{\sin A}=\frac {\sin B-\sin C}{\sin C}$$ $\implies$

$$\frac{2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}=\frac{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{B-C}{2}\right)}{2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right)}$$ $\implies$

$$\sin ^2\left(\frac{C}{2}\right)\left(2\cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)\right)=\sin ^2\left(\frac{A}{2}\right)\left(2\cos \left(\frac{A}{2}\right)\cos \left(\frac{B-C}{2}\right)\right)$$ $\implies$

$$\sin ^2\left(\frac{C}{2}\right) \left(\sin B+\sin A\right)=\sin ^2\left(\frac{A}{2}\right) \left(\sin B+\sin C\right)$$

Any way to proceed?

Answer 1

Hint:

Use cosine rule of triangle

and $\cos2x=1-2\sin^2x$ for $A,B,C$ to express $\sin^2$ half angles in terms of $a,b,c$

Answer 2

We need $\csc^2$ half angles in arithmetic progression

$\iff\cot^2$ in arithmetic progression

Use $\cot\dfrac A2=\dfrac{s(s-a)}\triangle$

Answer 3

Given \begin{align} b &= 2\,\frac{ac}{a+c} \tag{1}\label{1} ,\\ \end{align}
prove that \begin{align} \sin^2\tfrac\beta2&= 2\,\frac{\sin^2\tfrac\alpha2 \sin^2\tfrac\gamma2}{\sin^2\tfrac\alpha2+\sin^2\tfrac\gamma2} \tag{2}\label{2} \end{align}

Using the half-angle formulas \begin{align} \sin^2\tfrac\alpha2 &=\frac{(s-b)(s-c)}{bc}=\tfrac14\,\frac{(a-b+c)(a+b-c)}{bc} ,\\ \sin^2\tfrac\gamma2 &=\frac{(s-a)(s-b)}{ab}=\tfrac14\,\frac{(-a+b+c)(a-b+c)}{ab} ,\\ \sin^2\tfrac\beta2 &=\frac{(s-a)(s-c)}{ac}=\tfrac14\,\frac{(-a+b+c)(a+b-c)}{ac} \tag{3}\label{3} \end{align}

with $b$ from \eqref{1}, it follows that indeed \eqref{3}=\eqref{2}.

Tested in Maxima:

(%i1) display2d:false$
(%i2) declare([a,b,c,sa,sb,sc],real)$
(%i3) halfSinSqred(a,b,c):=1/4*(a-b+c)*(a+b-c)/b/c$
(%i4) b:2*a*c/(a+c)$
(%i5) sa:factor(halfSinSqred(a,b,c));
(%o5) -((c^2+a^2)*(c^2-2*a*c-a^2))/(8*a*c^2*(c+a))
(%i6) sc:factor(halfSinSqred(c,a,b));
(%o6) ((c^2+a^2)*(c^2+2*a*c-a^2))/(8*a^2*c*(c+a))
(%i7) sb:factor(halfSinSqred(b,c,a));
(%o7) -((c^2-2*a*c-a^2)*(c^2+2*a*c-a^2))/(4*a*c*(c+a)^2)
(%i8) sbharm:factor(2*sa*sc/(sa+sc));
(%o8) -((c^2-2*a*c-a^2)*(c^2+2*a*c-a^2))/(4*a*c*(c+a)^2)
(%i9) factor(sb-sbharm);
(%o9) 0
(%i10) quit();