Question

If the sides a, b, c of a triangle ABC are in H.P. Prove that ${{\sin }^{2}}\dfrac{A}{2},{{\sin }^{2}}\dfrac{B}{2},{{\sin }^{2}}\dfrac{C}{2}$ are in H.P.

Answer 1

Assume that ${{\sin }^{2}}\dfrac{A}{2},{{\sin }^{2}}\dfrac{B}{2}$ and ${{\sin }^{2}}\dfrac{C}{2}$ are in H.P and prove that $\dfrac{2}{b}:\dfrac{1}{a}+\dfrac{1}{c}$ by means of which we can say that $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in A.P. and hence a, b, c are in H.P. Use the formula: - $\sin \dfrac{A}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{bc}},\sin \dfrac{B}{2}=\sqrt{\dfrac{\left( s-a \right)\left( s-c \right)}{ac}}$ and $\sin \dfrac{C}{2}=\sqrt{\dfrac{\left( s-a \right)\left( s-b \right)}{ab}}$, where ‘s’ is the semi – perimeter of triangle and a, b, c are the sides to get the result.

Complete step-by-step solution
We have been given that a, b, c are in H.P. and we have to prove that ${{\sin }^{2}}\dfrac{A}{2},{{\sin }^{2}}\dfrac{B}{2},{{\sin }^{2}}\dfrac{C}{2}$.
Let us assume that ${{\sin }^{2}}\dfrac{A}{2},{{\sin }^{2}}\dfrac{B}{2}$ and ${{\sin }^{2}}\dfrac{C}{2}$ are in H.P, then we must prove that a, b, c are in H.P.
$\because {{\sin }^{2}}\dfrac{A}{2},{{\sin }^{2}}\dfrac{B}{2},{{\sin }^{2}}\dfrac{C}{2}$ are in H.P.
$\Rightarrow \dfrac{1}{{{\sin }^{2}}\dfrac{A}{2}},\dfrac{1}{{{\sin }^{2}}\dfrac{B}{2}},\dfrac{1}{{{\sin }^{2}}\dfrac{C}{2}}$ are in A.P.
Now, we know that: -

& \Rightarrow \sin \dfrac{A}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{bc}} \\\ & \Rightarrow \sin \dfrac{B}{2}=\sqrt{\dfrac{\left( s-a \right)\left( s-c \right)}{ac}} \\\ & \Rightarrow \sin \dfrac{C}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-a \right)}{ab}} \\\ \end{aligned}$$ Here, ‘s’ is the semi – perimeter and a, b, c are the sides of the triangle. Substituting these values in the expression, we get, $$\Rightarrow $$$$\dfrac{bc}{\left( s-b \right)\left( s-c \right)},\dfrac{ac}{\left( s-a \right)\left( s-c \right)},\dfrac{ab}{\left( s-a \right)\left( s-b \right)}$$, are in A.P Multiplying and dividing $${{1}^{st}}$$ term with $$s\left( s-a \right),{{2}^{nd}}$$ term with $$s\left( s-b \right)$$ and third term with $$s\left( s-c \right)$$, we get, $$\Rightarrow \dfrac{s\left( s-a \right)bc}{s\left( s-a \right)\left( s-b \right)\left( s-c \right)},\dfrac{s\left( s-b \right)ac}{s\left( s-a \right)\left( s-b \right)\left( s-c \right)},\dfrac{s\left( s-c \right)ab}{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$$ are in A.P. Canceling the common term from the above expression, we have, $$\Rightarrow \left( s-a \right)bc,\left( s-b \right)ac,\left( s-c \right)ab$$ are in A.P. $$\Rightarrow \left( sbc-abc \right),\left( sac-abc \right),\left( sab-abc \right)$$ are in A.P. $$\Rightarrow bc,ac,ab$$ are in A.P. Now, we can see that ac is the arithmetic mean of bc and ab, therefore using the formula of arithmetic mean between two terms, we get, $$\begin{aligned} & \Rightarrow 2ac=bc+ab \\\ & \Rightarrow 2ac=b\left( a+c \right) \\\ & \Rightarrow \dfrac{2}{b}=\dfrac{a+c}{ac} \\\ & \Rightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \\\ \end{aligned}$$ Clearly, we can see that $$\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$$ are in A.P. because $$\dfrac{1}{b}$$ is the arithmetic mean of $$\dfrac{1}{a}$$ and $$\dfrac{1}{c}$$. Now, $$\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$$ are in A.P., therefore its reciprocal must be in H.P. **Hence, a, b, c are in H.P.** **Note:** One may note that it is very difficult to prove the above result by assuming a, b, c are in H.P. and then proving $${{\sin }^{2}}\dfrac{A}{2},{{\sin }^{2}}\dfrac{B}{2},{{\sin }^{2}}\dfrac{C}{2}$$ are in H.P. Therefore, we have applied the reverse process. We must convert the given sine function in terms of sides of the triangle so that further simplification can take place. All the conversion formulas of the topic ‘properties of the triangle must be remembered.