Question

In a triangle \[\Delta ABC\], if a, b, c are in H.P., then \[{{\sin }^{2}}\left( \dfrac{A}{2} \right)\], \[{{\sin }^{2}}\left( \dfrac{B}{2} \right),{{\sin }^{2}}\left( \dfrac{C}{2} \right)\] are in: -
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.

Answer 1

Hint: Take the reciprocal of a, b, c and consider them to be in A. P. Now, form relations between \[{{\sin }^{2}}\left( \dfrac{A}{2} \right)\], \[{{\sin }^{2}}\left( \dfrac{B}{2} \right),{{\sin }^{2}}\left( \dfrac{C}{2} \right)\] to get a sequence of known form. Check which of the options is correct. Use the formulas: - \[{{\sin }^{2}}\left( \dfrac{A}{2} \right)=\dfrac{\left( s-b \right)\left( s-c \right)}{bc},{{\sin }^{2}}\left( \dfrac{B}{2} \right)=\dfrac{\left( s-a \right)\left( s-c \right)}{ac}\] and \[{{\sin }^{2}}\left( \dfrac{C}{2} \right)=\dfrac{\left( s-a \right)\left( s-b \right)}{ab}\] , where ‘s’ is the semi – perimeter of the triangle.

Complete step-by-step solution
Here, we have been given that in a \[\Delta ABC\], a, b, c are in H.P, i.e. harmonic progression. Here, a, b and c are the sides of the \[\Delta ABC\] lying opposite to the angles A, B, and C respectively.

Now, we know that if some terms are in H.P. then their reciprocal will be in A.P. So, we have,
\[\Rightarrow \] a, b, c are in H.P.
\[\Rightarrow \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] are in A.P.
Multiplying all the terms with abc, we get,
\[\Rightarrow \dfrac{abc}{a},\dfrac{abc}{b},\dfrac{abc}{c}\] are in A.P.
\[\Rightarrow \] bc, ac, ab are in A.P.
Multiplying all the terms with semi – perimeter s, we get,
\[\Rightarrow \] sbc, sac, sab are in A.P.
Subtracting abc from all the terms, we get,
\[\Rightarrow \] sbc – abc, sac – abc, sab – abc are in A.P.
\[\Rightarrow \] bc (s – a), ac (s – b), ab (s – c) are in A.P.
Now, we know that: -
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}\left( \dfrac{A}{2} \right)=\dfrac{\left( s-b \right)\left( s-c \right)}{bc} \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{B}{2} \right)=\dfrac{\left( s-a \right)\left( s-c \right)}{ac} \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{C}{2} \right)=\dfrac{\left( s-a \right)\left( s-b \right)}{ab} \\
\end{align}\]
So, substituting the values of bc, ac and ab, we get,
\[\Rightarrow \dfrac{\left( s-a \right)\left( s-b \right)\left( s-c \right)}{{{\sin }^{2}}\left( \dfrac{A}{2} \right)},\dfrac{\left( s-a \right)\left( s-b \right)\left( s-c \right)}{{{\sin }^{2}}\left( \dfrac{B}{2} \right)},\dfrac{\left( s-a \right)\left( s-b \right)\left( s-c \right)}{{{\sin }^{2}}\left( \dfrac{C}{2} \right)}\] are in A.P.
Canceling the common numerator from all the three terms, we get,
\[\Rightarrow \dfrac{1}{{{\sin }^{2}}\left( \dfrac{A}{2} \right)},\dfrac{1}{{{\sin }^{2}}\left( \dfrac{B}{2} \right)},\dfrac{1}{{{\sin }^{2}}\left( \dfrac{C}{2} \right)}\] are in A.P.
Taking reciprocal of the above terms, we can say that,
\[\Rightarrow {{\sin }^{2}}\left( \dfrac{A}{2} \right),{{\sin }^{2}}\left( \dfrac{B}{2} \right),{{\sin }^{2}}\left( \dfrac{C}{2} \right)\] are in H.P.
Hence, the option (c) is the correct answer.

Note: One may note that multiplying, dividing, adding, and subtracting does not change the sequence of terms in arithmetic progression. That means they remain in the sequence of an arithmetic progression. This is an important property of A.P. which is used above. Note that you must try not try to change sine of the given half angles into the cosine form using the identity: - \[\Rightarrow {{\sin }^{2}}\left( \dfrac{\theta }{2} \right)=\dfrac{1-\cos \theta }{2}\]. In this way also we will get the answer but the calculation will be too hard to handle.