Question

$\text{If the sides a, b, c of a}\Delta ABC\text{are in H.P., prove that}si{n}^2\frac{A}{2},si{n}^2\frac{B}{2},si{n}^2\frac{C}{2}\text{are in H.P.}$

Answer 1

$si{n}^2\frac{A}{2},si{n}^2\frac{B}{2}andsi{n}^2\frac{C}{2}\text{are in H.P.}\Rightarrow \frac{1}{si{n}^2\frac{A}{2}},\frac{1}{si{n}^2\frac{B}{2}}and\frac{1}{si{n}^2\frac{C}{2}}\text{are in A.P.}\Rightarrow \frac{1}{si{n}^2\frac{B}{2}}-\frac{1}{si{n}^2\frac{A}{2}}=\frac{1}{si{n}^2\frac{C}{2}}-\frac{1}{si{n}^2\frac{B}{2}}\Rightarrow \frac{si{n}^2\frac{A}{2}-si{n}^2\frac{B}{2}}{si{n}^2\frac{A}{2}si{n}^2\frac{B}{2}}=\frac{si{n}^2\frac{B}{2}-si{n}^2\frac{C}{2}}{si{n}^2\frac{B}{2}si{n}^2\frac{C}{2}}\Rightarrow \frac{sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)}{si{n}^2\frac{A}{2}}=\frac{sin\left(\frac{B+C}{2}\right)sin\left(\frac{B-C}{2}\right)}{si{n}^2\frac{C}{2}}\Rightarrow \frac{cos\left(\frac{C}{2}\right)sin\left(\frac{A-B}{2}\right)}{si{n}^2\frac{A}{2}}=\frac{cos\left(\frac{A}{2}\right)sin\left(\frac{B-C}{2}\right)}{si{n}^2\frac{C}{2}}[\because AsA+B+C=\pi ]\Rightarrow si{n}^2\frac{C}{2}cos\left(\frac{C}{2}\right)sin\left(\frac{A-B}{2}\right)=si{n}^{2}cos\left(\frac{A}{2}\right)sin\left(\frac{B-C}{2}\right)\Rightarrow 2sin\frac{C}{2}sin\frac{C}{2}cos\left(\frac{C}{2}\right)sin\left(\frac{A-B}{2}\right)=2sin\frac{A}{2}sin\frac{A}{2}cos\left(\frac{A}{2}\right)sin\left(\frac{B-C}{2}\right)\Rightarrow sin\frac{C}{2}sinCsin\left(\frac{A-B}{2}\right)=sin\frac{A}{2}sinAsin\left(\frac{B-C}{2}\right)[\because sin2\theta =2sin\theta cos\theta ]\Rightarrow sinCcos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)=sinAcos\left(\frac{B+C}{2}\right)sin\left(\frac{B-C}{2}\right)[AsA+B+C=\pi ]\Rightarrow sinC\left(\frac{sinA-sinB}{2}\right)=sin\left(\frac{sinB-sinC}{2}\right)[\because sinC-sinD=2cos\left(\frac{C+D}{2}\right)sin\left(\frac{C-D}{2}\right)]\Rightarrow sinC(sinA-sinB)=sinA(sinB-sinC)\Rightarrow ck(ak-bk)=ak(bk-ck)(\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k\left(say\right))\Rightarrow ca-cb=ab-ac\Rightarrow 2ac=ab+bc\Rightarrow \frac{2}{b}=\frac{1}{c}+\frac{1}{a}\left[\text{multiplying both the sides by abc}\right]\Rightarrow \text{a, b, c are in H.P.}\text{PQ. In any}\Delta ABC,\text{prove that :}a(cosC-cosB)=2(b-c)co{s}^2\frac{A}{2}Leta=ksinAa(cosC-cosB)=2(b-c)co{s}^2\frac{A}{2}\text{LHS}=a(cosC-cosB)=a2.sin\frac{C+B}{2}.sin\frac{B-C}{2}=2ksinAsin\frac{\pi -A}{2}.sin\frac{B-C}{2}=2k2sin\frac{A}{2}.cos\frac{A}{2}.cos\frac{A}{2}.sin\frac{B-C}{2}=2kco{s}^2\frac{A}{2}(2sin\frac{B-C}{2}.sin\frac{A}{2})=2kco{s}^2\frac{A}{2}(2sin\frac{B-C}{2}.sin\frac{\pi -(B+C)}{2})=2kco{s}^2\frac{A}{2}(2sin\frac{B-C}{2}.cos\frac{(B+C)}{2})=2kco{s}^2\frac{A}{2}(sinB-sinC)=2co{s}^2\frac{A}{2}(ksinB-ksinC)=2co{s}^2\frac{A}{2}(b-c)=RHS$