Question

In a triangle the angles $A,B,C$ are in $A.P.$, show that
$2\cos \frac{A-C}{2}=\frac{a+c}{\sqrt{(a^2-ac+c^2)}}$

Answer 1

Since $A,B,C$ are in $A.P$., we have $2B=A+C$

But $A+B+C={180}^o$

Hence $3B={180}^o$ or $B={180}^o$ and $A+C={120}^o$

Now $b^2=c^2+a^2-2ca\cos B$

$=c^2+a^2-2ca\cos {60}^o$

$=c^2+a^2-ca$

Hence $\frac{a+c}{\surd (a^2-ac+c^2)}=\frac{a+c}{b}$ form $\left(1\right)$

$=\frac{k(\sin A+\sin C)}{k\sin B}$

$\frac{2\sin [A+C/2]\cos [A-C/2]}{\sin B}$

$=\frac{2\sin {60}^o\cos \frac{A-C}{2}}{\sin {60}^o}=2\cos \left(\frac{A-C}{2}\right)$