Question

If A , B , C are angle of a triangle show that
$\cos \left(\frac{A+2B+3C}{2}\right)+\cos \left(\frac{A-C}{2}\right)=0$

Answer 1

Given,

To prove, $\cos \left(\frac{A+2B+3C}{2}\right)+\cos \left(\frac{A-C}{2}\right)=0$

$\Rightarrow \cos \left(\frac{A+2B+3C}{2}\right)=-\cos \left(\frac{A-C}{2}\right)$

RHS:

$-\cos \left(\frac{A-C}{2}\right)$

$=\cos (\pi -\frac{A-C}{2})$

$=\cos \left(\frac{2\pi -A+C}{2}\right)$

$=\cos \left(\frac{\pi +(\pi -A)+C}{2}\right)$

$A+B+C=\pi \Rightarrow \pi -A=B+C$

$=\cos \left(\frac{\pi +B+C+C}{2}\right)$

$=\cos \left(\frac{\pi +B+2C}{2}\right)$........(i)

LHS:

$\cos \left(\frac{A+2B+3C}{2}\right)$

$=\cos \left(\frac{[A+B+C]+B+2C}{2}\right)$

$=\cos \left(\frac{\pi +B+2C}{2}\right)$........(ii)

comparing (i) and (ii), we get,

LHS=RHS

$\cos \left(\frac{A+2B+3C}{2}\right)+\cos \left(\frac{A-C}{2}\right)=0$

Hence proved.