Question

If $A=4(\cos \frac{2\pi }{15}+\cos \frac{4\pi }{15}-\cos \frac{7\pi }{15}-\cos \frac{\pi }{15})$ and $B=8\cot (\alpha +\beta +\gamma )$, where $\tan \alpha ,\tan \beta ,\tan \gamma$ are the real roots of the equation $x^3-8(a-b)x^2+(2a-3b)x-4(b+1)=0$, then the value of $A-B$ is

Answer 1

$A=4(\cos \frac{2\pi }{15}+\cos \frac{4\pi }{15}-\cos \frac{7\pi }{15}-\cos \frac{\pi }{15})$
$=4\left[\right(\cos {24}^{\circ }+\cos {48}^{\circ })-(\cos {84}^{\circ }+\cos {12}^{\circ }\left)\right]=4[2\cos {36}^{\circ }\cos {12}^{\circ }-2\cos {48}^{\circ }\cos {36}^{\circ }]$
$=8\cos {36}^{\circ }[\cos {12}^{\circ }-\cos {48}^{\circ }]=8\cos {36}^{\circ }\left[2\sin {30}^{\circ }\sin {18}^{\circ }\right]$
$=16\times \frac{\sqrt{5}+1}{4}\times \frac{1}{2}\times \frac{\sqrt{5}-1}{4}=2$

$B=8\cot (\alpha +\beta +\gamma )$
$\tan \alpha ,\tan \beta ,\tan \gamma$ are the real roots of the equation $x^3-8(a-b)x^2+(2a-3b)x-4(b+1)=0$
$\therefore \sum \tan \alpha =8(a-b)$
$\sum \tan \alpha \tan \beta =(2a-3b)$
and $\prod \tan \alpha =4(b+1)$
Now, $\tan (\alpha +\beta +\gamma )=\frac{\sum \tan \alpha -\prod \tan \alpha }{1-\sum \tan \alpha \tan \beta }$
$=\frac{8(a-b)-4(b+1)}{1-(2a-3b)}=\frac{4(2a-2b-b-1)}{(1-2a+3b)}$
$=\frac{4(2a-3b-1)}{(1-2a+3b)}=-4$
$\Rightarrow \cot (\alpha +\beta +\gamma )=-\frac{1}{4}$
$\therefore B=8\cot (\alpha +\beta +\gamma )=-2$

Hence, $A-B=4$