Question

If in a $△ABC,\angle B=\frac{2\pi }{3}$ then $\cos A+\cos C$ lie in

Answer 1

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

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

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

$=2\sin \frac{\frac{2\pi }{3}}{2}\cos \left(\frac{A-C}{2}\right)$

$=2\sin \frac{\pi }{3}\cos \left(\frac{A-C}{2}\right)$

$=2\times \frac{\sqrt{3}}{2}\cos \left(\frac{A-C}{2}\right)$

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

$\Rightarrow 0\le A+C\le {60}^{\circ }$

$\Rightarrow 0<A-C<{60}^{\circ }$

$\Rightarrow \cos \left(\frac{A-C}{2}\right)\Rightarrow (\frac{\sqrt{3}}{2},1)$

$\Rightarrow \cos A+\cos C$ lie in $(\frac{3}{2},\sqrt{3}]$