Question

If in a $\Delta ABC$, $\angle B=\frac{2\pi }{3}$ then the $(cosA+cosC)$ lie in

• A. $[-\sqrt{3},\sqrt{3}]$
• B. $(-\sqrt{3},\sqrt{3}]$
• C. $(\frac{3}{2},\sqrt{3}]$
• D. $[\frac{3}{2},\sqrt{3}]$

Answer 1

The correct option is A $[-\sqrt{3},\sqrt{3}]$

In $△ABC,\angle B=\frac{2\pi }{3}⟹A+C=\pi -\frac{2\pi }{3}=\frac{\pi }{3}$

$\cos A+\cos C=2\cos \frac{A+C}{2}\cos \frac{A-C}{2}=2\cos \frac{\pi }{6}\cos \frac{A-C}{2}=\sqrt{3}\cos \frac{A-C}{2}$

As we know that

$\cos \frac{A-C}{2}\in [-1,1]⟹\sqrt{3}\cos \frac{A-C}{2}\in [-\sqrt{3},\sqrt{3}]$

So $\cos A+\cos C\in [-\sqrt{3},\sqrt{3}]$