Question

If in a $\Delta ABC$, $\angle B=\frac{2\pi }{3}$, then find the interval $\cos A+\cos C$ lies.

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

Answer 1

Answer: (A)

$[-\sqrt{3},\sqrt{3}]$

In a triangle ,we know that $A+B+C=\pi$ $=>A=\pi -B-C$

In $cosA+cosC$ putting $=>A=\pi -B-C$

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

$=\cos (\frac{\pi }{3}-C)+\cos \left(C\right)$

$=\frac{1}{2}\cos \left(C\right)+\frac{\sqrt{3}}{2}\sin \left(C\right)+\cos \left(C\right)$ $\left[cos\right(a-b)=cosa.cosb-sina.sinb]$

$=\frac{3}{2}\cos \left(C\right)+\frac{\sqrt{3}}{2}\sin \left(C\right)$

$=\sqrt{3}\left[\frac{\sqrt{3}}{2}\cos \right(C)+\frac{1}{2}\sin (C\left)\right]$

$=\sqrt{3}\left[\sin \left(\frac{\pi }{3}\right)\cos \right(C)+\cos \left(\frac{\pi }{3}\right)\sin (C\left)\right]$

$=\sqrt{3}\left[\sin (C+\frac{\pi }{3})\right]$

We know $-1\le \sin \left(\theta \right)\le 1$

Hence $-1\le \sin (C+\frac{\pi }{3})\le 1$

$-1.\sqrt{3}.\le \sqrt{3}\sin (60+C)\le \sqrt{3}.1$

$-\sqrt{3}.\le \sqrt{3}\sin (60+C)\le \sqrt{3}$

Hence,$\cos A+\cos C$ lies in the interval $[-\sqrt{3},\sqrt{3}]$