Trig2.17

Show that $\cos A+\cos (240-A)+\cos (240+A)=0\,$

LHS= $\cos A+\cos (180+(60-A))+\cos (180+(60+A))\,$

$\cos A-\cos (60-A)-\cos (60+A)\,$ (Since the ratios in the third quadrant and is negative)

$\cos A-[\cos 60 \cos A+\sin 60 \sin A+\cos 60 \cos A-\sin 60 \sin A]\,$

$\cos A-2\cos 60 \cos A\,$

$\cos A-2(\frac{1}{2} \cos A\,$

$\cos A-\cos A=0\,$ =RHS