Question

2cos 50°cos70°+cos160°
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Answer 1

$\displaystyle\bf\\2Cos50^\circ Cos70^\circ+Cos160^\circ=\\\\\\=2\cdot\frac{Cos(50^\circ+70^\circ)+Cos(50^\circ-70^\circ)}{2} +Cos(180^\circ-20^\circ)=\\\\\\=Cos120^\circ+Cos20^\circ-Cos20^\circ=Cos120^\circ=-0,5$