Chứng minh đẳng thức: $(\cos70^0+\cos 50^0)(\cos310^0+\cos290^0)+(\cos40^0+\cos160^0)(\cos210^0-\cos380^0)=1$

Ta có: $\cos310^o=\cos(360^o-50^o)=\cos(-50^o)=\cos 50^o$
            $\cos 290^o=\cos(360^o-70^o)=\cos70^o$
            $\cos160^o=\cos(180^o-20^o)=-\cos20^o=-\sin(90^o-20^o)=-\sin70^o$
            $\cos380^o=\cos(360^o+20^o)=\cos20^o=\sin70^o$
            $\cos320^o=\cos(360^o-40^o)=\cos(-40^o)=\cos 40^o=\sin50^o$

$\Rightarrow VT=(\cos70^o+\cos50^o)(\cos70^o+\cos50^o)$
                                                    $+(\sin50^o-\sin70^o)(\sin50^o-\sin70^o)$
              $=(\cos70^o+\cos50^o)^2+(\sin50^o-\sin70^o)^2$
              $=(\cos^270^o+\sin^270^o)+(\cos^250^o+\sin^250^o)$
                                                   $+2(\cos70^o\cos50^o-\sin70^o\sin50^o)$
              $=2+2\cos(50^o+70^o)$
              $=2+2\cos120^o$
              $=2-2.\frac{1}{2}=1$   (đpcm).