Ta có sin100=cos800(vì 100+800=900)⇒sin2100=cos2800
sin200=cos700(vì 200+700=900)⇒sin2200=cos2700
Ta có công thức sin2a+cos2a=1
\(P=cos^210^0+cos^220^0+cos^270^0+cos^280^0=cos^210^0+cos^220^0+sin^220^0+sin^210^0=\left(cos^210^0+sin^210^0\right)+\left(cos^220^0+sin^220^0\right)=1+1=2\)