∑7r=0tan2(πr16)=
35
I=∑7r=0tan2πr16)=tan2π16+tan22π16+tan23π16+tan24π16+tan25π16+tan26π16+tan27π16=(tan2π16+cot2π16)+(tan22π16+cot22π16)+(tan23π16+cot23π16)+1=(tanπ16+cotπ16)2−2+(tan2π16+cot2π16)2−2+(tan3π16+cot3π16)2−2+1=1sin2π16cot2π16+1sin22π16cos22π16+1sin23π16cot23π16−5.
=4[1sin2π8+1sin24π16+1sin26π16]−5=4[1sin2π8+1cos2π8]+3=4.4sin2π4+3=35.