7. $[\mathbf{3 0}]$ Evaluate
$$
\sum_{k=1}^{2010} \cos ^{2}(k)
$$
Answer: $1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$ We use the identity $\cos ^{2}(k)=\frac{1+\cos (2 x)}{2}$. Then our expression evalutes to $1005+\frac{(\cos (2)+\ldots+\cos (4020))}{2}$.
To evaluate $\cos (2)+\cdots+\cos (4020)$, let $y=\cos (2)+\cdots+\cos (4020) \Rightarrow y(\sin (1))=\cos (2) \sin (1)+$ $\ldots+\cos (4020) \sin (1)$. Observe that for any $x, \cos (x) \sin (1)=\frac{\sin (x+1)-\sin (x-1)}{2}$. Then $y(\sin (1))=$ $\frac{\sin (3)-\sin (1)}{2}+\frac{\sin (5)-\sin (3)}{2}+\ldots+\frac{\sin 4021-\sin (4019)}{2}$. This is clearly a telescoping sum; we get $y(\sin (1))=$ $\frac{\sin (4021)-\sin (1)}{2}$. Then we have the desired $y=\frac{\sin (4021)-\sin (1)}{2 \sin (1)}$. Then our original expression evaluates to $1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$.