HMMT February

Note that $$\begin{aligned} & \sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(x) \\ & \quad=\left(1-\cos ^{2}(x)\right)+\left(1-\cos ^{2}(2 x)\right) \cos ^{2}(x)+\cdots+\left(1-\cos ^{2}\left(2^{2010} x\right)\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(x) \end{aligned}$$ which telescopes to $1-\cos ^{2}(x) \cos ^{2}(2 x) \cos ^{2}(4 x) \cdots \cos ^{2}\left(2^{2010} x\right)$. To evaluate $\cos ^{2}(x) \cos ^{2}(2 x) \cdots \cos ^{2}\left(2^{2010} x\right)$, multiply and divide by $\sin ^{2}(x)$. We then get $$1-\frac{\sin ^{2}(x) \cos ^{2}(x) \cos ^{2}(2 x) \cdots \cos ^{2}\left(2^{2010} x\right)}{\sin ^{2}(x)}$$ Using the double-angle formula for sin, we get that $\sin ^{2}(y) \cos ^{2}(y)=\frac{\sin ^{2}(2 y)}{4}$. Applying this 2011 times makes the above expression $$1-\frac{\sin ^{2}\left(2^{2011} x\right)}{4^{2011} \sin ^{2}(x)}$$ which is in closed form.