jmc

By the Trivial Inequality, $(x-y{)}^2\ge 0$ for all real numbers $x$ and $y.$ We can re-arrange this as $$xy\le \frac{x^2+y^2}{2}.$$(This looks like AM-GM, but we need to establish it for all real numbers, not just nonnegative numbers.)

Hence, $$\begin{array}{rl}& \cos {\theta }_1\sin {\theta }_2+\cos {\theta }_2\sin {\theta }_3+\cos {\theta }_3\sin {\theta }_4+\cos {\theta }_4\sin {\theta }_5+\cos {\theta }_5\sin {\theta }_1\\ & \le \frac{{\cos }^2{\theta }_1+{\sin }^2{\theta }_2}{2}+\frac{{\cos }^2{\theta }_2+{\sin }^2{\theta }_3}{2}\\ & +\frac{{\cos }^2{\theta }_3+{\sin }^2{\theta }_4}{2}+\frac{{\cos }^2{\theta }_4+{\sin }^2{\theta }_5}{2}+\frac{{\cos }^2{\theta }_5+{\sin }^2{\theta }_1}{2}\\ & =\frac{{\cos }^2{\theta }_1+{\sin }^2{\theta }_1}{2}+\frac{{\cos }^2{\theta }_2+{\sin }^2{\theta }_2}{2}\\ & +\frac{{\cos }^2{\theta }_3+{\sin }^2{\theta }_3}{2}+\frac{{\cos }^2{\theta }_4+{\sin }^2{\theta }_4}{2}+\frac{{\cos }^2{\theta }_5+{\sin }^2{\theta }_5}{2}\\ & =\frac{5}{2}.\end{array}$$Equality occurs when all the ${\theta }_i$ are equal to $45^{\circ },$ so the maximum value is $\boxed{\frac{5}{2}}.$