Example 5 (11th "Hope Cup" Invitation Competition Question) The range of real number $a$ that makes the inequality $2^{x}-a>\arccos x$ have the solution $-\frac{1}{2}<x \leqslant 1$ is ( ).
A. $1-\frac{\pi}{2}$
B. $\frac{\sqrt{2}}{2}-\frac{2}{3} \pi$
C. $\frac{\sqrt{2}}{2}-\frac{5}{6} \pi$
D. $\frac{1}{2}-\pi$
Solution: Choose B. Reason: From the given, we know that the solution set of the inequality $2^{x}-\arccos x>a$ is $\left(-\frac{1}{2}, 1\right]$. On this interval, the function $f(x)=2^{x}-\arccos x$ is monotonically increasing, so the value of $a$ should satisfy the relation $f\left(-\frac{1}{2}\right)=a$. Therefore, $a=2^{-\frac{1}{2}}-\arccos \left(-\frac{1}{2}\right)=\frac{\sqrt{2}}{2}-\frac{2}{3} \pi$