4. [3] Suppose that $a, b, c, d$ are real numbers satisfying $a \geq b \geq c \geq d \geq 0, a^{2}+d^{2}=1, b^{2}+c^{2}=1$, and $a c+b d=1 / 3$. Find the value of $a b-c d$.
Answer: $\frac{2 \sqrt{2}}{3}$ We have
$$
(a b-c d)^{2}=\left(a^{2}+d^{2}\right)\left(b^{2}+c^{2}\right)-(a c+b d)^{2}=(1)(1)-\left(\frac{1}{3}\right)^{2}=\frac{8}{9} .
$$

Since $a \geq b \geq c \geq d \geq 0, a b-c d \geq 0$, so $a b-c d=\frac{2 \sqrt{2}}{3}$.
Comment: Another way to solve this problem is to use the trigonometric substitutions $a=\sin \theta$, $b=\sin \phi, c=\cos \phi, d=\cos \theta$