### Instruction: Compute $\cos 72^\circ.$.### Response: One possible way to solve this problem is to use trigonometric identities and the angle sum formula.
I know that $\cos 72^\circ = \cos (36^\circ + 36^\circ) = \cos 36^\circ \cos 36^\circ - \sin 36^\circ \sin 36^\circ.$
Now, if I can find the values of $\cos 36^\circ$ and $\sin 36^\circ$, I can plug them in and simplify.
I recall that there is a special formula for $\cos 36^\circ$ that involves the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}.$
Specifically, $\cos 36^\circ = \frac{\phi}{2}.$
To find $\sin 36^\circ$, I can use the Pythagorean identity $\sin^2 36^\circ + \cos^2 36^\circ = 1.$
Substituting $\cos 36^\circ = \frac{\phi}{2}$, I get $\sin^2 36^\circ + \frac{\phi^2}{4} = 1.$
Solving for $\sin 36^\circ$, I get $\sin 36^\circ = \pm \sqrt{\frac{4 - \phi^2}{4}}.$
Since $36^\circ$ is in the first quadrant, I take the positive root, so $\sin 36^\circ = \sqrt{\frac{4 - \phi^2}{4}}.$
Now, I can plug in these values into the original expression for $\cos 72^\circ$ and simplify.
I get $\cos 72^\circ = \frac{\phi^2}{4} - \frac{4 - \phi^2}{4} = \frac{\phi^2 - 4 + \phi^2}{4} = \frac{2\phi^2 - 4}{4}.$
I can factor out a $\frac{1}{2}$ and use the fact that $\phi^2 = \phi + 1$ to simplify further.
I get $\cos 72^\circ = \frac{1}{2} (\phi^2 - 2) = \frac{1}{2} (\phi + 1 - 2) = \frac{1}{2} (\phi - 1).$
Finally, I can substitute $\phi = \frac{1 + \sqrt{5}}{2}$ and get $\cos 72^\circ = \frac{1}{2} \left( \frac{1 + \sqrt{5}}{2} - 1 \right) = \frac{1}{4} (\sqrt{5} - 1).$