Compute $\cos 72^\circ.$
Let $a = \cos 36^\circ$ and $b = \cos 72^\circ.$  Then by the double angle formula,
\[b = 2a^2 - 1.\]Also, $\cos (2 \cdot 72^\circ) = \cos 144^\circ = -\cos 36^\circ,$ so
\[-a = 2b^2 - 1.\]Subtracting these equations, we get
\[a + b = 2a^2 - 2b^2 = 2(a - b)(a + b).\]Since $a$ and $b$ are positive, $a + b$ is nonzero.  Hence, we can divide both sides by $2(a + b),$ to get
\[a - b = \frac{1}{2}.\]Then $a = b + \frac{1}{2}.$  Substituting into $-a = 2b^2 - 1,$ we get
\[-b - \frac{1}{2} = 2b^2 - 1.\]Then $-2b - 1 = 4b^2 - 2,$ or $4b^2 + 2b - 1 = 0.$  By the quadratic formula,
\[b = \frac{-1 \pm \sqrt{5}}{4}.\]Since $b = \cos 72^\circ$ is positive, $b = \boxed{\frac{-1 + \sqrt{5}}{4}}.$
\frac{-1+\sqrt{5}}{4}
Level 2
Precalculus
Problem: Compute $\cos 72^\circ.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To compute $\cos 72^\circ$, recognize that $72^\circ$ is related to the angles in a regular pentagon, where each interior angle is $108^\circ$ and the exterior angle is $72^\circ$. The cosine of $72^\circ$ can be derived from the relationship between the angles in a pentagon and the properties of the golden ratio.

The golden ratio, denoted as $\phi$, is defined as $\phi = \frac{1 + \sqrt{5}}{2}$. The cosine of $72^\circ$ can be expressed in terms of the golden ratio. Specifically, it is known that $\cos 72^\circ = \frac{\phi - 1}{2}$. 

Substituting the expression for $\phi$, we have:
$$\cos 72^\circ = \frac{\frac{1 + \sqrt{5}}{2} - 1}{2}.$$

Simplifying this expression:
1. Start with $\frac{1 + \sqrt{5}}{2} - 1 = \frac{1 + \sqrt{5} - 2}{2} = \frac{\sqrt{5} - 1}{2}$.
2. Now divide by 2: 
$$\cos 72^\circ = \frac{\sqrt{5} - 1}{4}.$$

Rearranging gives:
$$\cos 72^\circ = \frac{-1 + \sqrt{5}}{4}.$$

Thus, the final answer is $\frac{-1+\sqrt{5}}{4}$.<|END_OF_LATENT|> The final answer is $\frac{-1+\sqrt{5}}{4}$.