A right-angled triangle has legs of length 1 and $\sqrt{2}$. The smallest angle of the triangle is $\alpha$. What is the exact value of $\cos 8 \alpha$?
Solution. Let's calculate the length of the third side of the right triangle:

$$
c=\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3}
$$

Since $1<\sqrt{2}<\sqrt{3}$, the smallest side of the triangle is 1, and the angle opposite to it is $\alpha$, thus

$$
\cos \alpha=\sqrt{\frac{2}{3}}
$$

According to the known identity:

$$
\cos 2 x=\cos ^{2} x-\sin ^{2} x=2 \cos ^{2} x-1
$$

In our case, $\cos 2 \alpha=2 \cdot \frac{2}{3}-1=\frac{1}{3}$. Using the identity repeatedly, we get $\cos 4 \alpha=-\frac{7}{9}$, and $\cos 8 \alpha=\frac{17}{81}$.