1. Considering the triple angle sine and cosine formulas $\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta, \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$, from the perspective of $\frac{\cos 3 \theta}{\cos \theta}$ $=4 \cos ^{2} \theta-3, \frac{\sin 3 \theta}{\sin \theta}=3-4 \sin ^{2} \theta$, what kind of trigonometric problems can you construct?
1. Since $\frac{\cos 3^{k+1} \theta}{\cos 3^{k} \theta}=4 \cos ^{2} 3^{k} \theta-3$, then $\prod_{k=0}^{n-1}\left(4 \cos ^{2} 3^{k} \theta-3\right)=\frac{\cos 3^{n} \theta}{\cos \theta}$. Also, since $\frac{\sin 3^{k+1} \theta}{\sin 3^{k} \theta}=3-4 \sin ^{2} 3^{k} \theta$, then $\prod_{k=0}^{n-1}\left(3-4 \sin ^{2} 3^{k} \theta\right)=\frac{\sin 3^{n} \theta}{\sin \theta}$.
If let $\theta=9^{\circ}, n=2$, then $\left(4 \cos ^{2} 9^{\circ}-3\right)\left(4 \cos ^{2} 27^{\circ}-3\right)=\tan 9^{\circ}$.
If let $\theta=9^{\circ}, n=4$, then $\left(4 \cos ^{2} 9^{\circ}-3\right)\left(4 \cos ^{2} 27^{\circ}-3\right)\left(4 \cos ^{2} 81^{\circ}-3\right)\left(4 \cos ^{2} 243^{\circ}-3\right)=1$.