prove that $q_n = p_n \, \sin(\frac{t}{2^{n-1}})$ is a geometric progression

Question

Given $$p_n = \prod_{n}^{p=1} \cos\left(\frac{t}{2^{n-1}}\right) \hspace{3mm} \text{and} \hspace{3mm} q_n = p_n \, \sin\left(\frac{t}{2^{n-1}}\right).$$ How to prove that $(q_n)_{n}$ is a geometric progression with $\frac{1}{2}$ as ratio ?

Answer 1

Hint:

$$\sin a=2\cos\frac a2\sin\frac a2=4\cos\frac a2\cos\frac a4\sin\frac a4=8\cos\frac a2\cos\frac a4\cos\frac a8\sin\frac a8=\cdots$$

Answer 2

Hint: Compute the Quotient $$\frac{q_{n+1}}{q_n}=\frac{p_{n+1}\sin\left(\frac{t}{2^n}\right)}{p_n\sin\left(\frac{t}{2^{n-1}}\right)}$$