MCQ
If n = 1,2,3, ..., then $\cos\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha$ is equal to:
- A$\frac{\sin2\text{n}\alpha}{2\text{n}\sin\alpha}$
- B$\frac{\sin2^\text{n}\alpha}{2^\text{n}\sin2^{\text{n}-1}\alpha}$
- C$\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$
- D$\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$