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
    $\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$
  • $\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$

Answer

Correct option: D.
$\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$
$\therefore\cos\alpha\cos2\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha=\frac{\sin2^​\text{n}\alpha​}{2^\text{n}\sin\alpha}$

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