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Question
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
Options
\[\frac{\sin 2n \alpha}{2n \sin \alpha}\]
- \[\frac{\sin 2^n \alpha}{2^n \sin 2^{n - 1} \alpha}\]
\[\frac{\sin 4^{n - 1} \alpha}{4^{n - 1} \sin \alpha}\]
- \[\frac{\sin 2^n \alpha}{2^n \sin \alpha}\]
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Solution
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