cos⁡Acos⁡2Acos⁡4A……cos⁡2n−1A=

=cos⁡Acos⁡2Acos⁡4A…cos⁡2n−1A =12sin⁡A(2sin⁡Acos⁡A)cos⁡2Acos⁡4A…cos⁡2n−1A=12sin⁡A(sin⁡2Acos⁡2A)cos⁡4A…cos⁡2n−1A=122sin⁡A(sin⁡4Acos⁡4A)…cos⁡2n−1A=123sin⁡Asin⁡8A…cos⁡2n−1A=12n−1sin⁡Asin⁡2n−1Acos⁡2n−1A=12nsin⁡Asin⁡2nA.

cos⁡Acos⁡2Acos⁡4A……cos⁡2n−1A=

=cos⁡Acos⁡2Acos⁡4A…cos⁡2n−1A =12sin⁡A(2sin⁡Acos⁡A)cos⁡2Acos⁡4A…cos⁡2n−1A=12sin⁡A(sin⁡2Acos⁡2A)cos⁡4A…cos⁡2n−1A=122sin⁡A(sin⁡4Acos⁡4A)…cos⁡2n−1A=123sin⁡Asin⁡8A…cos⁡2n−1A=12n−1sin⁡Asin⁡2n−1Acos⁡2n−1A=12nsin⁡Asin⁡2nA.

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$\cos Acos 2 Acos 4 A \dots \dots \cos 2^{n - 1} A =$

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$\frac{\sin 2^{n} A}{2^{n} \sin A}$

$\frac{2^{n} \sin 2^{n} A}{\sin A}$

$\frac{2^{n} \sin A}{\sin 2^{n} A}$

$\frac{\sin A}{2^{n} \sin 2^{n} A}$

answer is A.

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Detailed Solution

$= \cos Acos 2 Acos 4 A \dots \cos 2^{n - 1} A \begin{array}{l} = \frac{1}{2 \sin A} (2 \sin Acos A) \cos 2 Acos 4 A \dots \cos 2^{n - 1} A \\ = \frac{1}{2 \sin A} (\sin 2 Acos 2 A) \cos 4 A \dots \cos 2^{n - 1} A = \frac{1}{2^{2} \sin A} (\sin 4 Acos 4 A) \dots \cos 2^{n - 1} A \\ = \frac{1}{2^{3} \sin A} \sin 8 A \dots \cos 2^{n - 1} A = \frac{1}{2^{n - 1} \sin A} \sin 2^{n - 1} Acos 2^{n - 1} A = \frac{1}{2^{n} \sin A} \sin 2^{n} A . \end{array}$