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Question Number 18003 by Tinkutara last updated on 13/Jul/17

$$\mathrm{The}\mathrm{value}\mathrm{of}\cos A\cdot \cos 2A\cdot {\cos 2}^{2}A.....\cos \left(2^{n-1}A\right),$$$$\mathrm{where}A\in R\mathrm{may}\mathrm{be}$$$$\left(1\right)1$$$$\left(2\right)2$$$$\left(3\right)-1$$$$\left(4\right)\frac{\sin 2^{n}A}{2^n\sin A}$$

Answered by alex041103 last updated on 13/Jul/17

$$LetE=cos\left(A\right)cos\left(2A\right)...=\underset{k=0}{\prod ^{n-1}}cos\left(2^{k}A\right)$$$$Becausesin\left(2^{k+1}A\right)=2sin\left(2^{k}A\right)cos\left(2^{k}A\right)$$$$\Rightarrow cos\left(2^{k}A\right)=\frac{sin\left(2^{k+1}A\right)}{2sin\left(2^{k}A\right)}$$$$\Rightarrow E=\underset{k=0}{\prod ^{n-1}}\frac{sin\left(2^{k+1}A\right)}{2sin\left(2^{k}A\right)}=$$$$=\left[\underset{k=0}{\prod ^{n-1}}sin\left(2^{k+1}A\right)\right]\times \left[\underset{k=0}{\prod ^{n-1}}\frac{1}{2sin\left(2^{k}A\right)}\right]$$$$=\frac{sin\left(2^{n}A\right)}{2sin\left(A\right)}\left[\underset{k=0}{\prod ^{n-2}}sin\left(2^{k+1}A\right)\right]\left[\underset{k=1}{\prod ^{n-1}}\frac{1}{2sin\left(2^{k}A\right)}\right]$$$$=\frac{sin\left(2^{n}A\right)}{2sin\left(A\right)}\left[\underset{k=1}{\prod ^{n-1}}sin\left(2^{k}A\right)\right]\left[\underset{k=1}{\prod ^{n-1}}\frac{1}{2sin\left(2^{k}A\right)}\right]$$$$=\frac{sin\left(2^{n}A\right)}{2sin\left(A\right)}\left[\underset{k=1}{\prod ^{n-1}}\frac{sin\left(2^{k}A\right)}{2sin\left(2^{k}A\right)}\right]$$$$=\frac{sin\left(2^{n}A\right)}{2sin\left(A\right)}\left[\underset{k=1}{\prod ^{n-1}}\frac{1}{2}\right]$$$$=\frac{sin\left(2^{n}A\right)}{2sin\left(A\right)}\times \frac{1}{2^{n-1}}$$$$=\frac{sun\left(2^{n}A\right)}{2^{n}sin\left(A\right)}=cos\left(A\right)cos\left(2A\right)...$$

Commented by Tinkutara last updated on 13/Jul/17

$$\mathrm{But}\mathrm{which}\mathrm{options}\mathrm{are}\mathrm{correct}?\mathrm{It}\mathrm{can}$$$$\mathrm{have}\mathrm{more}\mathrm{than}\mathrm{one}\mathrm{option}\left(\mathrm{s}\right).$$

Commented by alex041103 last updated on 13/Jul/17

$${you}^{\prime }reright$$$$itcanbe1whenA=2\pi$$$$and-1forA=\pi andn=1$$$$but{i}^{\prime }mstillthinkingaboutthe2$$

Commented by Tinkutara last updated on 13/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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