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Question Number 143057 by ERA last updated on 09/Jun/21

$$\cos \left(\mathit{\alpha }\right)\times \cos \left(2\alpha \right)\times \cos \left(4\alpha \right)\times ....\times \cos \left(2^{\mathrm{n}}\mathit{\alpha }\right)=\frac{\mathbf{sin}\left(2^{\mathbf{n}+1}\mathit{\alpha }\right)}{2^{\mathrm{n}+1}\sin \left(\alpha \right)}$$$$\mathbf{prove}$$

Answered by Dwaipayan Shikari last updated on 09/Jun/21

$$cos\left(\alpha \right).cos\left(2\alpha \right).cos\left(4\alpha \right)...cos\left(2^n\alpha \right)$$$$=\frac{sin\left(2\alpha \right)}{2sin\left(\alpha \right)}.\frac{sin\left(4\alpha \right)}{2sin\left(2\alpha \right)}.\frac{sin\left(8\alpha \right)}{2sin\left(4\alpha \right)}...\frac{sin\left(2^{n+1}\alpha \right)}{2sin\left(2^n\alpha \right)}$$$$=\frac{sin\left(2^{n+1}\alpha \right)}{2^{n}sin\left(\alpha \right)}$$

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