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Mathematics Question on Trigonometric Functions

If n = 1 , 2, 3, ..... , then cos⁡ α cos⁡ 2α cos⁡ 22α cos⁡ 23 α......cos⁡ 2n−1α\cos \, \alpha \, \cos \, 2\alpha \, \cos \, 2^2 \alpha \, \cos \, 2^3 \, \alpha . ..... \cos \, 2^{n-1} \alpha is equal to

A

sin⁡ 2n α2n sin⁡α\frac{\sin \, 2n \, \alpha}{ 2 n \, \sin \alpha}

B

sin⁡ 2n α2n sin⁡ 2n−1 α\frac{\sin \, 2^n \, \alpha}{2^n \, \sin \, 2^{n -1} \,\alpha}

C

sin⁡ 4n−1 α4n−1sin⁡ α\frac{\sin \, 4^{n -1} \, \alpha}{4^{n -1} \sin \,\alpha}

D

sin⁡ 2n α2n sin⁡ α\frac{\sin \, 2^n \, \alpha}{2^n \, \sin \, \alpha}

Answer

sin⁡ 2n α2n sin⁡ α\frac{\sin \, 2^n \, \alpha}{2^n \, \sin \, \alpha}

Explanation

Solution

cos⁡αcos⁡2αcos⁡22α....cos⁡2n−1α\cos\alpha \cos 2 \alpha \cos 2^{2} \alpha .... \cos 2^{n-1} \alpha =2nsin⁡αcos⁡αcos⁡2αcos⁡22α...cos⁡2n−1α2nsin⁡α= \frac{2^{n} \sin\alpha \cos\alpha\cos 2 \alpha \cos 2^{2}\alpha ... \cos 2^{n-1} \alpha }{2^{n} \sin\alpha } = \frac{\sin \left\\{2\left(2^{n-1} \alpha \right)\right\\}}{2^{n}\sin\alpha } (using 2sin⁡θcos⁡θ=sin⁡2θagain and again) \left(\text{using} \, 2 \sin\theta \cos \theta =\sin2 \theta \text{again} \, \text{and} \, \text{again} \right)