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  • Prove that, cos θ cos 2θ cos22θ ….. cos 2n–1θ for all n ϵ N.

Prove that, \cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}, for all n ϵ N.

Answers (1)

Given:

\cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}

Now, we’ll substitute different values for n,

\\\begin{aligned} n=1 \\ \cos \theta= 2 \sin \theta \cos \theta / 2 \sin \theta \\ &=\sin 2^{1} \theta / 2^{1} \sin \theta \end{aligned} \\ \: \: Thus, \: \: it's\: \: true. \\Now, n=2

\begin{aligned} \cos \theta \cos 2 \theta=& \sin 2^{2} \theta / 2^{2} \sin \theta \\ &=2 \sin 2 \theta \cos 2 \theta / 4 \sin \theta \\ &=4 \sin \theta \cos \theta \cos 2 \theta / 4 \sin \theta \end{aligned} $$\\ Thus,\: \: it\: \: is\: \: true.

n=3\\ \begin{array}{l} \cos \theta \cos 2 \theta \cos 4 \theta=\sin 2^{3} \theta / 2^{3} \sin \theta \\ =2 \sin 4 \theta \cos 4 \theta / 8 \sin \theta \\ =4 \sin 2 \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta \\ =8 \sin \theta \cos \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta \end{array}

Now, let us assume that, 

n=k,\\ \cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta=\sin 2^{k} \theta / 2^{k} \sin \theta\ to \ be \ true.\\ Thus, at \: \: n=k+1,\\ \begin{aligned} \cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta \cos 2^{k+1-1} \theta &=\sin 2^{k} \theta / 2^{k} \sin \theta \cdot \cos 2^{k} \theta \\ &=2 \sin 2^{k} \theta \cos 2^{k} \theta / 2^{k+1} \sin \theta \\ &=\sin 2^{k+1} \theta / 2^{k+1} \sin \theta \end{aligned}

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

\cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}

Posted by

infoexpert21

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