First slide

Introduction to P.M.I

Question

By principle of mathematical induction $\cos θcos 2 θcos 4 θ \dots \cos (2^{n - 1} θ), \forall n \in N =$

Easy

A

$\frac{\sin 2^{n} θ}{2^{n} \sin θ}$
B

$\frac{\cos 2^{n} θ}{2^{n} \sin θ}$
C

$\frac{\sin 2^{n} θ}{2^{n - 1} \sin θ}$
D

None of these

Solution

Let $P (n) : \cos θcos 2 θcos 4 θ \dots \cos (2^{n - 1} θ) = \frac{\sin 2^{n} θ}{2^{n} \sin θ} \dots (i)$
Step l :For n=1,
$LHS = \cos θ and RHS = \frac{\sin 2 θ}{2 \sin θ} = \cos θ$ P(1 ) is true.
Step ll :Let P(n) is true, then
$P (k) : \cos θcos 2 θcos 4 θ \dots \cos (2^{k - 1} θ) = \frac{\sin 2^{k} θ}{2^{k} \sin θ}$
Step III: For n=k+1
$\begin{array}{l} P (k + 1) : \cos θcos 2 θ \dots \cos (2^{k} θ) = \frac{\sin 2^{k + 1} θ}{2^{k + 1} \sin θ} \\ LHS = \cos θcos 2 θ \dots \cos 2^{(k - 1)} θcos 2^{k} θ \\ = \frac{\sin (2^{k} θ)}{2^{k} \sin θ} \cdot \cos (2^{k} θ) \\ = \frac{2 \sin (2^{k} θ) \cdot \cos (2^{k} θ)}{2^{k + 1} \sin θ} \\ = \frac{\sin 2^{k + 1} θ}{2^{k + 1} \sin θ} = RHS \end{array}$
For n =k + 1, P(n) is true.
Hence, by principle for mathematical induction for all n $\in$ N,P(n) is true.

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