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Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = sin2nθ2nsinθ, for all n ∈ N.

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Question

Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.

Theorem
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Solution

Let P(n): cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, ∀ n ∈ N.

Step 1: P(1): cosθ = `(sin 2^1 theta)/(2^1 sin theta)`

= `(sin 2theta)/(2sin theta)`

= `(2sin theta cos theta)/(2sin theta)`

= cosθ

⇒ cosθ = cosθ which is true for P(1)

Step 2: P(k): cosθ.cos2θ.cos22θ ... cos2k – 1θ = `(sin 2^k theta)/(2^k sin theta)`

Let it be true for P(k).

Step 3: P(k + 1): cosθ.cos2θ.cos22θ ... cos2k – 1θ . cos`2^(("k"+1)–1`θ

= `(sin 2^k theta)/(2^k sin theta) . cos 2^((k + 1) - 1)theta`

=  `(sin 2^k theta)/(2^k sin theta) . cos 2^ktheta`

= `(2 sin 2^k theta . cos 2^k theta)/(2.2^k sin theta)`

= `(sin 2.2^k theta)/(2^(k + 1) sin theta)`  .....[∵ 2 sinθ cosθ = sin2θ]

= `(sin 2^(k + 1)theta)/(2^(k + 1) sin theta)` which is true for P(k + 1).

Hence, P(k + 1) is true whenever P(k) is true.

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Chapter 4: Principle of Mathematical Induction - Exercise [Page 71]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 4 Principle of Mathematical Induction
Exercise | Q 21 | Page 71

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