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