Question

Prove that, cosθ cos2θ cos2²θ ... cos2^n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.

Answer 1

Let P(n): cosθ cos2θ cos2²θ ... cos2^n–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 ... cos2^{k – 1}θ = `(sin 2^k theta)/(2^k sin theta)`

Let it be true for P(k).

Step 3: P(k + 1): cosθ.cos2θ.cos2²θ ... cos2^{k – 1}θ . 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.