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Question: (4\cot^2 9^\circ -1)(4\cos^2 27^\circ -1)(4\cos^2 81^\circ -1)(4\cos^2 243^\circ -1)...

(4\cot^2 9^\circ -1)(4\cos^2 27^\circ -1)(4\cos^2 81^\circ -1)(4\cos^2 243^\circ -1)

A

1

B

-1

C

2

D

None of these

Answer

1

Explanation

Solution

The problem asks for the value of the expression (4cot⁡29∘−1)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)(4\cot^2 9^\circ -1)(4\cos^2 27^\circ -1)(4\cos^2 81^\circ -1)(4\cos^2 243^\circ -1).

Observation and Assumption:

The first term in the given expression is (4cot⁡29∘−1)(4\cot^2 9^\circ -1). However, the subsequent terms are of the form (4cos⁡2θ−1)(4\cos^2 \theta -1). A very similar problem (provided as "similar question") has all terms of the form (4cos⁡2θ−1)(4\cos^2 \theta -1). It is highly probable that there is a typo in the first term of the given question, and it should be (4cos⁡29∘−1)(4\cos^2 9^\circ -1) instead of (4cot⁡29∘−1)(4\cot^2 9^\circ -1). If we proceed with (4cot⁡29∘−1)(4\cot^2 9^\circ -1), the expression becomes significantly more complex and does not follow the elegant pattern observed in such problems. Therefore, we will assume the first term is (4cos⁡29∘−1)(4\cos^2 9^\circ -1).

Core Identity:

We use the trigonometric identity: 4cos⁡2θ−1=sin⁡3θsin⁡θ4\cos^2\theta - 1 = \frac{\sin 3\theta}{\sin \theta}

Proof of the Identity:

We know that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta. Dividing by sin⁡θ\sin\theta (assuming sin⁡θ≠0\sin\theta \neq 0): sin⁡3θsin⁡θ=3sin⁡θ−4sin⁡3θsin⁡θ=3−4sin⁡2θ\frac{\sin 3\theta}{\sin \theta} = \frac{3\sin\theta - 4\sin^3\theta}{\sin\theta} = 3 - 4\sin^2\theta Using the identity sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta: 3−4(1−cos⁡2θ)=3−4+4cos⁡2θ=4cos⁡2θ−13 - 4(1 - \cos^2\theta) = 3 - 4 + 4\cos^2\theta = 4\cos^2\theta - 1. Thus, the identity 4cos⁡2θ−1=sin⁡3θsin⁡θ4\cos^2\theta - 1 = \frac{\sin 3\theta}{\sin \theta} is verified.

Applying the Identity to Each Term:

Now, let's apply this identity to each term in the product, assuming the corrected first term:

  1. For θ=9∘\theta = 9^\circ: 4cos⁡29∘−1=sin⁡(3×9∘)sin⁡9∘=sin⁡27∘sin⁡9∘4\cos^2 9^\circ - 1 = \frac{\sin(3 \times 9^\circ)}{\sin 9^\circ} = \frac{\sin 27^\circ}{\sin 9^\circ}

  2. For θ=27∘\theta = 27^\circ: 4cos⁡227∘−1=sin⁡(3×27∘)sin⁡27∘=sin⁡81∘sin⁡27∘4\cos^2 27^\circ - 1 = \frac{\sin(3 \times 27^\circ)}{\sin 27^\circ} = \frac{\sin 81^\circ}{\sin 27^\circ}

  3. For θ=81∘\theta = 81^\circ: 4cos⁡281∘−1=sin⁡(3×81∘)sin⁡81∘=sin⁡243∘sin⁡81∘4\cos^2 81^\circ - 1 = \frac{\sin(3 \times 81^\circ)}{\sin 81^\circ} = \frac{\sin 243^\circ}{\sin 81^\circ}

  4. For θ=243∘\theta = 243^\circ: 4cos⁡2243∘−1=sin⁡(3×243∘)sin⁡243∘=sin⁡729∘sin⁡243∘4\cos^2 243^\circ - 1 = \frac{\sin(3 \times 243^\circ)}{\sin 243^\circ} = \frac{\sin 729^\circ}{\sin 243^\circ}

Multiplying the Terms (Telescoping Product):

Let P be the product of these terms: P=(sin⁡27∘sin⁡9∘)×(sin⁡81∘sin⁡27∘)×(sin⁡243∘sin⁡81∘)×(sin⁡729∘sin⁡243∘)P = \left(\frac{\sin 27^\circ}{\sin 9^\circ}\right) \times \left(\frac{\sin 81^\circ}{\sin 27^\circ}\right) \times \left(\frac{\sin 243^\circ}{\sin 81^\circ}\right) \times \left(\frac{\sin 729^\circ}{\sin 243^\circ}\right) Notice that this is a telescoping product, where the numerator of each term cancels with the denominator of the next term: P=sin⁡729∘sin⁡9∘P = \frac{\sin 729^\circ}{\sin 9^\circ}

Simplifying the Result:

We need to simplify sin⁡729∘\sin 729^\circ. We know that sin⁡(n⋅360∘+θ)=sin⁡θ\sin(n \cdot 360^\circ + \theta) = \sin\theta for any integer nn. 729∘=2×360∘+9∘=720∘+9∘729^\circ = 2 \times 360^\circ + 9^\circ = 720^\circ + 9^\circ. So, sin⁡729∘=sin⁡(720∘+9∘)=sin⁡9∘\sin 729^\circ = \sin(720^\circ + 9^\circ) = \sin 9^\circ.

Substituting this back into the expression for P: P=sin⁡9∘sin⁡9∘=1P = \frac{\sin 9^\circ}{\sin 9^\circ} = 1

The final answer is 1\boxed{1}.