Trigonometry & Inverse Trigonometry
Trigonometric Products
Grade 11
Question:
<p>The value of \(\cos\left(\frac{\pi}{2^2}\right)\cdot\cos\left(\frac{\pi}{2^3}\right)\cdots\cos\left(\frac{\pi}{2^{10}}\right)\cdot\sin\left(\frac{\pi}{2^{10}}\right)\) is:</p>
<p>\(\dfrac{1}{512}\)</p>
<p>\(\dfrac{1}{1024}\)</p>
<p>\(\dfrac{1}{256}\)</p>
<p>\(\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the product-to-sum identity sin(2x) = 2sin(x)cos(x) repeatedly by multiplying and dividing by sin(π/2²), telescoping the product into a single sine term at the highest power.
<p><strong>Step 1:</strong> Let P = cos(π/2²)·cos(π/2³)···cos(π/2¹⁰)·sin(π/2¹⁰). Multiply and divide by sin(π/2²).</p><p><strong>Step 2:</strong> Apply sin(2x) = 2sin(x)cos(x) repeatedly. We have 2sin(π/2²)cos(π/2²) = sin(π/2¹), so cos(π/2²) = sin(π/2¹)/(2sin(π/2²)).</p><p><strong>Step 3:</strong> Continuing this telescoping pattern:</p><p>P = [sin(π/2¹)/(2sin(π/2²))] · [sin(π/2²)/(2sin(π/2³))] · ... · [sin(π/2⁹)/(2sin(π/2¹⁰))] · sin(π/2¹⁰)</p><p><strong>Step 4:</strong> All intermediate terms cancel (telescoping), leaving:</p><p>P = sin(π/2¹)/(2⁹·sin(π/2²)) · sin(π/2²)/(2⁸) · ... · sin(π/2¹⁰)/sin(π/2¹⁰)</p><p><strong>Step 5:</strong> After telescoping: P = sin(π/2)/(2⁹) = 1/512 or equivalently sin(π/4)/(2⁸) = (1/√2)/256 = 1/(256√2) = √2/512</p><p>∴ Answer: A</p>
Correct Answer: A