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Exercise 3.5 · Q11

Q.Prove that 323 sin⁡π48cos⁡π48cos⁡π24cos⁡π12cos⁡π6=332\sqrt3\,\sin\dfrac{\pi}{48}\cos\dfrac{\pi}{48}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6} = 3.

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Each adjacent pair sin⁡(angle)cos⁡(angle)\sin(\text{angle})\cos(\text{angle}) collapses to 12sin⁡(2⋅angle)\tfrac12\sin(2\cdot\text{angle}); applying this four times in a row, the angle doubles each time (π48→π24→π12→π6→π3\frac\pi{48}\to\frac\pi{24}\to\frac\pi{12}\to\frac\pi6\to\frac\pi3) until only sin⁡π3\sin\frac\pi3 remains, divided by 24=162^4=16.

Step 1. Combine sin⁡π48cos⁡π48\sin\frac\pi{48}\cos\frac\pi{48}.

sin⁡π48cos⁡π48=12sin⁡2π48=12sin⁡π24.\sin\frac\pi{48}\cos\frac\pi{48}=\frac12\sin\frac{2\pi}{48}=\frac12\sin\frac\pi{24}.

Step 2. Bring in cos⁡π24\cos\frac\pi{24}.

12sin⁡π24cos⁡π24=12⋅12sin⁡2π24=14sin⁡π12.\frac12\sin\frac\pi{24}\cos\frac\pi{24}=\frac12\cdot\frac12\sin\frac{2\pi}{24}=\frac14\sin\frac\pi{12}.

Step 3. Bring in cos⁡π12\cos\frac\pi{12}.

14sin⁡π12cos⁡π12=14⋅12sin⁡2π12=18sin⁡π6.\frac14\sin\frac\pi{12}\cos\frac\pi{12}=\frac14\cdot\frac12\sin\frac{2\pi}{12}=\frac18\sin\frac\pi6.

Step 4. Bring in cos⁡π6\cos\frac\pi6. …

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