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Miscellaneous Exercise 3 (II) · Q91

Q.Prove the following: cos⁡2π15cos⁡4π15cos⁡8π15cos⁡16π15=116\cos\dfrac{2\pi}{15}\cos\dfrac{4\pi}{15}\cos\dfrac{8\pi}{15}\cos\dfrac{16\pi}{15}=\dfrac{1}{16}

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✓ Free question

Step 1: Using cos⁡Acos⁡2Acos⁡4Acos⁡8A=sin⁡16A16sin⁡A\cos A\cos2A\cos4A\cos8A=\dfrac{\sin16A}{16\sin A} with A=2π15A=\dfrac{2\pi}{15}: the product =sin⁡32π1516sin⁡2π15=\dfrac{\sin\frac{32\pi}{15}}{16\sin\frac{2\pi}{15}}.

Step 2: sin⁡32π15=sin⁡(32π15−2π)=sin⁡2π15\sin\dfrac{32\pi}{15}=\sin\left(\dfrac{32\pi}{15}-2\pi\right)=\sin\dfrac{2\pi}{15}.

Step 3: So the product =sin⁡2π1516sin⁡2π15=116=\dfrac{\sin\frac{2\pi}{15}}{16\sin\frac{2\pi}{15}}=\dfrac{1}{16}.

✓Final answer

cos⁡2π15cos⁡4π15cos⁡8π15cos⁡16π15=116\cos\dfrac{2\pi}{15}\cos\dfrac{4\pi}{15}\cos\dfrac{8\pi}{15}\cos\dfrac{16\pi}{15}=\dfrac{1}{16}

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