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Prove that $\cos\left({\pi \over 11}\right)\cdot\cos\left({2\pi \over 11}\right)\cdot\cos\left({3\pi \over 11}\right)\cdot\cos\left({4\pi \over 11}\right)\cdot\cos\left({5\pi \over 11}\right)={1 \over 32}$

My solution starts here LHS $= \cos\left({\pi \over 11}\right)\cdot\cos\left({2\pi \over 11}\right)\cdot\cos\left({3\pi \over 11}\right)\cdot\cos\left(2\pi-{3\pi \over 11}\right)$

Then I changed the last expression to $\sin {3\pi \over 11}$ but Now what to do next?

lisyarus
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Iaamuser user
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  • I don't know if you can make use of sin 2x = 2 sin x cos x, based on your proposed simplification? – sjb Feb 03 '16 at 17:11
  • @Watson, though the questions appears to be same but I could not understand the solution given by lab bhattacharje as I am not familiar with it and the second answer is not understood by me.. – Iaamuser user Feb 03 '16 at 17:22
  • It's the same to the old post http://math.stackexchange.com/questions/1351337/product-of-cosines-prod-r-17-cos-fracr-pi15?lq=1 – mapping Feb 03 '16 at 17:24
  • @ Mapping, I could not understand the solutions there. – Iaamuser user Feb 03 '16 at 17:35

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