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Miscellaneous Exercise 3 (II) · Q92

Q.Prove the following: (1+cos⁡π8)(1+cos⁡3π8)(1+cos⁡5π8)(1+cos⁡7π8)=18\left(1+\cos\dfrac{\pi}{8}\right)\left(1+\cos\dfrac{3\pi}{8}\right)\left(1+\cos\dfrac{5\pi}{8}\right)\left(1+\cos\dfrac{7\pi}{8}\right)=\dfrac{1}{8}

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Step 1: Since cos⁡5π8=cos⁡(π−3π8)=−cos⁡3π8\cos\dfrac{5\pi}{8}=\cos\left(\pi-\dfrac{3\pi}{8}\right)=-\cos\dfrac{3\pi}{8} and cos⁡7π8=−cos⁡π8\cos\dfrac{7\pi}{8}=-\cos\dfrac{\pi}{8}, pair terms: (1+cos⁡π8)(1−cos⁡π8)(1+cos⁡3π8)(1−cos⁡3π8)\left(1+\cos\dfrac{\pi}{8}\right)\left(1-\cos\dfrac{\pi}{8}\right)\left(1+\cos\dfrac{3\pi}{8}\right)\left(1-\cos\dfrac{3\pi}{8}\right).

Step 2: =(1−cos⁡2π8)(1−cos⁡23π8)=sin⁡2π8sin⁡23π8=\left(1-\cos^2\dfrac{\pi}{8}\right)\left(1-\cos^2\dfrac{3\pi}{8}\right)=\sin^2\dfrac{\pi}{8}\sin^2\dfrac{3\pi}{8}. …

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