• # question_answer $\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1+\cos \frac{5\pi }{8} \right)\,\left( 1+\cos \frac{7\pi }{8} \right)=$  [IIT 1984; WB JEE 1992] A) $\frac{1}{2}$ B) $\frac{1}{4}$ C) $\frac{1}{8}$ D) $\frac{1}{16}$

$\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1+\cos \frac{5\pi }{8} \right)\,\left( 1+\cos \frac{7\pi }{8} \right)$ $=\left( 1+\cos \frac{\pi }{8}+\cos \frac{7\pi }{8}+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)$$\left( 1+\cos \frac{5\pi }{8}+\cos \frac{3\pi }{8}+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)$ $=\left( 1+\cos \frac{\pi }{8}-\cos \frac{\pi }{8}+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)$$\left( 1+\cos \frac{5\pi }{8}-\cos \frac{5\pi }{8}+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)$ $=\left( 1+\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)$ $=\frac{1}{4}\,\,\left( 2+2\cos \frac{\pi }{8}\cos \frac{7\pi }{8} \right)\,\,\left( 2+2\cos \frac{3\pi }{8}\cos \frac{5\pi }{8} \right)$ $=\frac{1}{4}\left( 2+\cos \frac{3\pi }{4}+\cos \pi \right)\left( 2+\cos \frac{\pi }{4}+\cos \pi \right)$ $=\frac{1}{4}\,\left( 1+\cos \frac{3\pi }{4} \right)\,\left( 1+\cos \frac{\pi }{4} \right)=\frac{1}{4}\left( 1-\cos \frac{\pi }{4} \right)\,\left( 1+\cos \frac{\pi }{4} \right)$ $=\frac{1}{4}\left( 1-{{\cos }^{2}}\frac{\pi }{4} \right)=\frac{1}{4}\left( 1-\frac{1}{2} \right)=\frac{1}{8}$. Aliter:$\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{7\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1+\cos \frac{5\pi }{8} \right)$ $=\left( 1+\cos \frac{\pi }{8} \right)\,\left( 1-\cos \frac{\pi }{8} \right)\,\left( 1+\cos \frac{3\pi }{8} \right)\,\left( 1-\cos \frac{3\pi }{8} \right)$ $=\left( 1-{{\cos }^{2}}\frac{\pi }{8} \right)\text{ }\left( 1-{{\cos }^{2}}\frac{3\pi }{8} \right)={{\sin }^{2}}\frac{\pi }{8}{{\sin }^{2}}\frac{3\pi }{8}$ $=\frac{1}{4}{{\left( 2\sin \frac{\pi }{8}.\sin \frac{3\pi }{8} \right)}^{2}}$$=\frac{1}{4}{{\left( \cos \frac{\pi }{4}-\cos \frac{\pi }{2} \right)}^{2}}=\frac{1}{8}$.