aangenaamyj

2022-07-02

What is the value of the expression $\left[1+\mathrm{cos}\left(\frac{\pi }{8}\right)\right]\left[1+\mathrm{cos}\left(\frac{3\pi }{8}\right)\right]\left[1+\mathrm{cos}\left(\frac{5\pi }{8}\right)\right]\left[1+\mathrm{cos}\left(\frac{7\pi }{8}\right)\right]$?

Lana Schwartz

Expert

$\left(1+\mathrm{cos}\left(\frac{\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\frac{3\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\frac{5\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\frac{7\pi }{8}\right)\right)=\phantom{\rule{0ex}{0ex}}=\left(1+\mathrm{cos}\left(\frac{\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\frac{3\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\pi -\frac{3\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\pi -\frac{\pi }{8}\right)\right)=\phantom{\rule{0ex}{0ex}}=\left(1+\mathrm{cos}\left(\frac{\pi }{8}\right)\right)\left(1+\mathrm{cos}\left(\frac{3\pi }{8}\right)\right)\left(1-\mathrm{cos}\left(\frac{3\pi }{8}\right)\right)\left(1-\mathrm{cos}\left(\frac{\pi }{8}\right)\right)=\phantom{\rule{0ex}{0ex}}=\left(1-{\mathrm{cos}}^{2}\frac{\pi }{8}\right)\left(1-{\mathrm{cos}}^{2}\frac{3\pi }{8}\right)={\mathrm{sin}}^{2}\frac{\pi }{8}{\mathrm{sin}}^{2}\frac{3\pi }{8}=\frac{\left(1-\mathrm{cos}\frac{\pi }{4}\right)}{2}\cdot \frac{\left(1-\mathrm{cos}\frac{3\pi }{4}\right)}{2}=\phantom{\rule{0ex}{0ex}}=\frac{\left(1-\mathrm{cos}\frac{\pi }{4}\right)\left(1-\mathrm{cos}\left(\pi -\frac{\pi }{4}\right)\right)}{4}=\frac{\left(1-\mathrm{cos}\frac{\pi }{4}\right)\left(1+\mathrm{cos}\frac{\pi }{4}\right)}{4}=\frac{1-{\mathrm{cos}}^{2}\frac{\pi }{4}}{4}=\frac{1-\frac{1}{2}}{4}=\frac{1}{8}$