Question Number 88503 by jagoll last updated on 11/Apr/20
$$\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$
Commented by john santu last updated on 11/Apr/20
$$\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}=\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{8}}\right)=\:−\mathrm{cos}\:\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\:=−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)=\:\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}=\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$\therefore\:\left[\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}.\mathrm{sin}\:\frac{\pi}{\mathrm{8}}\right]^{\mathrm{2}} =\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{8}}−\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\right]^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}}\: \\ $$
Commented by peter frank last updated on 11/Apr/20
$${thank}\:{you} \\ $$
Commented by peter frank last updated on 15/Apr/20
$${help}\:{Qn}\:\mathrm{88937} \\ $$