Question Number 224966 by Ismoiljon_008 last updated on 14/Oct/25
$$ \\ $$$$\:\:\:{cos}\frac{\pi}{\mathrm{7}}\:−\:{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\:{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}\:=\:? \\ $$$$\:\:\:{Help}\:{me}\:{please} \\ $$$$ \\ $$
Commented by fantastic last updated on 14/Oct/25
$$\mathrm{0}.\mathrm{5} \\ $$
Commented by Ismoiljon_008 last updated on 14/Oct/25
$$\:{how}? \\ $$
Answered by Frix last updated on 14/Oct/25
![cos (π/7) −cos ((2π)/7) +cos ((3π)/7) =x (cos (π/7) −cos ((2π)/7) +cos ((3π)/7))^2 =x^2 expand and use formulas cos^2 α =(1/2)(1+cos 2α) cos α cos β =(1/2)(cos (α−β) +cos (α+β)) to get (3/2)−(5/2)cos (π/7) +(5/2)cos ((2π)/7) −(5/2)cos ((3π)/7) =x^2 cos (π/7) −cos ((2π)/7) +cos ((3π)/7) =((3−2x^2 )/5) x=((3−2x^2 )/5) x^2 +((5x)/2)−(3/2)=0 x=−3 [rejected] x=(1/2)](https://www.tinkutara.com/question/Q224975.png)
$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:={x} \\ $$$$\left(\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{expand}\:\mathrm{and}\:\mathrm{use}\:\mathrm{formulas} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\left(\alpha−\beta\right)\:+\mathrm{cos}\:\left(\alpha+\beta\right)\right) \\ $$$$\mathrm{to}\:\mathrm{get} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:={x}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:=\frac{\mathrm{3}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${x}=\frac{\mathrm{3}−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$${x}=−\mathrm{3}\:\left[\mathrm{rejected}\right] \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Ismoiljon_008 last updated on 15/Oct/25
$$\:\:\:{thank}\:{you} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 14/Oct/25
$${let}\:\frac{\pi}{\mathrm{7}}={x}\Rightarrow\pi−\mathrm{4}{x}=\mathrm{3}{x} \\ $$$$\:{cosx}−{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sinx}}\left(\mathrm{2}{sinxcosx}−\mathrm{2}{sinxcos}\mathrm{2}{x}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{sinxcos}\mathrm{3}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sinx}}\left(\cancel{{sin}\mathrm{2}{x}}−{sin}\mathrm{3}{x}+{sinx}+{sin}\mathrm{4}{x}−\cancel{{sin}\mathrm{2}{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sinx}}×\left\{{sinx}−{sin}\mathrm{3}{x}+{sin}\left(\pi−\mathrm{3}{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sinx}}\left({sinx}−{sin}\mathrm{3}{x}+{sin}\mathrm{3}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sinx}}×{sinx}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Frix last updated on 14/Oct/25
$$\mathrm{Nice}! \\ $$
Commented by Ismoiljon_008 last updated on 15/Oct/25
$${thank}\:{you} \\ $$