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Question Number 186741 by cortano12 last updated on 09/Feb/23

 cos ((π/(18))).cos (((3π)/(18))).cos (((5π)/(18))).cos (((7π)/(18)))=?

$$\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=? \\ $$

Answered by pablo1234523 last updated on 09/Feb/23

(1/2)[cos ((8π)/(18))+cos ((6π)/(18))]∙(1/2)[cos ((8π)/(18))+cos ((2π)/(18))]  =(1/4)(cos^2  ((4π)/9)+cos ((4π)/9)cos (π/9)+cos ((3π)/9)cos ((4π)/9)+cos ((3π)/9)cos (π/9))  =(1/8)(1+cos ((8π)/9)+cos ((5π)/9)+cos ((3π)/9)+cos ((7π)/9)+cos (π/9)+cos ((4π)/9)+cos ((2π)/9))  =(1/8)(1−cos (π/9)−cos ((4π)/9)+cos ((3π)/9)−cos ((2π)/9)+cos (π/9)+cos ((4π)/9)+cos ((2π)/9))  =(1/8)(1+cos ((3π)/9))  =(1/8)(1+cos (π/3))  =(1/8)(1+(1/2))=(1/8)((3/2))  =(3/(16))

$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{18}}+\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{18}}\right]\centerdot\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{18}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{18}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}\mathrm{cos}\:\frac{\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}\mathrm{cos}\:\frac{\pi}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{9}}−\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}−\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}} \\ $$

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